Question 1

Question 1Table BOOK_INFORMATIONColumn NameBOOK_IDBOOK_TITLEPRICEWhich SQL statement allows you to find the highest price from the table BOOK_INFORMATION?SELECT BOOK_ID, BOOK_TITLE, MAX(PRICE) FROM BOOK_INFORMATION;SELECT MAX(PRICE) FROM BOOK_INFORMATION;SELECT MAXIMUM(PRICE) FROM BOOK_INFORMATION;SELECT PRICE FROM BOOK_INFORMATION ORDER BY PRICE DESC;

Accepted Answer

SELECT MAX(PRICE) FROM BOOK_INFORMATION;ExplanationSELECT BOOK_ID, BOOK_TITLE, MAX(PRICE) FROM BOOK_INFORMATION; — This query selects the BOOK_ID, BOOK_TITLE, and maximum PRICE from the BOOK_INFORMATION table. However, the requirement is to find the highest price only.SELECT MAX(PRICE) FROM BOOK_INFORMATION; — This query selects the maximum PRICE from the BOOK_INFORMATION table using the MAX() aggregate function. This option is correct because it directly retrieves the highest price among all the books listed in the BOOK_INFORMATION table, which is what the question asks for.SELECT MAXIMUM(PRICE) FROM BOOK_INFORMATION; — There is no MAXIMUM() function in SQL.SELECT PRICE FROM BOOK_INFORMATION ORDER BY PRICE DESC; — This query selects all prices from the BOOK_INFORMATION table and orders them in descending order using ORDER BY PRICE DESC but it doesn't directly give the highest price.

Question 2

Question 2Table SALESColumn NameSTORE_IDSALES_DATESALES_AMOUNTWhich SQL statement lets you find the sales amount for each store?SELECT STORE_ID, SUM(SALES_AMOUNT) FROM SALES;SELECT STORE_ID, SUM(SALES_AMOUNT) FROM SALES ORDER BY STORE_ID;SELECT STORE_ID, SUM(SALES_AMOUNT) FROM SALES GROUP BY STORE_ID;SELECT STORE_ID, SUM(SALES_AMOUNT) FROM SALES HAVING UNIQUE STORE_ID;

Accepted Answer

SELECT STORE_ID, SUM(SALES_AMOUNT) FROM SALES GROUP BY STORE_ID;ExplanationSELECT STORE_ID, SUM(SALES_AMOUNT) FROM SALES; — This statement selects the store_ID and calculates the total sales amount using SUM() aggregate function from the SALES table. It does not group the results by store ID, so it will return a single row with the total sales amount across all stores.SELECT STORE_ID, SUM(SALES_AMOUNT) FROM SALES ORDER BY STORE_ID; — This statement selects the store_ID and calculates the total sales amount using SUM() aggregate function from the SALES table and uses an ORDER BY clause to sort the results by store ID. However, it still doesn't group the results by store_ID.SELECT STORE_ID, SUM(SALES_AMOUNT) FROM SALES GROUP BY STORE_ID; — This statement selects the store_ID and calculates the total sales amount using SUM() aggregate function from the SALES table and uses the GROUP BY clause to group the results by store ID. It calculates the total sales amount for each store ID separately. As a result, it calculates the total sales amount for each unique store ID separately, providing a breakdown of sales amounts for each store in the dataset.SELECT STORE_ID, SUM(SALES_AMOUNT) FROM SALES HAVING UNIQUE STORE_ID; — This statement is incorrect because the HAVING clause is used for filtering grouped data based on a condition, not for identifying unique values. Also, "UNIQUE STORE_ID" is not a valid condition in SQL.

Question 3

Question 3Table SALESColumn NameSTORE_IDSALES_DATESALES_AMOUNTWhich SQL statement lets you list all stores whose total sales amount is over 5000 ?SELECT STORE_ID, SUM(SALES_AMOUNT) FROM SALES GROUP BY STORE_ID HAVING SUM(SALES_AMOUNT) > 5000;SELECT STORE_ID, SUM(SALES_AMOUNT) FROM SALES GROUP BY STORE_ID HAVING SALES_AMOUNT > 5000;SELECT STORE_ID, SUM(SALES_AMOUNT) FROM SALESWHERE SUM(SALES_AMOUNT) > 5000 GROUP BY STORE_ID;SELECT STORE_ID, SUM(SALES_AMOUNT) FROM SALESWHERE SALES_AMOUNT > 5000 GROUP BY STORE_ID;

Accepted Answer

SELECT STORE_ID, SUM(SALES_AMOUNT) FROM SALES
GROUP BY STORE_ID HAVING SUM(SALES_AMOUNT) > 5000;ExplanationSELECT STORE_ID, SUM(SALES_AMOUNT) FROM SALES GROUP BY STORE_ID HAVING SUM(SALES_AMOUNT) > 5000; — This statement selects the store ID and calculates the total sales amount for each store using the SUM() aggregate function. The GROUP BY STORE_ID clause ensures that the results are grouped by store ID. The HAVING SUM(SALES_AMOUNT) > 5000 condition then filters the grouped data, selecting only those stores whose total sales amount is over 5000.SELECT STORE_ID, SUM(SALES_AMOUNT) FROM SALES GROUP BY STORE_ID HAVING SALES_AMOUNT > 5000; — This option is incorrect because the HAVING clause cannot directly reference SALES_AMOUNT without an aggregate function like SUM() since SUM(SALES_AMOUNT) is used in the SELECT statement.SELECT STORE_ID, SUM(SALES_AMOUNT) FROM SALES WHERE SUM(SALES_AMOUNT) > 5000 GROUP BY STORE_ID; — This option is incorrect because aggregate functions like SUM(SALES_AMOUNT) cannot be used directly in the WHERE clause. The WHERE clause is used for filtering individual rows before grouping.SELECT STORE_ID, SUM(SALES_AMOUNT) FROM SALES WHERE SALES_AMOUNT > 5000 GROUP BY STORE_ID; — This option is incorrect because it tries to filter individual sales amounts (SALES_AMOUNT) directly without using the SUM() aggregate function to calculate the total sales amount for each store. The GROUP BY STORE_ID clause is used for grouping after the filtering, which is not the correct approach for filtering based on the total sales amount.

