More from Sunita Tanwar

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Chapter, , 1, , Querying and SQL, Functions, , “Any unique image that you desire, probably already exists on the, internet or in some database... The, problem today is no longer how to, create the right image, but how to, find an already existing one”, — Lev Manovich, , In this chapter, »» Introduction, , 1.1 Introduction, In Class XI, we have understood database, concepts and learned how to create databases, using MySQL. We have also learnt how to, populate, manipulate and retrieve data from, a database using SQL queries., In this chapter, we are going to learn, more SQL commands which are required, to perform various queries in a database., We will understand how to use single row, functions, multiple row functions, arranging, records in ascending or descending order,, grouping records based on some criteria,, and working on multiple tables using SQL., Let us create a database called, CARSHOWROOM, having the schema as, , Chapter 1.indd 1, , »» Functions in SQL, »» Group By in SQL, »» Operations on, Relations, »» Using Two Relations, in a Query, , 11/26/2020 12:31:29 PM

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2, , Informatics Practices, , shown in Figure 1.1. It has the following four relations:, • INVENTORY: Stores name, price, model, year, of manufacturing, and fuel type for each car in, inventory of the showroom,, • CUSTOMER: Stores customer Id, name, address,, phone number and email for each customer,, • SALE: Stores the invoice number, car Id, customer, id, sale date, mode of payment, sales person’s, employee Id, and selling price of the car sold,, • EMPLOYEE: Stores employee Id, name, date of, birth, date of joining, designation, and salary of, each employee in the showroom., Inventory, , Customer, , Car ID, CarName, Price, Model, YearManufacture, FuelType, , CustID, CustName, CustAdd, Phone, Email, , Sale, InvoiceNo, CarID, CustID, SaleDate, PaymentMode, EmpID, SalePrice, , Employee, EmpID, EmpName, DOB, DOJ, Designation, Salary, , Figure 1.1: Schema diagram of database CARSHOWROOM, , The records of the four relations are shown in Tables, 1.1, 1.2, 1.3, and 1.4 respectively., Table 1.1 INVENTORY, mysql> SELECT * FROM INVENTORY;, +-------+--------+-----------+-----------+-----------------+----------+, | CarId | CarName| Price, | Model, | YearManufacture | Fueltype |, +-------+--------+-----------+-----------+-----------------+----------+, | D001 | Car1, | 582613.00 | LXI, |, 2017 | Petrol, |, | D002 | Car1, | 673112.00 | VXI, |, 2018 | Petrol, |, | B001 | Car2, | 567031.00 | Sigma1.2 |, 2019 | Petrol, |, | B002 | Car2, | 647858.00 | Delta1.2 |, 2018 | Petrol, |, , Chapter 1.indd 2, , 11/26/2020 12:31:29 PM

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4, , Informatics Practices, , 1.2 Functions, , in, , SQL, , We know that a function is used to perform some, particular task and it returns zero or more values as a, result. Functions are useful while writing SQL queries, also. Functions can be applied to work on single or, multiple records (rows) of a table. Depending on their, application in one or multiple rows, SQL functions, are categorised as Single row functions and Aggregate, functions., 1.2.1 Single Row Functions, These are also known as Scalar functions. Single row, functions are applied on a single value and return, a single value. Figure 1.2 lists different single row, functions under three categories — Numeric (Math),, String, Date and Time., Math functions accept numeric value as input, and, return a numeric value as a result. String functions, accept character value as input, and return either, character or numeric values as output. Date and, time functions accept date and time values as input,, and return numeric or string, or date and time values, as output., Single Row Function, , Numeric Function, , String Function, , Date Function, , POWER(), , UCASE(), , NOW(), , ROUND(), , LCASE(), , DATE(), , MOD(), , MID(), , MONTH(), , LENGTH(), , MONTHNAME(), , LEFT(), , YEAR(), , RIGHT(), , DAY(), , INSTR(), , DAYNAME(), , LTRIM(), RTRIM(), TRIM(), Figure 1.2: Three categories of single row functions in SQL, , Chapter 1.indd 4, , 11/26/2020 12:31:30 PM

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Querying, , and, , SQL Functions, , 5, , (A) Numeric Functions, Three commonly used numeric functions are POWER(),, ROUND() and MOD(). Their usage along with syntax is, given in Table 1.5., Table 1.5 Math Functions, Function, , Description, , Example with output, , POWER(X,Y), Calculates X to the power Y., can also be written as, POW(X,Y), , mysql> SELECT POWER(2,3);, Output:, 8, , ROUND(N,D), , Rounds off number N to D, number of decimal places., Note: If D=0, then it rounds, off the number to the nearest, integer., , mysql>SELECT ROUND(2912.564, 1);, Output:, 2912.6, mysql> SELECT ROUND(283.2);, Output:, 283, , MOD(A, B), , Returns, the, remainder mysql> SELECT MOD(21, 2);, after dividing number A by Output:, number B., , 1, , Example 1.1, , In order to increase sales, suppose the car dealer decides, to offer his customers to pay the total amount in 10, easy EMIs (equal monthly installments). Assume that, EMIs are required to be in multiples of 10,000. For that,, the dealer wants to list the CarID and Price along with, the following data from the Inventory table:, a) Calculate GST as 12% of Price and display the result, after rounding it off to one decimal place., mysql> SELECT ROUND(12/100*Price,1) "GST", FROM INVENTORY;, +---------+, | GST, |, +---------+, | 69913.6 |, | 80773.4 |, | 68043.7 |, | 77743.0 |, | 42624.6 |, | 78589.7 |, | 61680.0 |, | 73680.0 |, +---------+, 8 rows in set (0.00 sec), , b) Add a new column FinalPrice to the table inventory,, which will have the value as sum of Price and 12%, of the GST., , Chapter 1.indd 5, , 11/26/2020 12:31:30 PM

