Sql wksht-5

SQL Simple Queries – WK5

SQL Queries - Basics
Worksheet - 5
1 Get the last name and first name of all authors whose lastname has 5 or more
characters.
SELECT LNAme, FName
FROM Authors
WHERE LEN(Lname) >= 5

2 Find the Employee number, Sales, and DeptID of each employee. If the employee has
no sales, then four dashes should be printed
SELECT EmpID,
IFNULL(Sales, ‘----‘), DeptID
FROM Employee

3 Find the cities that repeat in the Authors table.
SELECT CITY
FROM AUTHORS
GROUP BY CITY
HAVING COUNT(*) > 1
ORDER BY CITY

4 Find all the cities where the company has customers.
SELECT CITY
FROM AUTHORS
GROUP BY CITY
ORDER BY CITY

5 Find the names of employees who joined the company 22 years or later after they were
born.
SELECT EMPID, LNAME
FROM EMPLOYEE
WHERE YEAR(DOB) + 22 >= YEAR(DOJ)

6 Find the EmpID of all male workers born before Jan 1, 1990.
SLEECT EmpID
FROM Employee
WHERE Gender = ‘M’ AND YEAR(DOB) < 1990

7 Find the names of all employees who do not live in ‘Mumbai’.
SELECT LName
FROM Employee
WHERE City <> ‘Mumbai’

8 Find the employee number of all employees born between 1-1-1980 and 1-1-1990
SELECT EmpID
FROM Employee
WHERE DOB >= ’01-01-1980’ AND DOB <= ’01-01-1990’
Prof. Mukesh N. Tekwani Page 1 of 14

SQL Simple Queries – WK 5

This can also be written as :

SELECT EmpID
FROM Employee
WHERE DOB BETWEEN ’01-01-1980’ AND ’01-01-1990’

9 Find the names of employees who have worked in one of the following depts.: Sales,
Mktg, Accounts, PR.
SELECT LNAme
FROM Employee
WHERE Dept IN (‘Sales’, ‘Mktg’, ‘Accounts’, ‘PR’)

10 Find the names of all employees whose lastname starts with C.
SELECT LName
FROM Employee
WHERE LNAME LIKE ‘C%’

11 Find the names of all employees whose lastname ends with C.
SELECT LName
FROM Employee
WHERE LNAME LIKE ‘%C’

12 Find the names of all employees whose lastname has the penultimate letter ‘e’.
SELECT LName
FROM Employee
WHERE LNAME LIKE ‘%e_’

13 Find the names of all employees whose lastname has the letter ‘e’.
SELECT LName
FROM Employee
WHERE LNAME LIKE ‘%#e%’ ESCAPE #

14 Wildcard explained
The LIKE operator allows us to perform basic pattern-matching using wildcard characters.

Wildcard Description
_ (underscore) matches any single character
% matches a string of one or more characters
matches any single character within the specified range (e.g. [a-f]) or
[]
set (e.g. [abcdef]).
matches any single character not within the specified range (e.g. [^a-
[^]
f]) or set (e.g. [^abcdef]).

Examples:

• WHERE FirstName LIKE '_im' finds all three-letter first names that end with 'im' (e.g.
Jim, Tim).
• WHERE LastName LIKE '%stein' finds all employees whose last name ends with 'stein'
Page 2 of 14 mukeshtekwani@hotmail.com


• WHERE LastName LIKE '%stein%' finds all employees whose last name includes
'stein' anywhere in the name.
• WHERE FirstName LIKE '[JT]im' finds three-letter first names that end with 'im' and
begin with either 'J' or 'T' (that is, only Jim and Tim)
• WHERE LastName LIKE 'm[^c]%' finds all last names beginning with 'm' where the
following (second) letter is not 'c'.