Question 4

Question 4Table SALESColumn NameSTORE_IDSALES_DATESALES_AMOUNTWhich SQL statement lets you find the total number of stores in the SALES table?SELECT COUNT(STORE_ID) FROM SALES;SELECT COUNT(DISTINCT STORE_ID) FROM SALES;SELECT DISTINCT STORE_ID FROM SALES;SELECT COUNT(STORE_ID) FROM SALES GROUP BY STORE_ID;

Accepted Answer

SELECT COUNT(DISTINCT STORE_ID) FROM SALES;ExplanationSELECT COUNT(STORE_ID) FROM SALES; — This query uses the COUNT() aggregate function with the STORE_ID column in the SELECT statement. It counts the number of non-null values in the STORE_ID column, and this count includes duplicates.SELECT COUNT(DISTINCT STORE_ID) FROM SALES; — This option uses COUNT(DISTINCT STORE_ID) to count the number of unique store IDs in the SALES table. The DISTINCT keyword ensures that only distinct (unique) values are counted, avoiding overcounting due to duplicates.SELECT DISTINCT STORE_ID FROM SALES; — This option selects distinct (unique) store IDs from the SALES table but does not count or provide the total number of stores.SELECT COUNT(STORE_ID) FROM SALES GROUP BY STORE_ID; — This option attempts to count the number of occurrences of each store ID by using COUNT(STORE_ID) and grouping by store ID with GROUP BY STORE_ID. However, this results in a count for each unique store ID separately, not the total number of stores in the table.

Question 5

Question 5Table SALESColumn NameSTORE_IDSALES_DATESALES_AMOUNTWhich SQL statement allows you to find the total sales amount for Store ID 25 and the total sales amount for Store ID 45?SELECT STORE_ID, SUM(SALES_AMOUNT) FROM SALESWHERE STORE_ID IN (25, 45) GROUP BY STORE_ID;SELECT STORE_ID, SUM(SALES_AMOUNT) FROM SALESGROUP BY STORE_ID HAVING STORE_ID IN (25, 45);SELECT STORE_ID, SUM(SALES_AMOUNT) FROM SALES WHERE STORE_ID IN (25, 45);SELECT STORE_ID, SUM(SALES_AMOUNT) FROM SALESWHERE STORE_ID = 25 AND STORE_ID = 45 GROUP BY STORE_ID;

Accepted Answer

SELECT STORE_ID, SUM(SALES_AMOUNT) FROM SALES   
WHERE STORE_ID IN (25, 45) GROUP BY STORE_ID;ExplanationSELECT STORE_ID, SUM(SALES_AMOUNT) FROM SALES WHERE STORE_ID IN (25, 45) GROUP BY STORE_ID; — This query uses the IN operator to filter rows where the STORE_ID is either 25 or 45. It then calculates the total sales amount for each store ID using SUM(SALES_AMOUNT) and groups the results by STORE_ID. This query correctly finds the total sales amount for Store ID 25 and Store ID 45 separately.SELECT STORE_ID, SUM(SALES_AMOUNT) FROM SALES GROUP BY STORE_ID HAVING STORE_ID IN (25, 45); — This query will also give the required output but it is inefficient because it first retrieves all rows from the "SALES" table, then groups the results by store ID, and finally filters the result set to include only store IDs 25 and 45. The inefficiency arises from the fact that it processes all rows in the "SALES" table before filtering out the unnecessary data. This means that it processes more data than necessary, which can be wasteful in terms of computational resources and time. A more efficient approach would be to select only the rows having store IDs 25 and 45 first (using WHERE clause), and then perform the aggregation.SELECT STORE_ID, SUM(SALES_AMOUNT) FROM SALES WHERE STORE_ID IN (25, 45); — This query filters rows where the STORE_ID is either 25 or 45 and calculates the total sales amount for these store IDs using SUM(SALES_AMOUNT). However, it doesn't include a GROUP BY clause, so it would return a single row with the total sales amount for both Store ID 25 and Store ID 45 combined.SELECT STORE_ID, SUM(SALES_AMOUNT) FROM SALES WHERE STORE_ID = 25 AND STORE_ID = 45 GROUP BY STORE_ID; — This query filter rows where the STORE_ID is both 25 and 45 simultaneously using STORE_ID = 25 AND STORE_ID = 45. However, this condition is impossible to satisfy because a single value cannot be both 25 and 45 at the same time. Therefore, this query would not return any results.