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6, , Informatics Practices, , mysql> ALTER TABLE INVENTORY ADD(FinalPrice, Numeric(10,1));, Query OK, 8 rows affected (0.03 sec), Records: 8 Duplicates: 0 Warnings: 0, mysql> UPDATE INVENTORY SET, FinalPrice=Price+Round(Price*12/100,1);, Query OK, 8 rows affected (0.01 sec), Rows matched: 8 Changed: 8 Warnings: 0, mysql> SELECT * FROM INVENTORY;, +-------+--------+-----------+----------+---------------+----------+-------------+, | Model, |YearManufacture| FuelType | FinalPric |, | CarId |CarName | Price, +-------+--------+-----------+----------+---------------+----------+-------------+, | D001, , |Car1, , | 582613.00 | LXI, , |, , 2017, , | Petrol, , |, , 652526.6, , |, , | D002, , |Car1, , | 673112.00 | VXI, , |, , 2018, , | Petrol, , |, , 753885.4, , |, , | B001, , |Car2, , | 567031.00 | Sigma1.2 |, , 2019, , | Petrol, , |, , 635074.7, , |, , | B002, , |Car2, , | 647858.00 | Delta1.2 |, , 2018, , | Petrol, , |, , 725601.0, , |, , | E001, , |Car3, , | 355205.00 | 5STR STD |, , 2017, , | CNG, , |, , 397829.6, , |, , | E002, , |Car3, , | 654914.00 | CARE, , |, , 2018, , | CNG, , |, , 733503.7, , |, , | S001, , |Car4, , | 514000.00 | LXI, , |, , 2017, , | Petrol, , |, , 575680.0, , |, , | S002, , |Car4, , | 614000.00 | VXI, , |, , 2018, , | Petrol, , |, , 687680.0, , |, , +-------+--------+-----------+----------+---------------+----------+-------------+, 8 rows in set (0.00 sec), , c), , Calculate and display the amount to be paid, each month (in multiples of 1000) which is to be, calculated after dividing the FinalPrice of the car, into 10 instalments., d) After dividing the amount into EMIs, find out the, remaining amount to be paid immediately, by, performing modular division., Following SQL query can be used to solve the above, mentioned problems:, mysql> select CarId, FinalPrice, ROUND((FinalPriceMOD(FinalPrice,10000))/10,0) "EMI", MOD(FinalPrice,10000) "Remaining Amount", FROM INVENTORY;, +-------+------------+-------+------------------+, | CarId | FinalPrice | EMI, | Remaining Amount |, +-------+------------+-------+------------------+, | D001 |, 652526.6 | 65000 |, 2526.6 |, | D002 |, 753885.4 | 75000 |, 3885.4 |, | B001 |, 635074.7 | 63000 |, 5074.7 |, | B002 |, 725601.0 | 72000 |, 5601.0 |, | E001 |, 397829.6 | 39000 |, 7829.6 |, | E002 |, 733503.7 | 73000 |, 3503.7 |, | S001 |, 575680.0 | 57000 |, 5680.0 |, | S002 |, 687680.0 | 68000 |, 7680.0 |, +-------+------------+-------+------------------+, 8 rows in set (0.00 sec), , Chapter 1.indd 6, , 11/26/2020 12:31:30 PM

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Querying, , and, , SQL Functions, , 7, , Example 1.2, , a) Let us now add a new column Commission to the, SALE table. The column Commission should have, a total length of 7 in which 2 decimal places to, be there., mysql> ALTER TABLE SALE ADD(Commission, Numeric(7,2));, Query OK, 6 rows affected (0.34 sec), Records: 6 Duplicates: 0 Warnings: 0, , b) Let us now calculate commission for sales agents, as 12 per cent of the SalePrice, insert the values, to the newly added column Commission and then, display records of the table SALE where commission, > 73000., mysql> UPDATE SALE SET, Commission=12/100*SalePrice;, Query OK, 6 rows affected (0.06 sec), Rows matched: 6 Changed: 6 Warnings: 0, mysql> SELECT * FROM SALE WHERE Commission > 73000;, +---------------+------+----------+------------+------+-----------+-----------+, |invoiceno|carid|custid| saledate |paymentmode |empid | saleprice |Commission |, +---------------+------+----------+------------+------+-----------+-----------+, |I00001, |D001 |C0001 |2019-01-24|Credit Card |E004 | 613247.00 | 73589.64 |, |I0000, |D002 |C0001 |2018-10-15|Bank Finance|E007 | 659982.00 | 79197.84 |, |I00006, |S002 |C0002 |2019-01-30|Bank Finance|E007 | 620214.00 | 74425.68 |, +---------------+------+-----------+------------+------+----------+-----------+, 3 rows in set (0.02 sec), , c), , Display InvoiceNo, SalePrice and Commission such, that commission value is rounded off to 0., , mysql> SELECT InvoiceNo, SalePrice,, Round(Commission,0) FROM SALE;, +-----------+-----------+---------------------+, | InvoiceNo | SalePrice | Round(Commission,0) |, +-----------+-----------+---------------------+, | I00001, | 613247.00 |, 73590 |, | I00002, | 590321.00 |, 70839 |, | I00003, | 604000.00 |, 72480 |, | I00004, | 659982.00 |, 79198 |, | I00005, | 369310.00 |, 44317 |, | I00006, | 620214.00 |, 74426 |, +-----------+-----------+---------------------+, 6 rows in set (0.00 sec), , (B) String Functions, String functions can perform various operations on, alphanumeric data which are stored in a table. They, can be used to change the case (uppercase to lowercase, , Chapter 1.indd 7, , Activity 1.1, Using the table SALE, of CARSHOWROOM, database, write, SQL queries for the, following:, a) Display the InvoiceNo, and commission, value rounded off to, zero decimal places., b) Display the details of, SALE where payment, mode is credit card.., , 11/26/2020 12:31:30 PM