For Q 15 to Q 27, the table structure is :

Sales_Header (OrderNo , OrderDate , CustomerID ,
Salesmanid , PaymentStatus , TransactionType)

Sales_Detail(OrderNo , ProductID , Qty , Rate, Amt)

15 Display the orders that were issued in the first quarter of the current year. (Nov 04)
SELECT Orderno
FROM Orders
WHERE MONTH(Orderdate) < 4
AND YEAR(Orderdate) = YEAR(GETDATE())

16 Display the order number, order date , customer name and order amount for
orders having value of Rs 500 and above. (Nov 04)
SELECT Orderno, Orderdate, Customer_Name, Amt
FROM Sales_Header H, Sales_Detail D
WHERE H.Orderno = D.Orderno AND Amt >= 500

17 Display the order detail where RIN001 soap is sold for minimum of Rs 50 (Nov 04)
SELECT Sales_Header.*
FROM Sales_Header H, Sales_Detail D
WHERE H.Orderno = D.Orderno AND Productid = ’RIN001’ AND
Rate >= 50

18 Display the order collected by Executive no. ‘S120’ Nov 04
SELECT O.Orderid , D.Productid, D.Quantity
FROM Orders O, [Order Details] D
WHERE O.Orderid = D.Orderid AND O.Employeeid = ‘S120’

19 Assign the privileges to see the data from both tables to ‘Raj’. Nov 04
GRANT SELECT
ON SALES_HEADER, SALES_DETAILS
TO RAJ

20 Grant the INSERT and UPDATE privileges to Raj, for all columns in the
SALES_DETAILS table.
GRANT INSERT, UPDATE
ON SALES_DETAILS
TO RAJ

21 Display the various types of product sold to Customers. Nov 04



SELECT DISTINCT Productid
FROM SalesDetails

22 Display all Credit transactions , with their payment status done in Dec 2003.
SELECT *
FROM SALES_HEADER
WHERE TRANSACTIONTYPE = ‘CREDIT’
AND PAYMENTSTATUS = ’PAID’
AND MONTH(ORDERDATE) = 12
AND YEAR(ORDERDATE) = 2003

23 Display the details of total cash transactions done by each sales executive.
SELECT *
FROM Sales_Header
GROUP BY Salesmanid
HAVING TransactionType = ’Cash’

24 Update salary of employee ‘Raj’ by giving him the salary of ‘Radha’ working in
same company. Nov 04
UPDATE Emp
SET salary =
(SELECT salary FROM Emp WHERE Ename = ’Radha’)
WHERE Ename = ’Raj’ AND
Compid = (SELECT compid FROM emp WHERE ename=’Radha’)

25 Display how many male and female members have joined in January 2004. Nov 04
SELECT COUNT (ename)
FROM emp
GROUP BY gender
HAVING TO_CHAR(joindate,’mon-yyyy’) = ’JAN-2004’

26 Display the total number of customers.
SELECT COUNT(customerId)
FROM customers

27 Display the total number of customers located in each city. Nov 04
The city with maximum number of customers should be at the top of the list.
SELECT count(customerid) , city
FROM customers
GROUP BY city
ORDER BY 1 DESC

Assume the following table structure:
Emp (Empid , ename, city, CompID , salary , Joindate )
Company (CompID , CompName , City)

28 Give name of companies located in those cities where ‘TATA’ is located. Nov 04
SELECT Companyname
FROM company


WHERE city IN
(SELECT city
FROM company
WHERE companyName = ’TATA’)

29 Give the names of the employees living in the same city where their company is
located. Nov 04
SELECT ename
FROM Emp e, company c
WHERE e.city = c.city

30 Display all customers who stay in the same city as the employees.
SELECT c.customerid, e.employeeid
FROM customers c, employees e
WHERE c.city = e.city

Consider the following table structure for questions below:
Book (BookID , Title , Author , Publisher , Year , Price , DistID)
Distributors (DistID , Name , City) Nov 04

31 Find the title and price of the most expensive book(s).
SELECT Title, Price
FROM Book
WHERE price = (SELECT MAX(Price) FROM Titles)