Question 6

Question 6Table EXAM_RESULTSSTU IDFNAMELNAMEEXAM IDEXAM_SCORE10LAURALYNCH19010LAURALYNCH28511GRACEBROWN17811GRACEBROWN27212JAYJACKSON19512JAYJACKSON29213WILLIAMBISHOP17013WILLIAMBISHOP210014CHARLESPRADA285What SQL statement do we use to find the average exam score for EXAM_ID = 1?SELECT AVG(EXAM_SCORE) FROM EXAM_RESULTS;SELECT AVG(EXAM_SCORE) FROM EXAM_RESULTS GROUP BY EXAM_ID WHERE EXAM_ID = 1;SELECT AVG(EXAM_SCORE) FROM EXAM_RESULTS GROUP BY EXAM_ID HAVING EXAM_ID = 1;SELECT COUNT(EXAM_SCORE) FROM EXAM_RESULTS WHERE EXAM_ID = 1;

Accepted Answer

SELECT AVG(EXAM_SCORE) FROM EXAM_RESULTS GROUP BY EXAM_ID HAVING EXAM_ID = 1;Output+-----------------+
| AVG(EXAM_SCORE) |
+-----------------+
|         83.2500 |
+-----------------+
ExplanationSELECT AVG(EXAM_SCORE) FROM EXAM_RESULTS; — This statement calculates the average exam score across all exam IDs in the EXAM_RESULTS table.SELECT AVG(EXAM_SCORE) FROM EXAM_RESULTS GROUP BY EXAM_ID WHERE EXAM_ID = 1; — This statement is incorrect because the WHERE clause should come before the GROUP BY clause. Additionally, grouping by EXAM_ID and then trying to filter by EXAM_ID = 1 within the GROUP BY clause will result in an error because grouping should be done before filtering.SELECT AVG(EXAM_SCORE) FROM EXAM_RESULTS GROUP BY EXAM_ID HAVING EXAM_ID = 1; — This query groups the exam results by EXAM_ID and then calculates the average exam score for each group. The HAVING clause filters the groups and returns only those where the EXAM_ID is equal to 1, giving us the average exam score for the exam with EXAM_ID equal to 1.SELECT COUNT(EXAM_SCORE) FROM EXAM_RESULTS WHERE EXAM_ID = 1; — This statement calculates the count of exam scores for EXAM_ID = 1, but it doesn't calculate the average score.

Question 7

Question 7Table EXAM_RESULTSSTU IDFNAMELNAMEEXAM IDEXAM_SCORE10LAURALYNCH19010LAURALYNCH28511GRACEBROWN17811GRACEBROWN27212JAYJACKSON19512JAYJACKSON29213WILLIAMBISHOP17013WILLIAMBISHOP210014CHARLESPRADA285Which SQL statement do we use to find out how many students took each exam?SELECT COUNT(DISTINCT STU_ID) FROM EXAM_RESULTS GROUP BY EXAM_ID;SELECT EXAM_ID, MAX(STU_ID) FROM EXAM_RESULTS GROUP BY EXAM_ID;SELECT EXAM_ID, COUNT(DISTINCT STU_ID) FROM EXAM_RESULTS GROUP BY EXAM_ID;SELECT EXAM_ID, MIN(STU_ID) FROM EXAM_RESULTS GROUP BY EXAM_ID;

Accepted Answer

SELECT EXAM_ID, COUNT(DISTINCT STU_ID) FROM EXAM_RESULTS GROUP BY EXAM_ID;Output+---------+------------------------+
| EXAM_ID | COUNT(DISTINCT STU_ID) |
+---------+------------------------+
|       1 |                      4 |
|       2 |                      5 |
+---------+------------------------+
ExplanationSELECT COUNT(DISTINCT STU_ID) FROM EXAM_RESULTS GROUP BY EXAM_ID; — It groups the EXAM_RESULTS table by EXAM_ID and uses the COUNT(DISTINCT STU_ID) function to count the number of distinct student IDs for each exam. However, the result set does not include EXAM_ID.SELECT EXAM_ID, MAX(STU_ID) FROM EXAM_RESULTS GROUP BY EXAM_ID; — This query groups the results by EXAM_ID and then selects the maximum STU_ID for each exam. However, this doesn't provide the count of students who took each exam, as it gives the maximum student ID instead of counting the distinct student IDs.SELECT EXAM_ID, COUNT(DISTINCT STU_ID) FROM EXAM_RESULTS GROUP BY EXAM_ID; — It groups the EXAM_RESULTS table by EXAM_ID and uses the COUNT(DISTINCT STU_ID) function to count the number of distinct student IDs for each exam. The result set includes the EXAM_ID and the count of students who took each exam.SELECT EXAM_ID, MIN(STU_ID) FROM EXAM_RESULTS GROUP BY EXAM_ID; — This query groups the results by EXAM_ID and selects the minimum STU_ID for each exam. It does not provide information about the number of students who took each exam.