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8, , Informatics Practices, , or vice-versa), extract a substring, calculate the length, of a string and so on. String functions and their usage, are shown in Table 1.6., Table 1.6 String Functions, Function, , Description, , Example with output, , UCASE(string), OR, UPPER(string), , Converts string into uppercase., , mysql> SELECT, UCASE(“Informatics, Practices”);, Output:, INFORMATICS PRACTICES, , LOWER(string), OR, LCASE(string), , Converts string into lowercase., , mysql> SELECT, LOWER(“Informatics, Practices”);, Output:, informatics practices, , MID(string, pos, n), OR, SUBSTRING(string,, pos, n), OR, SUBSTR(string, pos, n), , Returns a substring of size n, starting from the specified position, (pos) of the string. If n is not, specified, it returns the substring, from the position pos till end of the, string., , mysql> SELECT, MID(“Informatics”, 3, 4);, Output:, form, , LENGTH(string), , Return the number of characters, in the specified string., , mysql> SELECT, LENGTH(“Informatics”);, Output:, 11, , LEFT(string, N), , Returns N number of characters, from the left side of the string., , mysql> SELECT, LEFT(“Computer”, 4);, Output:, Comp, , RIGHT(string, N), , Returns N number of characters, from the right side of the string., , mysql> SELECT, RIGHT(“SCIENCE”, 3);, Output:, NCE, , INSTR(string,, substring), , Returns the position of the first, occurrence of the substring in, the given string. Returns 0, if the, substring is not present in the, string., , mysql> SELECT, INSTR(“Informatics”, “ma”);, Output:, 6, , LTRIM(string), , Returns the given string after mysql> SELECT LENGTH(“, removing leading white space DELHI”), LENGTH(LTRIM(“, DELHI”));, characters., , mysql> SELECT, MID(‘Informatics’,7);, Output:, atics, , Output:, +--------+--------+, | 7, | 5, |, +--------+--------+, 1 row in set (0.00 sec), , Chapter 1.indd 8, , 11/26/2020 12:31:30 PM

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Querying, , RTRIM(string), , and, , SQL Functions, , Returns the given string after mysql>SELECT LENGTH(“PEN, removing trailing white space LENGTH(RTRIM(“PEN “));, characters., Output:, , 9, , “), , +--------+--------+, | 5, | 3, |, +--------+--------+, 1 row in set (0.00 sec), TRIM(string), , Returns the given string after mysql> SELECT LENGTH(“ MADAM, removing both leading and trailing “),LENGTH(TRIM(“ MADAM “));, white space characters., Output:, , +--------+--------+, | 9, | 5, |, +--------+--------+, 1 row in set (0.00 sec), , Example 1.3, , Let us use CUSTOMER relation shown in Table 1.2 to, understand the working of string functions., a) Display customer name in lower case and customer, email in upper case from table CUSTOMER., mysql> SELECT LOWER(CustName), UPPER(Email) FROM, CUSTOMER;, +-----------------+---------------------+, | LOWER(CustName) | UPPER(Email), |, +-----------------+---------------------+, | amitsaha, | [email protected] |, | [email protected] |, | rehnuma, | charvinayyar, | [email protected] |, | gurpreet, | [email protected] |, +-----------------+---------------------+, 4 rows in set (0.00 sec), , b) Display the length of the email and part of the email, from the email ID before the character ‘@’. Note - Do, not print ‘@’., mysql> SELECT LENGTH(Email), LEFT(Email, INSTR(Email,, "@")-1) FROM CUSTOMER;, +---------------+----------------------------------+, | LENGTH(Email) | LEFT(Email, INSTR(Email, "@")-1) |, +---------------+----------------------------------+, |, 19 | amitsaha2, |, |, 19 | rehnuma, |, |, 19 | charvi123, |, |, 19 | gur_singh, |, +---------------+----------------------------------+, 4 rows in set (0.03 sec), , Activity 1.2, Using the table, INVENTORY from, CARSHOWROOM, database, write, sql queries for the, following:, a) Convert the CarMake, to uppercase if its, value starts with the, letter ‘B’., b) If the length of, the car’s model is, greater than 4 then, fetch the substring, starting from position, 3 till the end from, attribute Model., , The function INSTR will return the position of “@”, in the email address. So to print email id without, “@” we have to use position -1., , Chapter 1.indd 9, , 11/26/2020 12:31:30 PM

