1. 2 tables
2. Select all employees who doesn't have phone?
SELECT empname
FROM Employee
WHERE (empid NOT IN
(SELECT DISTINCT empid
FROM phone))
3. Select the employee names who is having more than one phone numbers.
SELECT empname
FROM employee
WHERE (empid IN
(SELECT empid
FROM phone
GROUP BY empid
HAVING COUNT(empid) > 1))
4. Select the details of 3 max salaried employees from employee table.SELECT TOP 3 empid, salary
FROM employee
ORDER BY salary DESC
5. Display all managers from the table. (manager id is same as emp id)
SELECT empname
FROM employee
WHERE (empid IN
(SELECT DISTINCT mgrid
FROM employee))
6. Write a Select statement to list the Employee Name, Manager Name under a particular manager?
SELECT e1.empname AS EmpName, e2.empname AS ManagerName
FROM Employee e1 INNER JOIN
Employee e2 ON e1.mgrid = e2.empid
ORDER BY e2.mgrid
7. 2 tables emp and phone.
emp fields are - empid, name
Ph fields are - empid, ph (office, mobile, home). Select all employees who doesn't have any ph nos.
SELECT *
FROM employee LEFT OUTER JOIN
phone ON employee.empid = phone.empid
WHERE (phone.office IS NULL OR phone.office = ' ')
AND (phone.mobile IS NULL OR phone.mobile = ' ')
AND (phone.home IS NULL OR phone.home = ' ')
8. Find employee who is living in more than one city.
Two Tables:
9. SELECT empname, fname, lname
FROM employee
WHERE (empid IN
(SELECT empid
FROM city
GROUP BY empid
HAVING COUNT(empid) > 1))
10. Find all employees who is living in the same city. (table is same as above)
SELECT fname
FROM employee
WHERE (empid IN
(SELECT empid
FROM city a
WHERE city IN
(SELECT city
FROM city b
GROUP BY city
HAVING COUNT(city) > 1)))
11. There is a table named MovieTable with three columns - moviename, person and role. Write a query which gets the movie details where Mr. Amitabh and Mr. Vinod acted and their role is actor.
SELECT DISTINCT m1.moviename
FROM MovieTable m1 INNER JOIN
MovieTable m2 ON m1.moviename = m2.moviename
WHERE (m1.person = 'amitabh' AND m2.person = 'vinod' OR
m2.person = 'amitabh' AND m1.person = 'vinod') AND (m1.role = 'actor') AND (m2.role = 'actor')
ORDER BY m1.moviename
12. There are two employee tables named emp1 and emp2. Both contains same structure (salary details). But Emp2 salary details are incorrect and emp1 salary details are correct. So, write a query which corrects salary details of the table emp2
update a set a.sal=b.sal from emp1 a, emp2 b where a.empid=b.empid
13. Given a Table named “Students” which contains studentid, subjectid and marks. Where there are 10 subjects and 50 students. Write a Query to find out the Maximum marks obtained in each subject.
14. In this same tables now write a SQL Query to get the studentid also to combine with previous results.
15. Three tables – student , course, marks – how do go at finding name of the students who got max marks in the diff courses.
SELECT student.name, course.name AS coursename, marks.sid, marks.mark
FROM marks INNER JOIN
student ON marks.sid = student.sid INNER JOIN
course ON marks.cid = course.cid
WHERE (marks.mark =
(SELECT MAX(Mark)
FROM Marks MaxMark
WHERE MaxMark.cID = Marks.cID))
16. There is a table day_temp which has three columns dayid, day and temperature. How do I write a query to get the difference of temperature among each other for seven days of a week?
SELECT a.dayid, a.dday, a.tempe, a.tempe - b.tempe AS Difference
FROM day_temp a INNER JOIN
day_temp b ON a.dayid = b.dayid + 1
OR
Select a.day, a.degree-b.degree from temperature a, temperature b where a.id=b.id+1
17. There is a table which contains the names like this. a1, a2, a3, a3, a4, a1, a1, a2 and their salaries. Write a query to get grand total salary, and total salaries of individual employees in one query.