Question 8

Question 8Table EXAM_RESULTSSTU IDFNAMELNAMEEXAM IDEXAM_SCORE10LAURALYNCH19010LAURALYNCH28511GRACEBROWN17811GRACEBROWN27212JAYJACKSON19512JAYJACKSON29213WILLIAMBISHOP17013WILLIAMBISHOP210014CHARLESPRADA285What SQL statement do we use to print out the record of all students whose last name starts with 'L'?SELECT * FROM EXAM_RESULTS WHERE LNAME LIKE 'L%' ;SELECT * FROM EXAM_RESULTS WHERE LNAME LIKE 'L';SELECT * FROM EXAM_RESULTS WHERE LNAME 'L';SELECT * FROM EXAM_RESULTS WHERE LNAME <> 'L';

Accepted Answer

SELECT * FROM EXAM_RESULTS WHERE LNAME LIKE 'L%' ;Output+--------+-------+-------+---------+------------+
| stu_id | fname | lname | exam_id | exam_score |
+--------+-------+-------+---------+------------+
|     10 | LAURA | LYNCH |       1 |         90 |
|     10 | LAURA | LYNCH |       2 |         85 |
+--------+-------+-------+---------+------------+
ExplanationSELECT * FROM EXAM_RESULTS WHERE LNAME LIKE 'L%'; — The LIKE operator is used for pattern matching in SQL. '%' is a wildcard character that matches zero or more characters. 'L%' specifies that the last name (LNAME) should start with 'L' followed by zero or more characters. The SELECT * statement retrieves all columns from the EXAM_RESULTS table for the matching records.SELECT * FROM EXAM_RESULTS WHERE LNAME LIKE 'L'; — This query attempts to select all columns (*) from the EXAM_RESULTS table where the last name (LNAME) is exactly equal to 'L'. However, when using the LIKE operator in SQL for pattern matching, we use wildcard characters (%) to represent unknown parts of a string.SELECT * FROM EXAM_RESULTS WHERE LNAME 'L'; — This statement contains a syntax error. In SQL, when using the WHERE clause to filter records based on a specific condition, we need to use comparison operators or functions to define the condition properly.SELECT * FROM EXAM_RESULTS WHERE LNAME <> 'L'; — This query retrieves records where the last name is not equal to 'L'. It does not specifically look for last names starting with 'L', so it's not the correct option for the given requirement.

Question 9

Question 9Table EXAM_RESULTSSTU IDFNAMELNAMEEXAM IDEXAM_SCORE10LAURALYNCH19010LAURALYNCH28511GRACEBROWN17811GRACEBROWN27212JAYJACKSON19512JAYJACKSON29213WILLIAMBISHOP17013WILLIAMBISHOP210014CHARLESPRADA285What is the result of the following SQL statement ?SELECT MAX(EXAM_SCORE) FROM EXAM_RESULTS GROUP BY EXAM_ID HAVING EXAM_ID = 1;908510095

Accepted Answer

Output+-----------------+
| MAX(EXAM_SCORE) |
+-----------------+
|              95 |
+-----------------+
ExplanationThe above SQL query calculates the maximum exam score for EXAM_ID 1 from the EXAM_RESULTS table, by grouping results based on EXAM_ID and filtering using HAVING.

Question 10

Question 10Given the following table :Table : CLUBCOACH-IDCOACHNAMEAGESPORTSDATOFAPPPAYSEX1KUKREJA35KARATE27/03/19961000M2RAVINA34KARATE20/01/19981200F3KARAN34SQUASH19/02/19982000M4TARUN33BASKETBALL01/01/19981500M5ZUBIN36SWIMMING12/01/1998750M6KETAKI36SWIMMING24/02/1998800F7ANKITA39SQUASH20/02/19982200F8ZAREEN37KARATE22/02/19981100F9KUSH41SWIMMING13/01/1998900M10SHAILYA37BASKETBALL19/02/19981700MGive the output of following SQL statements :SELECT COUNT(DISTINCT SPORTS) FROM Club ;SELECT MIN(Age) FROM CLUB WHERE Sex = 'F' ;SELECT AVG(Pay) FROM CLUB WHERE Sports = 'KARATE' ;SELECT SUM(Pay) FROM CLUB WHERE Datofapp > '1998-01-31' ;

Accepted Answer

1.Output+------------------------+
| COUNT(DISTINCT SPORTS) |
+------------------------+
|                      4 |
+------------------------+
ExplanationThe SQL query SELECT COUNT(DISTINCT SPORTS) FROM Club ; calculates the count of unique values in the 'SPORTS' column of the 'Club' table. This query helps us to get information about the number of sports offered by the club.2.Output+----------+
| MIN(Age) |
+----------+
|       34 |
+----------+
ExplanationThe SQL query SELECT MIN(Age) FROM CLUB WHERE Sex = 'F' ; retrieves the minimum Age from the 'CLUB' table where the 'Sex' column has the value 'F'. This query gives us the age of the youngest female coach in the club.3.Output+-----------+
| AVG(Pay)  |
+-----------+
| 1100.0000 |
+-----------+
ExplanationThe SQL query SELECT AVG(Pay) FROM CLUB WHERE Sports = 'KARATE' ; calculates the average value of the 'Pay' column from the 'CLUB' table where the 'Sports' column has the value 'KARATE'. This query helps us to get information about the average pay of karate coaches in the club.4.Output+----------+
| SUM(Pay) |
+----------+
|     7800 |
+----------+
ExplanationThe SQL query SELECT SUM(Pay) FROM CLUB WHERE Dateofapp > '1998-01-31'; calculates the sum of the 'Pay' column from the 'CLUB' table where the 'Dateofapp' column has a date value greater than '1998-01-31'. This query gives us the total pay of all the coaches who joined after 31/01/1998.