32 Display the details of books whose price is more than the average price of all
books. Nov 04
SELECT *
FROM book
WHERE Price > (SELECT AVG (Price) FROM book )

33 Display the details of books written by ‘Groff’ and supplied by ‘TMH’ Nov 04
SELECT *
FROM book
WHERE Author = ’Groff’ AND Publisher = ’TMGH’

34 Create a view to show the Title, Author, Publisher and Distributer’s Name &
name this view as ShowDetails. Display the records held in this view. Nov 04
CREATE VIEW ShowDetails
AS
SELECT title, author, publisher, distributor
FROM book
35 Create the table RIVERS according to the description shown below.
Attribute Data type Constraint
Name Upto 20 characters long Unique and not null
Length Upto 4 digits Values between 100 and 4160
Outflow Upto 20 characters long Required

CREATE TABLE river


(Name VARCHAR(20) UNIQUE NOT NULL,
Length NUMERIC CHECK (LENGTH BETWEEN 101 AND 4160),
Outflow VARCHAR(20) NOT NULL)

36 Add a new column called Maxdepth to the rivers table. Make sure that the values
of this column range from 100 to 250 feet.
ALTER TABLE river
ADD maxdepth NUMERIC
CHECK (maxdepth BETWEEN 100 AND 250)

37 Create table ‘CITY’ With the following fields:
Field Data type Constraint
CID INTEGER PRIMARY KEY
CNAME Upto 20 char
State 2 char
Latitude Number
Longitude Number
CREATE TABLE city
(
CId INTEGER PRIMARY KEY,
CName VARCHAR(20),
State CHAR(2),
Latitude NUMERIC(5),
Longitude NUMERIC(5)
)

38 Create table DATA with following fields:

Field Data type Constraint
CID Integer Foreign Key Matching City Id
MONTH Integer Must Be Between 1 And 12
TEMP_F Number Must Be Between -80 And 150
Rain_I Number Must be between 0 and 100
(ID+MONTH) Primary key
CREATE TABLE data
(
CID INTEGER,
Month INTEGER CHECK(month BETWEEN 1 AND 12),
Temp_f NUMERIC(3) CHECK(temp_f BETWEEN -80 AND 150),
Rain_i NUMERIC(3) CHECK(rain_i BETWEEN 0 AND 100),
PRIMARY KEY (CID, MONTH),
CONSTRAINT fkey FOREIGN KEY(CID) REFERENCES city
)

39 List the temperatures for July from table Data, Lowest temperatures first, picking
up city name and latitude by joining with the table city.
SELECT Temp_f, cname, Latitude
FROM Data d, City c
WHERE d.cId = c.cId AND month = 7
ORDER BY Temp_f ASC


40 List (using Subquery) the cities with year round average temperature above 50
degrees.
SELECT name
FROM City
WHERE Id IN
(
SELECT cid FROM Data WHERE 50 < All
(SELECT avg(temp_f) FROM data GROUP BY CID)
)

Note:
When the ANY clause is used, a value from the current row is compared to each result
set of the subquery using the comparison operator. If any of the comparisons are true,
the entire comparison is considered to be true.

The ALL clause is similar to the ANY clause except that after the value is compared to
all the items returned by the subquery, if any of the comparisons is false, the entire
clause is false. That is, if there is a comparison based on equality, the value on the left
of the comparison must be equal to every item in the result set for the ALL clause to
evaluate to true.