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10, , Informatics Practices, , c), , Activity 1.3, Using the table, EMPLOYEE from, CARSHOWROOM, database, write, SQL queries for the, following:, a) Display employee, name and the last, 2 characters of his, EmpId., b) Display designation, of employee and the, position of character, ‘e’ in designation, if, present., , Let us assume that four digit area code is reflected, in the mobile number starting from position number, 3. For example, 1851 is the area code of mobile, number 9818511338. Now, write the SQL query to, display the area code of the customer living in Rohini., mysql> SELECT MID(Phone,3,4) FROM CUSTOMER WHERE, CustAdd like ‘%Rohini%’;, +----------------+, | MID(Phone,3,4) |, +----------------+, | 1163, |, +----------------+, 1 row in set (0.00 sec), , d) Display emails after removing the domain name, extension “.com” from emails of the customers., mysql> SELECT TRIM(“.com” from Email) FROM, CUSTOMER;, +-------------------------+, | TRIM(".com" FROM Email) |, +-------------------------+, | amitsaha2@gmail, |, | rehnuma@hotmail, |, | charvi123@yahoo, |, | gur_singh@yahoo, |, +-------------------------+, 4 rows in set (0.00 sec), , e), , Display details of all the customers having yahoo, emails only., , mysql> SELECT * FROM CUSTOMER WHERE Email LIKE, "%yahoo%";, +-------+-------------+----------------------+-----------+--------------------+, |CustID | CustName, | CustAdd, | Phone, | Email, |, +-------+-------------+----------------------+-----------+--------------------+, |C0003 |CharviNayyar |10/9, FF, Rohini, |6811635425 |[email protected] |, |C0004 |Gurpreet, | A-10/2,SF, MayurVihar|3511056125 | [email protected]|, +-------+-------------+----------------------+-----------+--------------------+, 2 rows in set (0.00 sec)t, , (C) Date and Time Functions, There are various functions that are used to perform, operations on date and time data. Some of the operations, include displaying the current date, extracting each, element of a date (day, month and year), displaying day, of the week and so on. Table 1.7 explains various date, and time functions., , Chapter 1.indd 10, , 11/26/2020 12:31:30 PM

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Querying, , and, , SQL Functions, , 11, , Table 1.7 Date Functions, Function, , Description, , Example with output, , NOW(), , It returns the current, system date and time., , mysql> SELECT NOW();, Output:, 2019-07-11 19:41:17, , DATE(), , It returns the date part, from the given date/, time expression., , mysql> SELECT DATE(NOW());, Output:, 2019-07-11, , MONTH(date), , It returns the month in, numeric form from the, date., , mysql> SELECT MONTH(NOW());, Output:, 7, , MONTHNAME(date), , It returns the month, name from the specified, date., , mysql> SELECT, MONTHNAME(“2003-11-28”);, Output:, November, , YEAR(date), , It returns the year from, the date., , mysql> SELECT YEAR(“2003-10-03”);, Output:, 2003, , DAY(date), , It returns the day part, from the date., , mysql> SELECT DAY(“2003-03-24”);, Output:, 24, , DAYNAME(date), , It returns the name of, the day from the date., , mysql> SELECT, DAYNAME(“2019-07-11”);, Output:, Thursday, , Example 1.4, , Let us use the EMPLOYEE table of CARSHOWROOM, database to illustrate the working of some of the date, and time functions., a) Select the day, month number and year of joining of, all employees., mysql> SELECT DAY(DOJ), MONTH(DOJ), YEAR(DOJ) FROM, EMPLOYEE;, +----------+------------+-----------+, | DAY(DOJ) | MONTH(DOJ) | YEAR(DOJ) |, +----------+------------+-----------+, |, 12 |, 12 |, 2017 |, |, 5 |, 6 |, 2016 |, |, 8 |, 1 |, 1999 |, |, 2 |, 12 |, 2010 |, |, 1 |, 7 |, 2012 |, |, 1 |, 1 |, 2017 |, |, 23 |, 10 |, 2013 |, +----------+------------+-----------+, 7 rows in set (0.03 sec), , Activity 1.4, Using the table, EMPLOYEE of, CARSHOWROOM, database, list the, day of birth for all, employees whose, salary is more than, 25000., , b) If the date of joining is not a Sunday, then display it, in the following format "Wednesday, 26, November,, 1979.", , Chapter 1.indd 11, , 11/26/2020 12:31:30 PM

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12, , Informatics Practices, , mysql> SELECT DAYNAME(DOJ), DAY(DOJ),, MONTHNAME(DOJ), YEAR(DOJ) FROM EMPLOYEE WHERE, DAYNAME(DOJ)!='Sunday';, +------------+---------+---------------+---------+, |DAYNAME(DOJ)| DAY(DOJ)|MONTHNAME(DOJ) |YEAR(DOJ)|, +------------+---------+---------------+---------+, |Tuesday, |, 12 | December, |, 2017 |, |Friday, |, 8 | January, |, 1999 |, |Thursday, |, 2 | December, |, 2010 |, |Wednesday, |, 23 | October, 2013 |, +------------+---------+---------------+---------+, 4 rows in set (0.00 sec), , Think and Reflect, Can we use arithmetic, operators (+, -. *, or /), on date functions?, , 1.2.2 Aggregate Functions, Aggregate functions are also called multiple row functions., These functions work on a set of records as a whole,, and return a single value for each column of the records, on which the function is applied. Table 1.8 shows the, differences between single row functions and multiple, row functions. Table 1.9 describes some of the aggregate, functions along with their usage. Note that column, must be of numeric type., Table 1.8 Differences between Single row and Multiple row Functions, Single_row Functions, , Multiple_row functions, , 1. It operates on a single row at a time., , 1. It operates on groups of rows., , 2. It returns one result per row., , 2. It returns one result for a group of rows., , 3. It can be used in Select, Where, and Order 3. It can be used in the select clause only., by clause., 4. Math, String and Date functions, examples of single row functions., , are 4. Max(), Min(), Avg(), Sum(), Count() and Count(*), are examples of multiple row functions., , Table 1.9 Aggregate Functions in SQL, Function, , Chapter 1.indd 12, , Description, , Example with output, , MAX(column), , Returns the largest value from, the specified column., , mysql> SELECT MAX(Price) FROM, INVENTORY;, Output:, 673112.00, , MIN(column), , Returns the smallest value from, the specified column., , mysql> SELECT MIN(Price) FROM, INVENTORY;, Output:, 355205.00, , AVG(column), , Returns the average of the values, in the specified column., , mysql> SELECT AVG(Price) FROM, INVENTORY;, Output:, 576091.625000, , 11/26/2020 12:31:30 PM