Question 11

Question 11In a Database, there are two tables given below :Table : EMPLOYEEEMPLOYEEIDNAMESALESJOBIDE1SUMIT SINHA1100000102E2VIJAY SINGH TOMAR1300000101E3AJAY RAJPAL1400000103E4MOHIT RAMNANI1250000102E5SHAILJA SINGH1450000103Table : JOBJOBIDJOBTITLESALARY101President200000102Vice President125000103Administration Assistant80000104Accounting Manager70000105Accountant65000106Sales Manager80000Write SQL Queries for the following :To display employee ids, names of employees, job ids with corresponding job titles.To display names of employees, sales and corresponding job titles who have achieved sales more than 1300000.To display names and corresponding job titles of those employees who have 'SINGH' (anywhere) in their names.Identify foreign key in the table EMPLOYEE.Write SQL command to change the JOBID to 104 of the EMPLOYEE with ID as E4 in the table 'EMPLOYEE'.

Accepted Answer

1.SELECT EMPLOYEE.EMPLOYEEID, EMPLOYEE.NAME, EMPLOYEE.JOBID, JOB.JOBTITLE
FROM EMPLOYEE, JOB
WHERE EMPLOYEE.JOBID = JOB.JOBID;Output+------------+-------------------+-------+--------------------------+
| EMPLOYEEID | NAME              | JOBID | JOBTITLE                 |
+------------+-------------------+-------+--------------------------+
| E1         | SUMIT SINHA       |   102 | VICE PRESIDENT           |
| E2         | VIJAY SINGH TOMAR |   101 | PRESIDENT                |
| E3         | AJAY RAJPAL       |   103 | ADMINISTARTION ASSISTANT |
| E4         | MOHIT RAMNANI     |   102 | VICE PRESIDENT           |
| E5         | SHAILJA SINGH     |   103 | ADMINISTARTION ASSISTANT |
+------------+-------------------+-------+--------------------------+
2.SELECT EMPLOYEE.NAME, EMPLOYEE.SALES, JOB.JOBTITLE
FROM EMPLOYEE, JOB 
WHERE EMPLOYEE.JOBID = JOB.JOBID
AND EMPLOYEE.SALES > 1300000;Output+---------------+---------+--------------------------+
| NAME          | SALES   | JOBTITLE                 |
+---------------+---------+--------------------------+
| AJAY RAJPAL   | 1400000 | ADMINISTARTION ASSISTANT |
| SHAILJA SINGH | 1450000 | ADMINISTARTION ASSISTANT |
+---------------+---------+--------------------------+
3.SELECT EMPLOYEE.NAME, JOB.JOBTITLE
FROM EMPLOYEE, JOB 
WHERE EMPLOYEE.JOBID = JOB.JOBID
AND EMPLOYEE.NAME LIKE '%SINGH%';Output+-------------------+--------------------------+
| NAME              | JOBTITLE                 |
+-------------------+--------------------------+
| VIJAY SINGH TOMAR | PRESIDENT                |
| SHAILJA SINGH     | ADMINISTARTION ASSISTANT |
+-------------------+--------------------------+
4. In the given tables, EMPLOYEE and JOB, the JOBID column in the EMPLOYEE table is a foreign key referencing the JOBID column in the JOB table.5.UPDATE EMPLOYEE
SET JOBID = 104
WHERE EMPLOYEEID = 'E4';OutputSELECT * FROM EMPLOYEE ;+------------+-------------------+---------+-------+
| EMPLOYEEID | NAME              | SALES   | JOBID |
+------------+-------------------+---------+-------+
| E1         | SUMIT AINHA       | 1100000 |   102 |
| E2         | VIJAY SINGH TOMAR | 1300000 |   101 |
| E3         | AJAY RAJPAL       | 1400000 |   103 |
| E4         | MOHIT RAMNANI     | 1250000 |   104 |
| E5         | SHAILJA SINGH     | 1450000 |   103 |
+------------+-------------------+---------+-------+

Question 12

Question 12Consider the following tables Employee and Salary. Write SQL commands for the statements (i) to (iv) and give outputs for SQL queries (v) to (vii)Table : EmployeeEidNameDepidQualificationSec1Deepali Gupta101MCAF2Rajat Tyagi101BCAM3Hari Mohan102B.A.M4Harry102M.A.M5Sumit Mittal103B.Tech.M6Jyoti101M.Tech.FTable : SalaryEidBasicD.AHRABonus1600020002300200220003003003031000300300404150039049030580009009008061000030049089To display the frequency of employees department wise.To list the names of those employees only whose name starts with 'H'To add a new column in salary table. The column name is Total_Sal.To store the corresponding values in the Total_Sal column.Select max(Basic) from Salary where Bonus > 40 ;Select count(*) from Employee group by Sex ;Select Distinct Depid from Employee ;

Accepted Answer

1.SELECT Depid, COUNT(*) AS Frequency 
FROM Employee 
GROUP BY Depid;2.SELECT Name
FROM Employee
WHERE Name LIKE 'H%';3.ALTER TABLE Salary
ADD COLUMN Total_Sal FLOAT;4.UPDATE Salary
SET Total_Sal = Basic + `D.A.` + HRA + Bonus;5.SELECT MAX(Basic)
FROM Salary
WHERE Bonus > 40;Output+------------+
| MAX(Basic) |
+------------+
|      10000 |
+------------+
6.SELECT Sec as sex, COUNT(*) AS Count 
FROM Employee 
GROUP BY Sec ;Output+-----+-------+
| sex | Count |
+-----+-------+
| F   |     2 |
| M   |     4 |
+-----+-------+
7.SELECT DISTINCT Depid FROM Employee ;Output+-------+
| Depid |
+-------+
|   101 |
|   102 |
|   103 |
+-------+