Consider the following tables and answer the queries that follow:

CUSTOMER
Attribute Data Type Constraints
Custid Varchar(2) Primary key
Lname Varchar(15)
Fname Varchar(15)
Area Varchar(2)
Phone Number(8)

MOVIE
Attribute Datatype Constraints
Mvno Number(2) Primary key
Title Varchar(25)
Type Varchar(10)
Star Varchar(25)
Price Number(8,2)

INVOICE
Attribute Datatype Constraints
Invno Varchar(3) Primary key
Mvno Number(2) Foreign key movie(mvno)
Custid Varchar(3) Foreign Key customer(custid)
Issue date Date
Return date Date

41 Find out the movies that cost more than 159 and also find the new cost as original
cost*15
SELECT mvno, price*15 “NEW COST”
FROM movie


WHERE price > 159

42 Print the names and types of all the movies except horror movies.
SELECT title, type
FROM movie
WHERE type NOT IN(‘horror’)

43 List the various movie types available
SELECT DISTINCT type
FROM movie

44 List the mvno, title of movies whose stars begin with the letter ’m’
SELECT mvno, title, type
FROM movie
WHERE star LIKE ‘m%’

45 Determine the maximum and minimum of price. Rename the columns as max-
price and min_price respectively.
SELECT MAX(price) “max_price”, MIN(price) “min_price”
FROM movie

46 Find out number of movies in each type,
SELECT COUNT (type)
FROM movie
GROUP BY type

47 Print the information of invoice table in the following format for all records.
The invoice no of customer id.{ custid} is { invno} and movie no is { movno}
SELECT ‘The invoice no of customer id : ’, custid, ’is ‘,
invno, ‘ and movie No. is ‘, movno
FROM invoice

48 Display the month(in alphabets) in which customers are supposed to return the
movies
SELECT MONTH(returnDate)
FROM Invoice

49 Delete all records having return date before 10th July ‘05
DELETE FROM invoice
WHERE returndate < ’10-JUL-05’

50 Find out if the movie starring ‘Tom Cruise’ is issued to any customer and list the
custid to whom it is issued
SELECT custid
FROM invoice, movie
WHERE invoice.mvno = movie.mvno AND
movie.star LIKE ‘Tom Cruise’

51 Find the names of customers who have been issued movie of type ’drama’.


SELECT Fname, ’ ‘, lname, “Name”
FROM customer c, movie M, invoice i
WHERE c.custid = i.custid AND m.mvno = i.mvno
AND m.type LIKE ‘drama’

52 Find out the title of the movies that have been issued to the customer whose
Fname is’ Mary’
SELECT title
FROM customer, movie, invoice
WHERE customer.custid = invoice.custid AND
Movie.mvno = invoice.mvno AND Customer.Fname LIKE ‘Mary’

53 Add a column Remark of type varchar and size 25 to the invoice table
ALTER TABLE invoice
ADD (Remark VARCHAR(25))

54 Find the names of all customers having ‘a’ in the second letter in their fname.
SELECT Fname, lname
FROM customer
WHERE fname LIKE ’_a%’

55 Find out the movie number which has been issued to customer whose first name
is ‘IVAN’
SELECT i.mvno
FROM customer c, invoice I
WHERE c.custid = i.custid AND fname LIKE ‘IVAN’

56 Display the title, Lname, Fname for customers having movie number greater than
or equal to three in the following format.
The movie taken by { Fname} { Lname} is { Title}
SELECT ‘The movie taken by’, Fname, ‘ ‘, Lname, ’ is ‘, Title
FROM customer c, movie m, invoice i
WHERE c.custid = i.custid AND m.mvno = i.mvno AND m.mvno >= 3

57 Find the second highest sales.
SELECT MAX(sales)
WHERE sales NOT IN
(SELECT MAX(sales) FROM sales)

Answer the following questions based on the following tables:
Student ( sid number, sname varchar, previous_percentage number,
aptitude_test_status number, grade char)
58 Write an ALTER TABLE statement that defines a single CHECK constraints for the
following: Column previous_percentage must be a value between 40 to 100.
ALTER TABLE Student
MODIFY Previous_percentage NUMERIC(5)
CHECK (Previous_percentage between 40 and 100)

59 Make the sid column unique.


ALTER TABLE Student
ADD CONSTRAINT ID_UNIQ UNIQUE (sid)