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Querying, , and, , SQL Functions, , SUM(column), , Returns the sum of the values, for the specified column., , mysql> SELECT SUM(Price) FROM, INVENTORY;, Output:, 4608733.00, , COUNT(column), , Returns the number of values, in the specified column ignoring, the NULL values., , mysql> SELECT * from MANAGER;, Output:, +------+---------+, | MNO | MEMNAME |, +------+---------+, |, 1 | AMIT, |, |, 2 | KAVREET |, |, 3 | KAVITA |, |, 4 | NULL, |, +------+---------+, 4 rows in set (0.00 sec), , Note:, In this example, let us consider, a MANAGER table having two, attributes and four records., , 13, , mysql> SELECT COUNT(MEMNAME), FROM MANAGER;, Output:, +----------------+, | COUNT(MEMNAME) |, +----------------+, 3, |, |, +----------------+, 1 row in set (0.01 sec), COUNT(*), , Returns the number of records, in a table., , mysql> SELECT COUNT(*) from, MANAGER;, , Note: In order to display the, number of records that matches, a particular criteria in the table,, we have to use COUNT(*) with, WHERE clause., , Output:, +----------+, | count(*) |, +----------+, |, 4 |, +----------+, 1 row in set (0.00 sec), , Example 1.5, , a) Display the total number of records from table, INVENTORY having a model as VXI., mysql> SELECT COUNT(*) FROM INVENTORY WHERE, Model=”VXI”;, +----------+, | COUNT(*) |, +----------+, |, 2 |, +----------+, 1 row in set (0.00 sec), , b) Display the total number of different types of Models, available from table INVENTORY., , Chapter 1.indd 13, , 11/26/2020 12:31:30 PM

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14, , Informatics Practices, , mysql> SELECT COUNT(DISTINCT Model) FROM, INVENTORY;, +-----------------------+, | COUNT(DISTINCT MODEL) |, +-----------------------+, |, 6 |, +-----------------------+, 1 row in set (0.09 sec), , Activity 1.5, a) Find sum of Sale, Price of the cars, purchased by the, customer having ID, C0001 from table, SALE., , c), , b) Find the maximum, and minimum, commission from the, SALE table., , Display the average price of all the cars with Model, LXI from table INVENTORY., mysql> SELECT AVG(Price) FROM INVENTORY WHERE, Model="LXI";, +---------------+, | AVG(Price), |, +---------------+, | 548306.500000 |, +---------------+, 1 row in set (0.03 sec), , 1.3 GROUP BY, , in, , SQL, , At times we need to fetch a group of rows on the, basis of common values in a column. This can be, done using a GROUP BY clause. It groups the rows, together that contain the same values in a specified, column. We can use the aggregate functions (COUNT,, MAX, MIN, AVG and SUM) to work on the grouped, values. HAVING Clause in SQL is used to specify, conditions on the rows with GROUP BY clause., Consider the SALE table from the CARSHOWROOM, database:, mysql> SELECT * FROM SALE;, , +-----------+------+-------+------------+------------------+-----+------------+------------+, , |InvoiceNo|CarId|CustId| SaleDate | PaymentMode, , |EmpID| SalePrice|Commission|, , +-----------+------+-------+------------+------------------+------+------------+------------+, |I00001, |I00002, |I00003, |I00004, |I00005, |I00006, , |D001, |S001, |S002, |D002, |E001, |S002, , |C0001, |C0002, |C0004, |C0001, |C0003, |C0002, , |2019-01-24|, |2018-12-12|, |2019-01-25|, |2018-10-15|, |2018-12-20|, |2019-01-30|, , Credit Card, Online, Cheque, Bank Finance, Credit Card, Bank Finance, , |, |, |, |, |, |, , E004|, E001|, E010|, E007|, E002|, E007|, , 613247.00|, 590321.00|, 604000.00|, 659982.00|, 369310.00|, 620214.00|, , 73589.64, 70838.52, 72480.00, 79197.84, 44317.20, 74425.68, , |, |, |, |, |, |, , +-----------+------+-------+------------+------------------+------+------------+------------+, 6 rows in set (0.11 sec), , Chapter 1.indd 14, , CarID, CustID, SaleDate, PaymentMode, EmpID,, SalePrice are the columns that can have rows with the, same values in it. So, GROUP BY clause can be used, , 11/26/2020 12:31:30 PM

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Querying, , and, , SQL Functions, , 15, , in these columns to find the number of records of a, particular type (column), or to calculate the sum of the, price of each car type., Example 1.6, , a) Display the number of cars purchased by each, customer from the SALE table., mysql> SELECT CustID, COUNT(*) "Number of Cars", FROM SALE GROUP BY CustID;, +--------+----------------+, | CustID | Number of Cars |, +--------+----------------+, | C0001 |, 2 |, | C0002 |, 2 |, | C0003 |, 1 |, | C0004 |, 1 |, +--------+----------------+, 4 rows in set (0.00 sec), , b) Display the customer Id and number of cars, purchased if the customer purchased more than 1, car from SALE table., mysql> SELECT CustID, COUNT(*) FROM SALE GROUP BY, CustID HAVING Count(*)>1;, +--------+----------+, | CustID | COUNT(*) |, +--------+----------+, | C0001 |, 2 |, | C0002 |, 2 |, +--------+----------+, 2 rows in set (0.30 sec), , c), , Display the number of people in each category of, payment mode from the table SALE., mysql> SELECT PaymentMode, COUNT(PaymentMode) FROM, SALE GROUP BY Paymentmode ORDER BY Paymentmode;, +--------------+--------------------+, | PaymentMode | Count(PaymentMode) |, +--------------+--------------------+, | Bank Finance |, 2 |, | Cheque, |, 1 |, | Credit Card |, 2 |, | Online, |, 1 |, +--------------+--------------------+, 4 rows in set (0.00 sec), , Activity 1.6, a) List the total number, of cars sold by each, employee., b) List the maximum, sale made by each, employee., , d) Display the PaymentMode and number of payments, made using that mode more than once., mysql> SELECT PaymentMode, Count(PaymentMode) FROM, SALE GROUP BY Paymentmode HAVING COUNT(*)>1 ORDER, , Chapter 1.indd 15, , 11/26/2020 12:31:30 PM