Question 13

Question 13With reference to following relations PERSONAL and JOB answer the questions that follow :Create following tables such that Empno and Sno are not null and unique, date of birth is after '12-Jan-1960', name is never blank, Area and Native place is valid, hobby, dept is not empty, salary is between 4000 and 10000.Table : PersonalEmpnoNameDobirthNative-placeHobby123Amit23-Jan-1965DelhiMusic127Manoj12-dec-1976MumbaiWriting124Abhai11-aug-1975AllahabadMusic125Vinod04-apr-1977DelhiSports128Abhay10-mar-1974MumbaiGardening129Ramesh28-oct-1981PuneSportsTable : JobSnoAreaApp_dateSalaryRetd_dateDept123Agra25-jan-2006500025-jan-2026Marketing127Mathura22-dec-2006600022-dec-2026Finance124Agra19-aug-2007550019-aug-2027Marketing125Delhi14-apr-2004850014-apr-2018Sales128Pune13-mar-2008750013-mar-2028Sales129Bangalore21-july-2003700021-july-2023Finance(a) Show empno, name and salary of those who have Sports as hobby.(b) Show name of the eldest employee.(c) Show number of employee area wise.(d) Show youngest employees from each Native place.(e) Show Sno, Name, Hobby and Salary in descending order of Salary.(f) Show the hobbies of those whose name pronounces as 'Abhay'.(g) Show the appointment date and native place of those whose name starts with 'A' or ends in 'd'.(h) Show the salary expense with suitable column heading of those who shall retire after 20-jan-2006.(i) Show additional burden on the company in case salary of employees having hobby as sports, is increased by 10%.(j) Show the hobby of which there are 2 or more employees.(k) Show how many employee shall retire today if maximum length of service is 20 years.(l) Show those employee name and date of birth who have served more than 17 years as on date.(m) Show names of those who earn more than all of the employees of Sales dept.(n) Increase salary of the employees by 5 % of their present salary with hobby as Music or they have completed atleast 3 years of service.(o) Write the output of :Select distinct hobby from personal ;Select avg(salary) from personal, job where Personal.Empno = Job.Sno and Area in ('Agra','Delhi') ;Select count(distinct Native_place) from personal.Select name, max(salary) from Personal, Job where Personal.Empno = Job.Sno;(p) Add a new tuple in the table Personal essentially with hobby as Music.(q) Insert a new column email in Job table(r) Create a table with values of columns empno, name, and hobby.(s) Create a view of Personal and Job details of those who have served less than 15 years.(t) Erase the records of employee from Job table whose hobby is not Sports.(u) Remove the table Personal.