60 Add a column called TestDate with default value as today’s
date.
ALTER TABLE Student
ADD TestDate datetime DEFAULT (GETDATE())

61 Aptitude_test_status must be a value in between 60 to 100.
ALTER Table Student
ADD Check (Aptitude_test_status BETWEEN 60 AND 100)

62 The grade column should allow values ‘a’, ’b’ and ‘c’ only.
ALTER TABLE Student
ADD CHECK (grade IN ('a', 'b', 'c'))

Following queries are based on these tables:
Supplier (sno, sname, status, city)
Parts (pno, pname, color, weight, city)
Shipment (sno, pno, quantity, sdate)
63 Change the status of each supplier to its double for suppliers in ‘Mumbai’.
UPDATE supplier
SET status = status * 2
WHERE city = 'Mumbai'

64 Get the sno for suppliers with status less than current maximum status in the
supplier table.
First we form the subquery to find the maximum status, as follows:
SELECT MAX(status)
FROM supplier

Now, the complete query is :
SELECT sno
FROM supplier
WHERE status < (SELECT MAX(status)FROM supplier)

65 Create a view of Suppliers whose status is above 2.
CREATE VIEW suppliers_view (sno,sname,status,city)
AS
SELECT * FROM supplier WHERE status > 2

66 Get all parts whose weight is 10, 20 or 30.
SELECT pno, pname
FROM parts
WHERE weight IN (10,20,30)

67 Get sname for suppliers who do not supply at least one part.
SELECT sname
FROM supplier s, shipment sh
WHERE s.sno = sh.sno AND s.sno NOT IN
(SELECT DISTINCT sno FROM shipment)


The following queries are based on the following tables:
Employee (ecode, ename, salary, status, deptno, city)
Dept(deptno, dname, city)
Product (pno, sno, pname, unitsinstock, unitprice)
Order(ono, pno, quantity, odate)
68 Create the Employee table
CREATE TABLE employee
(
ecode INTEGER,
ename VARCHAR(50),
salary MONEY,
status CHAR,
deptno INTEGER,
city VARCHAR(20)
)

69 Alter the table to add a column for qualifications.
ALTER TABLE Employee
ADD qual VARCHAR(20)

70 Alter the table to drop the qual column.
DROP COLUMN qual

71 Alter the table to add a column pfnum which is unique (add a column with a
constraint)
ADD pfnum INTEGER
CONSTRAINT ct_pfnum UNIQUE

72 Alter the table to add a column totalsalary with the constraint that this cannot be
greater than 10000.
ALTER TABLE employee
ADD totalsalary MONEY
CONSTRAINT maxsal CHECK (totalsalary <= 10000)

73 Alter the employee table so that the ecode is less than 100
ADD CONSTRAINT validecode CHECK (ecode <= 100)

74 Remove the constraint on ecode.
DROP CONSTRAINT validecode

75 Add a constraint that the dept code can be one of 11, 22, 33, 44 or 55.
ADD CONSTRAINT validdept CHECK
(deptno in(11, 22, 33, 44, 55))


76 Multiple check constraints:
Assume there is a column called ‘SalaryType’ (Hourly, Monthly, Annual). Create a
single constraint that checks both the salary and the salary type
ALTER TABLE Payroll
ADD CONSTRAINT CK_Payroll
CHECK
(SalaryType in ('Hourly','Monthly','Annual')
AND
(Salary > 100 AND Salary < 150000)

77 Add the following constraint in a single ALTER statement:
If SalaryType is Hourly, salary must be less than 100.
If SalaryType is Monthly, salary must not be more than 10000
If SalaryType is Annual, any salary amount is valid.
ALTER TABLE Payroll
CHECK
(
(SalaryType = 'Hourly' AND Salary < 100.00)
OR
(SalaryType = 'Monthly' AND Salary < 10000.00)
OR
(SalaryType = 'Annual')
)