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16, , Informatics Practices, , Notes, , BY Paymentmode;, +--------------+--------------------+, | PaymentMode | Count(PaymentMode) |, +--------------+--------------------+, | Bank Finance |, 2 |, | Credit Card |, 2 |, +--------------+--------------------+, 2 rows in set (0.00 sec), , 1.4 Operations, , on, , Relations, , We can perform certain operations on relations like, Union, Intersection, and Set Difference to merge the, tuples of two tables. These three operations are binary, operations as they work upon two tables. Note here, that, these operations can only be applied if both the relations, have the same number of attributes, and corresponding, attributes in both tables have the same domain., 1.4.1 UNION (U), This operation is used to combine the selected rows of, two tables at a time. If some rows are the same in both, the tables, then the result of the Union operation will, show those rows only once. Figure 1.3 shows union of, two sets., Music, , Dance, , Figure 1.3: Union of two sets, , Let us consider two relations DANCE and MUSIC, shown in Tables 1.10 and 1.11 respectively., Table 1.10, , DANCE, , +------+--------+-------+, | SNo | Name, | Class |, +------+--------+-------+, |, 1| Aastha | 7A, |, |, 2| Mahira | 6A, |, |, 3| Mohit | 7B, |, |, 4| Sanjay | 7A, |, +------+--------+-------+, , Chapter 1.indd 16, , 11/26/2020 12:31:30 PM

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Querying, , Table 1.11, , and, , SQL Functions, , 17, , Notes, , MUSIC, , +------+---------+-------+, | SNo | Name, | Class |, +------+---------+-------+, |, 1| Mehak, | 8A, |, |, 2| Mahira | 6A, |, |, 3| Lavanya | 7A, |, |, 4| Sanjay | 7A, |, |, 5| Abhay, | 8A, |, +------+---------+-------+, , If we need the list of students participating in either, of events, then we have to apply UNION operation, (represented by symbol U) on relations DANCE and MUSIC., The output of UNION operation is shown in Table 1.12., Table 1.12, , DANCE, , MUSIC, , +-------+------+------+, |SNo | Name, |Class |, +-------+------+------+, | 1 | Aastha | 7A, |, | 2 | Mahira | 6A, |, | 3 | Mohit, | 7B, |, | 4 | Sanjay | 7A, |, | 1 | Mehak, | 8A, |, | 3 | Lavanya | 7A |, | 5 | Abhay, | 8A, |, +-------+------+------+, , 1.4.2 INTERSECT (∩), Intersect operation is used to get the common tuples, from two tables and is represented by the symbol ∩., Figure 1.4 shows intersection of two sets., Music, , Dance, , Figure 1.4: Intersection of two sets, , Suppose we have to display the list of students, who are participating in both the events (DANCE and, MUSIC), then intersection operation is to be applied on, these two tables. The output of INTERSECT operation is, shown in Table 1.13., Table 1.13, , DANCE ∩ MUSIC, , +------+---------+-------+, | SNo | Name, | Class |, +------+---------+-------+, 2| Mahira | 6A, |, |, |, 4| Sanjay | 7A, |, +------+---------+-------+, , Chapter 1.indd 17, , 11/26/2020 12:31:30 PM

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18, , Informatics Practices, , Notes, , 1.4.3 MINUS (-), This operation is used to get tuples/rows which are, in the first table but not in the second table, and the, operation is represented by the symbol - (minus). Figure, 1.5 shows minus operation (also called set difference), between two sets., Music, , Figure 1.5:, , Dance, , Difference of two sets, , Suppose, we want the list of students who are only, participating in MUSIC and not in DANCE event. Then,, we will use the MINUS operation, whose output is given, in Table 1.14., Table 1.14, , DANCE - MUSIC, , +------+---------+-------+, | SNo | Name, | Class |, +------+---------+-------+, 1| Mehak, | 8A, |, |, |, 3| Lavanya | 7A, |, |, 5| Abhay, | 8A, |, +------+---------+-------+, , 1.4.4 Cartesian Product, Cartesian product operation combines tuples from two, relations. It results in all pairs of rows from the two input, relations, regardless of whether or not they have the, same values on common attributes. It is denoted as ‘X’., The degree of the resulting relation is calculated, as the sum of the degrees of both the relations under, consideration. The cardinality of the resulting relation is, calculated as the product of the cardinality of relations, on which cartesian product is applied. Let us use, the relations DANCE and MUSIC to show the output, of cartesian product. Note that both relations are of, degree 3. The cardinality of relations DANCE and MUSIC, is 4 and 5 respectively. Applying cartesian product on, these two relations will result in a relation of degree 6, and cardinality 20, as shown in Table 1.15., , Chapter 1.indd 18, , 11/26/2020 12:31:31 PM