Accepted Answer

(a)SELECT P.EMPNO, P.NAME, J.Salary FROM PERSONAL P, JOB J WHERE P.EMPNO = J.Sno AND P.Hobby = 'Sports';Output+-------+--------+--------+ | EMPNO | NAME | Salary | +-------+--------+--------+ | 125 | Vinod | 8500 | | 129 | Ramesh | 7000 | +-------+--------+--------+ (b)SELECT name FROM personal WHERE dobirth = ( SELECT MIN(dobirth) FROM personal );Output+------+ | name | +------+ | Amit | +------+ (c)SELECT Area, COUNT(Sno) AS Employee_Count FROM Job GROUP BY Area;Output+-----------+----------------+ | Area | Employee_Count | +-----------+----------------+ | Agra | 2 | | Delhi | 1 | | Mathura | 1 | | Pune | 1 | | Bangalore | 1 | +-----------+----------------+ (d)SELECT Name, `Native-place`, dobirth FROM personal WHERE dobirth = (SELECT MAX(dobirth) FROM personal p2 WHERE personal.`Native-place` = p2.`Native-place` ) ;Output+--------+--------------+------------+ | Name | Native-place | dobirth | +--------+--------------+------------+ | Abhai | Allahabad | 1975-08-11 | | Vinod | Delhi | 1977-04-04 | | Manoj | Mumbai | 1976-12-12 | | Ramesh | Pune | 1981-10-28 | +--------+--------------+------------+ (e)SELECT SNO, NAME, HOBBY, SALARY FROM PERSONAL, JOB WHERE PERSONAL.EMPNO = JOB.SNO ORDER BY SALARY DESC;Output+-----+--------+-----------+--------+ | SNO | NAME | HOBBY | SALARY | +-----+--------+-----------+--------+ | 125 | Vinod | Sports | 8500 | | 128 | Abhay | Gardening | 7500 | | 129 | Ramesh | Sports | 7000 | | 127 | Manoj | Writing | 6000 | | 124 | Abhai | Music | 5500 | | 123 | Amit | Music | 5000 | +-----+--------+-----------+--------+ (f)SELECT HOBBY FROM PERSONAL WHERE Name = 'abhay' or Name = 'abhai' ;Output+-----------+ | HOBBY | +-----------+ | Music | | Gardening | +-----------+ (g)SELECT App_date, nativeplace FROM personal, job WHERE personal.empno = job.sno AND (Name LIKE 'A%' OR Name LIKE '%d') ;Output+------------+--------------+ | App_date | native-place | +------------+--------------+ | 2006-01-25 | Delhi | | 2007-08-19 | Allahabad | | 2004-04-14 | Delhi | | 2008-03-13 | Mumbai | +------------+--------------+ (h)SELECT Salary AS "Salary Expense" FROM Job WHERE `Retd_date` > '2006-01-20';Output+----------------+ | Salary Expense | +----------------+ | 5000 | | 5500 | | 8500 | | 6000 | | 7500 | | 7000 | +----------------+ (i)SELECT SUM(Salary * 0.1) AS "Additional Burden" FROM PERSONAL, JOB WHERE PERSONAL.EMPNO = JOB.SNO AND HOBBY = 'SPORTS' ;Output+-------------------+ | Additional Burden | +-------------------+ | 1550.0 | +-------------------+ (j)SELECT Hobby FROM PERSONAL GROUP BY Hobby HAVING COUNT(*) >= 2;Output+--------+ | Hobby | +--------+ | Music | | Sports | +--------+ (k)SELECT COUNT(*) FROM Job WHERE DATEDIFF(CURDATE(), App_date) >= 20 * 365;Output+----------+ | COUNT(*) | +----------+ | 0 | +----------+ (l)SELECT P.Name, P.Dobirth FROM Personal P, Job J WHERE P.Empno = J.Sno AND J.Retd_date > CURDATE() AND DATEDIFF(CURDATE(), J.App_date) > 17 * 365;Output+-------+------------+ | Name | Dobirth | +-------+------------+ | Amit | 1965-01-23 | | Manoj | 1976-12-12 | +-------+------------+ (m)SELECT Name FROM Personal, job where personal.Empno = job.Sno and job.Salary > ( select max(salary) from job where dept = 'sales');ExplanationThere will be no output because there are no employees whose salary is greater than the highest salary in the Sales department.(n)UPDATE Job J, Personal P SET J.Salary = (J.Salary * 0.05 ) + J.Salary WHERE J.Sno = P.Empno AND (P.Hobby = 'Music' OR DATEDIFF(CURDATE(), J.App_date) >= 3 * 365);Output+-----+-----------+------------+--------+------------+-----------+ | sno | area | app_date | salary | retd_date | dept | +-----+-----------+------------+--------+------------+-----------+ | 123 | Agra | 2006-01-25 | 5250 | 2026-01-25 | Marketing | | 124 | Agra | 2007-08-19 | 5775 | 2027-08-19 | Marketing | | 125 | Delhi | 2004-04-14 | 8925 | 2018-04-14 | Sales | | 127 | Mathura | 2006-12-22 | 6300 | 2026-12-22 | Finance | | 128 | Pune | 2008-03-13 | 7875 | 2028-03-13 | Sales | | 129 | Bangalore | 2003-07-21 | 7350 | 2023-07-21 | Finance | +-----+-----------+------------+--------+------------+-----------+ (o)1.Select distinct hobby from personal ; Output+-----------+ | hobby | +-----------+ | Music | | Sports | | Writing | | Gardening | +-----------+ 2.Select avg(salary) from personal, job where Personal.Empno = Job.Sno and Area in ('Agra','Delhi') ; Output+-------------+ | AVG(SALARY) | +-------------+ | 6650.0000 | +-------------+ 3.Select count(distinct Native_place) from personal;Output+--------------------------------+ | COUNT(DISTINCT `NATIVE-PLACE`) | +--------------------------------+ | 4 | +--------------------------------+ 4.Select name, max(salary) from Personal, Job where Personal.Empno = Job.Sno ;Output+------+-------------+ | name | max(salary) | +------+-------------+ | Amit | 8500 | +------+-------------+ ExplanationThe given query retrieves the maximum salary from the 'Job' table and pairs it with the corresponding employee name from the 'Personal' table based on the matching Empno and Sno values. However, including a non-group field like 'name' in the select-list means it will return the value from the first record of the group for the 'name' field. Therefore, 'Amit' is the first record in the 'Personal' table, the query returns 'Amit' as the value for the 'name' field.(p)INSERT INTO Personal (Empno, Name, Dobirth, `Native-place`, Hobby) VALUES (130, 'Amruta', '1990-05-15', 'Chennai', 'Music');Outputselect * from personal; +-------+--------+------------+--------------+-----------+ | empno | name | dobirth | native-place | hobby | +-------+--------+------------+--------------+-----------+ | 123 | Amit | 1965-01-23 | Delhi | Music | | 124 | Abhai | 1975-08-11 | Allahabad | Music | | 125 | Vinod | 1977-04-04 | Delhi | Sports | | 127 | Manoj | 1976-12-12 | Mumbai | Writing | | 128 | Abhay | 1974-03-10 | Mumbai | Gardening | | 129 | Ramesh | 1981-10-28 | Pune | Sports | | 130 | Amruta | 1990-05-15 | Chennai | Music | +-------+--------+------------+--------------+-----------+ (q)ALTER TABLE Job ADD COLUMN Email VARCHAR(55); (r)insert into empdetails(empno, name, hobby) select empno, name, hobby from personal ;Outputselect * from empdetails ;+-------+--------+-----------+ | Empno | Name | Hobby | +-------+--------+-----------+ | 123 | Amit | Music | | 124 | Abhai | Music | | 125 | Vinod | Sports | | 127 | Manoj | Writing | | 128 | Abhay | Gardening | | 129 | Ramesh | Sports | | 130 | Amruta | Music | +-------+--------+-----------+ (s)CREATE VIEW LessThan15YearsView AS SELECT * FROM Personal p, Job j WHERE p.Empno = j.Sno AND DATEDIFF(CURDATE(), J.App_date) < 15 * 365;(t)DELETE j FROM Job j, Personal p WHERE j.Sno = p.Empno AND p.Hobby <> 'Sports';(u)DROP TAbLE Personal;