78 Alter the table so that the SalaryType is not null and SalaryType is one of these:
Hourly, Monthly, Annual.
ALTER TABLE Payroll
CHECK
(
(SalaryType IN ('Hourly','Monthly','Annual'))
AND
SalaryType IS NOT NULL
)

79 Write an ALTER table statement that allows only the following cities: ‘BOSTON’,
‘DALLAS’, and ‘AUSTIN’
ADD CONSTRAINT validcity
CHECK (CITY IN ('BOSTON', 'DALLAS', 'AUSTIN'))

80 Define a view for Sue (employee number 102) containing only orders placed by customers
assigned to her. (horizontal view)
CREATE VIEW SUEORDERS AS
SELECT *
FROM ORDERS
WHERE CUST IN
(SELECT CUST_NUM FROM CUSTOMERS WHERE CUST_REP = 102)



82 A vertical view restricts a user’s access to only certain columns of a table.
CREATE VIEW STUD_ADDRESS AS
SELECT ROLLNO, NAME, ADD1, ADD2, CITY
FROM STUDENTS

83 Define a view that contains summary order data for each salesperson. (grouped view)
CREATE VIEW ORD_BY_REP (WHO, HOW_MANY, TOTAL, LOW, HIGH, AVERAGE)
AS
SELECT REP, COUNT(*), SUM(AMOUNT), MIN(AMOUNT), MAX(AMOUNT), AVG(AMOUNT)
FROM ORDERS
GROUP BY REP

84 Create a view to print names of all movies in capital letters.
CREATE VIEW Movies_upper(title)
AS
SELECT UPPER(movie_title)
FROM Movies

85 Create a view to find the total revenue from all movies.
CREATE VIEW TRev (total_revenue)
AS
SELECT SUM(gross)
FROM Movies

86 Create a view to find all people who live in a state where a movie studio is located.
CREATE VIEW PS
AS
SELECT FName, LName, StudioName, Person_State
FROM People, Studio
WHERE Person_State = Studio_State

87 Create a view that returns the list of all the studios with an average budget of over $50 million.
(view with a sub-query)
CREATE VIEW Big_Movies
AS
SELECT movie_title, budget, gross
FROM Movies
WHERE studio_id IN
(SELECT studio_id
FROM Movies
GROUP BY studio_id
HAVING AVG(budget) > 50)

88 Add a single field to the 'Sales' table by specifying the new field name and data
type.
ALTER TABLE Sales
ADD UnitPrice MONEY

89 The following example adds the reserved words NOT NULL to require valid data
to be added to that field. However, you must also specify a default value for any
data already inserted in the table.
ALTER TABLE Sales
ADD UnitPrice MONEY


DEFAULT 3.00 NOT NULL

90 This example changes the data type of the column 'ItemCount' to an integer (INT)
ALTER TABLE Sales
ALTER COLUMN ItemCount INT

91 This example demonstrates how to use DROP COLUMN to delete a single field.
ALTER TABLE Sales
DROP COLUMN UnitPrice

92 Display all students whose surname is between ‘N’ and ‘R’
SELECT *
FROM STUDENTS
WHERE SURNAME BETWEEN ‘N’ and ‘R’

93 Display the highest salary paid in each department.
SELECT max(salary), dept
FROM employee
GROUP BY dept

Tables:
items_ordered
customerid order_date item quantity price

Cutsomer

customerid firstname lastname city state

94 How many people are in each unique state in the customers table? Select the
state and display the number of people in each. Hint: count is used to count
rows in a column, sum works on numeric data only.
SELECT state, COUNT(state)
FROM customers
GROUP BY state

95 Write a query using a join to determine which items were ordered by each of
the customers in the customers table. Select the customerid, firstname,
lastname, order_date, item, and price for everything each customer purchased
in the items_ordered table.
SELECT c.customerid, c.firstname, c.lastname,
i.order_date, i.item, i.price
FROM customers c, items_ordered i
WHERE c.customerid = i.customerid


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