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20, , Informatics Practices, , Notes, , b) From the all possible combinations of tuples of, relations DANCE and MUSIC, display only those, rows such that the attribute name in both have the, same value., mysql> SELECT * FROM DANCE D, MUSIC M WHERE D.Name, = M.Name;, , Table 1.16, , Tuples with same name, , +------+--------+-------+------+--------+-------+, | Sno | Name, | Class | Sno | Name, | class |, +------+--------+-------+------+--------+-------+, |, 2 | Mahira | 6A, |, 2 | Mahira | 6A, |, |, 4 | Sanjay | 7A, |, 4 | Sanjay | 7A, |, +------+--------+-------+------+--------+-------+, 2 rows in set (0.00 sec), , Note that in this query we have used table aliases, (D for DANCE and M for MUSIC), just like column, aliases to refer to tables by shortened names. It is, important to note that table alias is valid only for, current query and the original table name cannot be, used in the query if its alias is given in FROM clause., 1.5.2 JOIN on two tables, JOIN operation combines tuples from two tables on, specified conditions. This is unlike cartesian product,, which make all possible combinations of tuples. While, using the JOIN clause of SQL, we specify conditions on, the related attributes of two tables within the FROM, clause. Usually, such an attribute is the primary key, in one table and foreign key in another table. Let us, create two tables UNIFORM (UCode, UName, UColor), and COST (UCode, Size, Price) in the SchoolUniform, database. UCode is Primary Key in table UNIFORM., UCode and Size is the Composite Key in table COST., Therefore, Ucode is a common attribute between the, two tables which can be used to fetch the common data, from both the tables. Hence, we need to define Ucode as, foreign key in the Price table while creating this table., Table 1.17, , Uniform table, , +-------+-------+--------+, | Ucode | Uname | Ucolor |, +-------+-------+--------+, | 1, | Shirt | White |, | 2, | Pant | Grey, |, | 3, | Tie, | Blue, |, +-------+-------+--------+, , Chapter 1.indd 20, , 11/26/2020 12:31:31 PM

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Querying, , Table 1.18, , and, , SQL Functions, , 21, , Notes, , Cost table, , +-----+------+-------+, |Ucode| Size | Price |, +-----+------+-------+, | 1, | L, |, 580 |, | 1, | M, |, 500 |, | 2, | L, |, 890 |, | 2, | M, |, 810 |, +-------+----+-------+, , Example 1.7, , List the UCode, UName, UColor, Size and Price of related, tuples of tables UNIFORM and COST., The given query may be written in three different ways, as given below:, a) Using condition in where clause, mysql> SELECT * FROM UNIFORM U, COST C WHERE, U.UCode = C.UCode;, , Table 1.19, , Output of the query, , +-------+-------+--------+-------+---------+-------+, | UCode | UName | UColor | Ucode | Size, | Price |, +-------+-------+--------+-------+---------+-------+, |, 1, | Shirt | White | 1, | L, | 580 |, |, 1, | Shirt | White | 1, | M, | 500 |, |, 2, | Pant | Grey, | 2, | L, | 890 |, |, 2, | Pant | Grey, | 2, | M, | 810 |, +-------+-------+--------+-------+---------+-------+, 4 rows in set (0.08 sec), , As the attribute Ucode is in both tables, we need, to use table alias to remove ambiguity. Hence, we, have used qualifier with attribute UCode in SELECT, and FROM clauses to indicate its scope., b) Explicit use of JOIN clause, mysql> SELECT * FROM UNIFORM U JOIN COST C ON, U.Ucode=C.Ucode;, , c), , Chapter 1.indd 21, , The output of the query is the same as shown, in Table 1.19. In this query, we have used JOIN, clause explicitly along with condition in From, clause. Hence, no condition needs to be given in, where clause., Explicit use of NATURAL JOIN clause, The output of queries (a) and (b) shown in Table, 1.19 has a repetitive column Ucode having exactly, the same values. This redundant column provides, no additional information. There is an extension, of JOIN operation called NATURAL JOIN which, works similar to JOIN clause in SQL, but removes, the redundant attribute. This operator can be used, , 11/26/2020 12:31:31 PM

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22, , Informatics Practices, , Notes, , to join the contents of two tables iff there is one, common attribute in both the tables. The above, SQL query using NATURAL JOIN is shown below:, mysql> SELECT * FROM UNIFORM NATURAL JOIN COST;, +-------+-------+--------+------+-------+, | UCode | UName | UColor | Size | Price |, +-------+-------+--------+------+-------+, | 1, | Shirt | White | L, |, 580 |, | 1, | Shirt | White | M, |, 500 |, | 2, | Pant | Grey, | L, |, 890 |, | 2, | Pant | Grey, | M, |, 810 |, +-------+-------+--------+------+-------+, 4 rows in set (0.17 sec), , It is clear from the output that the result of this, query is same as that of queries written in (a) and (b),, except that the attribute Ucode appears only once., Following are some of the points to be considered, while applying JOIN operations on two or more relations:, • If two tables are to be joined on equality condition, on the common attribute, then one may use JOIN, with ON clause or NATURAL JOIN in FROM clause., If three tables are to be joined on equality condition,, then two JOIN or NATURAL JOIN are required., • In general, N-1 joins are needed to combine N tables, on equality condition., • With JOIN clause, we may use any relational, operators to combine tuples of two tables., , S u mmar y, • A Function is used to perform a particular task, and return a value as a result., • Single row functions work on a single row to, return a single value., • Multiple row functions work on a set of records as, a whole and return a single value., • Numeric functions perform operations on numeric, values and return numeric values., • String functions perform operations on character, type values and return either character or numeric, values., , Chapter 1.indd 22, , 11/26/2020 12:31:31 PM