Question 14

Question 14aWith reference to the table below, answer the questions that follow :Table : EmployeesEmpidFirstnameLastnameAddressCity010RaviKumarRaj nagarGZB105HarryWaltorGandhi nagarGZB152SamTones33 Elm St.Paris215SarahAckerman440 U.S. 110Upton244ManilaSengupta24 Friends streetNew Delhi300RobertSamuel9 Fifth CrossWashington335RituTondonShastri NagarGZB400RachelLee121 Harrison St.New York441PeterThompson11 Red RoadParisTable : EmpSalaryEmpidSalaryBenefitsDesignation0107500015000Manager1056500015000Manager1528000025000Director2157500012500Manager2445000012000Clerk3004500010000Clerk3354000010000Clerk400320007500Salesman441280007500SalesmanWrite the SQL commands for the following using above tables :(i) To show firstname, lastname, address and city of all employees living in Pairs.(ii) To display the content of Employees table in descending order of Firstname.(iii) To display the firstname, lastname and total salary of all managers from the tables Employes and EmpSalary, where total salary is calculated as Salary + Benefits.(iv) To display the maximum salary among managers and clerks from the table EmpSalary.

Accepted Answer

Question 15

Question 14bWith reference to the table below, answer the questions that follow :Table : EmployeesEmpidFirstnameLastnameAddressCity010RaviKumarRaj nagarGZB105HarryWaltorGandhi nagarGZB152SamTones33 Elm St.Paris215SarahAckerman440 U.S. 110Upton244ManilaSengupta24 Friends streetNew Delhi300RobertSamuel9 Fifth CrossWashington335RituTondonShastri NagarGZB400RachelLee121 Harrison St.New York441PeterThompson11 Red RoadParisTable : EmpSalaryEmpidSalaryBenefitsDesignation0107500015000Manager1056500015000Manager1528000025000Director2157500012500Manager2445000012000Clerk3004500010000Clerk3354000010000Clerk400320007500Salesman441280007500SalesmanGive the Output of following SQL commands :(i) Select firstname, Salary from Employees, Empsalary where Designation = 'Salesman' and Employees.Empid = Empsalary.Empid ;(ii) Select count(distinct designation) from EmpSalary ;(iii) Select designation, sum(salary) from EmpSalary group by designation having count(*) > 2 ;(iv) Select sum(Benefits) from EmpSalary where Designation = 'Clerk' ;

Accepted Answer

(i)Output+-----------+--------+
| FIRSTNAME | SALARY |
+-----------+--------+
| RACHEL    |  32000 |
| PETER     |  28000 |
+-----------+--------+
(ii)Output+-----------------------------+
| COUNT(DISTINCT DESIGNATION) |
+-----------------------------+
|                           4 |
+-----------------------------+
(iii)Output+-------------+-------------+
| DESIGNATION | SUM(SALARY) |
+-------------+-------------+
| MANAGER     |      215000 |
| CLERK       |      135000 |
+-------------+-------------+
(iv)Output+---------------+
| SUM(BENEFITS) |
+---------------+
|         32000 |
+---------------+

Question 16

Question 15Show the average salary for all departments with more than 3 people for a job.

Accepted Answer

SELECT job, AVG(Sal) AS AvgSalary 
FROM Empl 
GROUP BY job HAVING COUNT(*) > 3;Output+----------+-----------+
| job      | AvgSalary |
+----------+-----------+
| CLERK    |    1037.5 |
| SALESMAN |      1400 |
+----------+-----------+

Question 17

Question 16Display only the jobs with maximum salary greater than or equal to 3000.

Accepted Answer

SELECT Job 
FROM Empl 
GROUP BY Job HAVING MAX(Sal) >= 3000;Output+-----------+
| Job       |
+-----------+
| PRESIDENT |
| ANALYST   |
+-----------+

Question 18

Question 17Find out number of employees having "Manager" as Job.

Accepted Answer

SELECT COUNT(*) AS NumManagers
FROM EmpSalary
WHERE Designation = 'Manager';Output+-------------+
| NumManagers |
+-------------+
|           3 |
+-------------+

Question 19

Question 18List the count of employees grouped by deptno. (table EMPL)

Accepted Answer

SELECT deptno, COUNT(*) AS employee_count
FROM EMPL
GROUP BY deptno;Output+--------+----------------+
| deptno | employee_count |
+--------+----------------+
|     20 |              5 |
|     30 |              6 |
|     10 |              3 |
+--------+----------------+

Question 20

Question 19List the sum of employees' salaries grouped by department. (table EMPL)

Accepted Answer

SELECT deptno, SUM(sal) AS total_salary
FROM EMPL
GROUP BY deptno;Output+--------+--------------+
| deptno | total_salary |
+--------+--------------+
|     20 |        10885 |
|     30 |         9400 |
|     10 |         8750 |
+--------+--------------+

Grouping Records, Joins in SQL