SELECT empid, SUM(salary) AS salary
FROM employee
GROUP BY empid WITH ROLLUP
ORDER BY empid
18. How to know how many tables contains empno as a column in a database?
SELECT COUNT(*) AS Counter
FROM syscolumns
WHERE (name = 'empno')
19. Find duplicate rows in a table? OR I have a table with one column which has many records which are not distinct. I need to find the distinct values from that column and number of times it’s repeated.
SELECT sid, mark, COUNT(*) AS Counter
FROM marks
GROUP BY sid, mark
HAVING (COUNT(*) > 1)
20. How to delete the rows which are duplicate (don’t delete both duplicate records).
SET ROWCOUNT 1
DELETE yourtable
FROM yourtable a
WHERE (SELECT COUNT(*) FROM yourtable b WHERE b.name1 = a.name1 AND b.age1 = a.age1) > 1
WHILE @@rowcount > 0
DELETE yourtable
FROM yourtable a
WHERE (SELECT COUNT(*) FROM yourtable b WHERE b.name1 = a.name1 AND b.age1 = a.age1) > 1
SET ROWCOUNT 0
21. How to find 6th highest salary
SELECT TOP 1 salary
FROM (SELECT DISTINCT TOP 6 salary
FROM employee
ORDER BY salary DESC) a
ORDER BY salary
22. Find top salary among two tables
SELECT TOP 1 sal
FROM (SELECT MAX(sal) AS sal
FROM sal1
UNION
SELECT MAX(sal) AS sal
FROM sal2) a
ORDER BY sal DESC
23. Write a query to convert all the letters in a word to upper case
SELECT UPPER('test')
24. Write a query to round up the values of a number. For example even if the user enters 7.1 it should be rounded up to 8.
SELECT CEILING (7.1)
25. Write a SQL Query to find first day of month?
SELECT DATENAME(dw, DATEADD(dd, - DATEPART(dd, GETDATE()) + 1, GETDATE())) AS FirstDay
26. Table A contains column1 which is primary key and has 2 values (1, 2) and Table B contains column1 which is primary key and has 2 values (2, 3). Write a query which returns the values that are not common for the tables and the query should return one column with 2 records.
27. There are 3 tables Titles, Authors and Title-Authors (check PUBS db). Write the query to get the author name and the number of books written by that author, the result should start from the author who has written the maximum number of books and end with the author who has written the minimum number of books.
28.
29. List all products with total quantity ordered, if quantity ordered is null show it as 0.
30. ANY, SOME, or ALL?
SELECT tbla.a
FROM tbla, tblb
WHERE tbla.a <>
(SELECT tblb.a
FROM tbla, tblb
WHERE tbla.a = tblb.a)
UNION
SELECT tblb.a
FROM tbla, tblb
WHERE tblb.a <>
(SELECT tbla.a
FROM tbla, tblb
WHERE tbla.a = tblb.a)
OR (better approach)
SELECT a
FROM tbla
WHERE a NOT IN
(SELECT a
FROM tblb)
UNION ALL
SELECT a
FROM tblb
WHERE a NOT IN
(SELECT a
FROM tbla)
SELECT authors.au_lname, COUNT(*) AS BooksCount
FROM authors INNER JOIN
titleauthor ON authors.au_id = titleauthor.au_id INNER JOIN
titles ON titles.title_id = titleauthor.title_id
GROUP BY authors.au_lname
ORDER BY BooksCount DESC
UPDATE emp_master
SET emp_sal =
CASE
WHEN emp_sal > 0 AND emp_sal <= 20000 THEN (emp_sal * 1.01) WHEN emp_sal > 20000 THEN (emp_sal * 1.02)
END
SELECT name, CASE WHEN SUM(qty) IS NULL THEN 0 WHEN SUM(qty) > 0 THEN SUM(qty) END AS tot
FROM [order] RIGHT OUTER JOIN
product ON [order].prodid = product.prodid
GROUP BY name
Result:
coke 60
mirinda 0
pepsi 10
ALL means greater than every value--in other words, greater than the maximum value. For example, >ALL (1, 2, 3) means greater than 3.
ANY means greater than at least one value, that is, greater than the minimum. So >ANY (1, 2, 3) means greater than 1. SOME is an SQL-92 standard equivalent for ANY.