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Querying, , • Date and time functions allow us to deal with date, type data values., • GROUP BY function is used to group the rows, together that contain similar values in a specified, column. Some of the group functions are COUNT,, MAX, MIN, AVG and SUM., • Join is an operation which is used to combine, rows from two or more tables based on one or, more common fields between them., , and, , SQL Functions, , 23, , Notes, , Exercise, 1. Answer the following questions:, a) Define RDBMS. Name any two RDBMS software., b) What is the purpose of the following clauses in a, select statement?, i) ORDER BY, ii) HAVING, c) Site any two differences between Single_row, functions and Aggregate functions., d) What do you understand by Cartesian Product?, e) Write the name of the functions to perform the, following operations:, i) To display the day like “Monday”, “Tuesday”,, from the date when India got independence., ii) To display the specified number of characters, from a particular position of the given string., iii) To display the name of the month in which, you were born., iv) To display your name in capital letters., 2. Write the output produced by the following SQL, commands:, a) SELECT POW(2,3);, b) SELECT ROUND(123.2345, 2),, ROUND(342.9234,-1);, c) SELECT LENGTH("Informatics Practices");, d) SELECT YEAR(“1979/11/26”),, MONTH(“1979/11/26”),, DAY(“1979/11/26”),, , Chapter 1.indd 23, , 11/26/2020 12:31:31 PM

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24, , Informatics Practices, , Notes, , MONTHNAME(“1979/11/26”);, e) SELECT LEFT("INDIA",3), RIGHT("Computer, Science",4);, f) SELECT MID("Informatics",3,4),, SUBSTR("Practices",3);, 3. Consider the following table named “Product”,, showing details of products being sold in a grocery, shop., PCode, , PName, , UPrice, , Manufacturer, , 120, , Surf, , P01, , Washing Powder, , P02, , Tooth Paste, , 54, , Colgate, , P03, , Soap, , 25, , Lux, , P04, , Tooth Paste, , 65, , Pepsodant, , P05, , Soap, , 38, , Dove, , P06, , Shampoo, , 245, , Dove, , a) Write SQL queries for the following:, i. Create the table Product with appropriate, data types and constraints., ii. Identify the primary key in Product., iii. List the Product Code, Product name and, price in descending order of their product, name. If PName is the same then display the, data in ascending order of price., iv. Add a new column Discount to the table, Product., v. Calculate the value of the discount in the, table Product as 10 per cent of the UPrice, for all those products where the UPrice is, more than 100, otherwise the discount will, be 0., vi. Increase the price by 12 per cent for all the, products manufactured by Dove., vii.Display the total number of products, manufactured by each manufacturer., b) Write the output(s) produced by executing the, following queries on the basis of the information, given above in the table Product:, i. SELECT PName, Average(UPrice) FROM, Product GROUP BY Pname;, ii. SELECT DISTINCT Manufacturer FROM, Product;, , Chapter 1.indd 24, , 11/26/2020 12:31:31 PM

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Querying, , iii. SELECT COUNT(DISTINCT PName) FROM, Product;, , and, , SQL Functions, , 25, , Notes, , iv. SELECT PName, MAX(UPrice), MIN(UPrice), FROM Product GROUP BY PName;, 4. Using the CARSHOWROOM database given in the, chapter, write the SQL queries for the following:, a) Add a new column Discount in the INVENTORY, table., b) Set appropriate discount values for all cars, keeping in mind the following:, (i) No discount is available on the LXI model., (ii) VXI model gives a 10% discount., (iii) A 12% discount is given on cars other than, LXI model and VXI model., c) Display the name of the costliest car with fuel, type “Petrol”., d) Calculate the average discount and total discount, available on Car4., e) List the total number of cars having no discount., 5. Consider the following tables Student and Stream in, the Streams_of_Students database. The primary key, of the Stream table is StCode (stream code) which is, the foreign key in the Student table. The primary key, of the Student table is AdmNo (admission number)., AdmNo, , Name, , StCode, , 211, , Jay, , NULL, , 241, , Aditya, , S03, , 290, , Diksha, , S01, , 333, , Jasqueen, , S02, , 356, , Vedika, , S01, , 380, , Ashpreet, , S03, , StCode, , S01, S02, S03, , Stream, Science, Commerce, Humanities, , Write SQL queries for the following:, a) Create the database Streams_Of_Students., , Chapter 1.indd 25, , 11/26/2020 12:31:31 PM

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26, , Informatics Practices, , Notes, , b) Create the table Student by choosing appropriate, data types based on the data given in the table., c) Identify the Primary keys from tables Student, and Stream. Also, identify the foreign key from, the table Stream., d) Jay has now changed his stream to Humanities., Write an appropriate SQL query to reflect this, change., e) Display the names of students whose names end, with the character ‘a’. Also, arrange the students, in alphabetical order., f) Display the names of students enrolled in Science, and Humanities stream, ordered by student name, in alphabetical order, then by admission number, in ascending order (for duplicating names)., g) List the number of students in each stream having, more than 1 student., h) Display the names of students enrolled in, different streams, where students are arranged, in descending order of admission number., i) Show the Cartesian product on the Student, and Stream table. Also mention the degree and, cardinality produced after applying the Cartesian, product., j) Add a new column ‘TeacherIncharge” in the, Stream table. Insert appropriate data in each row., k) List the names of teachers and students., l) If Cartesian product is again applied on Student, and Stream tables, what will be the degree and, cardinality of this modified table?, , Chapter 1.indd 26, , 11/26/2020 12:31:31 PM