MySQL
Revision Sheet
1. Consider
the following tables GAMES. Write SQL commands for the statements (a)
& (c):
GCode |
GameName |
Number |
PrizeMoney |
ScheduleDate |
101 |
Garom Board |
2 |
5000 |
2007-01-23 |
102 |
Badminton |
2 |
12000 |
2008-12-12 |
103 |
Table Tennis |
4 |
8000 |
2007-02-14 |
105 |
Chess |
2 |
9000 |
2008-01-01 |
108 |
Lawn Tennis |
4 |
25000 |
2007-03-19 |
(a) To
display the name of all Games.
(b) To
display details of those games which are having PrizeMoney more than 7000.
(c) To
display the content of the Games table in ascending order of ScheduleDate.
(d) To
display day of week those Prize money is less than 10000.
(e) To
Display the all Record those prizeMoney in 4000,5000,9000,10000,15000.
(f) To
display the prizeMoney with Rs. 100 increment.
(g) To
display the GameName last 5 characters.
2. Write
the commands for the queres (a) to (f) and write the output of the SQL commands
given in (g) bases on relation EMPNEW:
RELATION
: EMPNEW
No |
Name |
Salary |
Area |
Age |
Grade |
Dept |
1 |
Keshar store |
40000 |
West |
45 |
C |
Civil |
2 |
Kirti arts |
35000 |
South |
38 |
A |
Elec |
3 |
Kripple |
60000 |
North |
52 |
B |
Civil |
4 |
Aryan stall |
38000 |
North |
29 |
B |
Civil |
5 |
Samsons |
42000 |
East |
35 |
A |
Comp |
6 |
Biswal |
29000 |
South |
34 |
A |
Mech |
(a) To
display the names of all employees who are in the area south.
(b) To
display name and age of all employees whose age>40.
(c) To
display list of all employees whose salary>=30000 and <=40000.
(d) To
display the employee’s names on ascending order of age.
(e) To
display the record those age between 33 and 45.
(f) To
display name, Salary and show to calculate 33% of Salary in new column HRA.
(g) Update
the salary for Grade B employee by increasing Rs. 1000.
(h) To
display the current date & Time.
(i) To
insert a new row with the following data:
7,
‘Tarik Singh’, 45000, ‘South’, 45, ‘C’, ‘Elec’ .
3. Create
database named as FAMILY using SQL command of the following column:
No,
Name, FemaleMemebers,MaleMembers,Income
(i) Take appropriate data type and constraints while
creating the table.
(ii) Insert 3 records into table.
(iii) Add a new field name as Occupation of type 2
characters.
4. Create
database named as LAB with appropriate data types and constraint using
SQL command of the following columns:
No,
ItemNAme, CostPerItem, Quantity, DateofPruchase, Warranty, Operational
Take appropriate data type and constraints while
creating the table.
5. Write
SQL commands for the queries (a) to (g) of the following relation:
RELATION:
FRIENDS
No |
Name |
Age |
Dept |
Dateofadm |
Fee |
Sex |
1 |
Pankaj |
24 |
Computer |
2007-01-10 |
120 |
M |
2 |
Shalini |
21 |
History |
2008-03-24 |
200 |
F |
3 |
Sanjay |
22 |
Hindi |
2006-12-12 |
300 |
M |
4 |
Sudha |
25 |
History |
2009-07-01 |
400 |
F |
5 |
Rakesh |
22 |
Hindi |
2007-09-05 |
250 |
M |
6 |
Shakeel |
30 |
History |
2008-01-27 |
300 |
M |
7 |
Surya |
34 |
Computer |
2007-02-25 |
210 |
M |
8 |
Shikha |
23 |
Hindi |
2007-07-31 |
200 |
F |
(a) To
show all the information about the student of History dept.
(b) To
list the names of female students who are in Hindi dept.
(c) To
list names of all students with their date of admission in ascending order.
(d) To
display student’s name, fee, age for male students only.
(e) To
count the number of students with age<23.
(f) Modify
the rows for computer department’s fees by increasing Rs. 100.
(g) To
display name of day those Fees is less than or equal 300.
(h) To
display the Month name first three characters those who are female.
(i) To
show the output of following Query:-
(a) SELECT name, age,dateofadm,dateofadm+2
FROM Friends Where Sex= ‘M’;
(b) SELECT name,dateofadm, dateofadm +565/24
FROM Friends Where Sex= ‘F’
(j) To
insert a new row in the table with following data:
9,
‘Zaheer’, 36, ‘Computer’, ’12-Mar-95’, 230, ‘M’.
6. Write
a SQL commands for the queries (a) to (e) of the following relation:
RELATION:
TOY_SHOPPE
SNo |
Toy_Name |
Category |
Price |
Quantity |
Statting_Age |
Ending_Age |
1 |
Popeye |
Stuff Toy |
450 |
12 |
0 |
1 |
2 |
Rapid Fire |
Two Players |
600 |
25 |
5 |
7 |
3 |
Teddy Bear |
Stuff Toy |
300 |
40 |
1 |
5 |
4 |
Creative Mind |
Building
Blocks |
800 |
18 |
5 |
8 |
5 |
Ping Pong |
Two Players |
500 |
53 |
5 |
9 |
6 |
Race-Trace |
Mechanical |
1800 |
17 |
9 |
12 |
(a) Display
the names of all Tous, which are for two players.
(b) To
display the names of toys and their Price having quantity more than 20
(c) To
display the list of all those Toys with
starting_age>2 in descending order of price.
(d) To
display a report with Toy_name, Category, Price and Total Value for only those
toys for which the Starting_Age is above 1 and Ending_Age is below 9.
(e) To
display the current Date & Time.
(f) To
calculate the average of price.
(g) To
calculate the sum of Quantity those toys categories are Stuff Toy.
(h) To
insert a new item in the existing table with the following data:
7,
Jerry, Stuff Toy, 600, 15, 1, 6
7. Write
SQL Commands for the queries (a) to (g) & output (h) to (i) for relation
STIPEND :
RELATION:
STIPEND
No |
Name |
Stipend |
Stream |
Avgmark |
Grade |
Class |
1 |
Neha |
450.00 |
Medical |
89.2 |
A |
11C |
2 |
Damini |
400.00 |
Commerce |
78.5 |
B |
12B |
3 |
Gaurav |
250.00 |
Humanities |
64.4 |
C |
11A |
4 |
Anu |
300.00 |
Commerce |
67.5 |
C |
12B |
5 |
Vikas |
500.00 |
Non-Medical |
92.0 |
A |
12A |
6 |
Rubina |
450.00 |
Non-Medical |
88.5 |
A |
12A |
(a) To
display the name of all students who are in Medical Stream.
(b) To
display name and average of all the students having Avg<70.0.
(c) To
display list of all students with stipend>350.00 in ascending order of Name.
(d) To
display a report with Name, Marks for each student in the table. Marks are
calculated as AvgMarks*5.
(e) To
count the Stream but do not show duplicate value(s).
(f) To
display the Stream and the number of students of each stream in the table
STUDENT.
(g) Update
stipend for the rows of Graqde A students by increasing Rs. 100.
(h) To
insert a new student in the table STUDENT with the following data:
7, ‘Sabina’, 500.00, ‘Non-medical’, 90.6, ‘A’, ‘11A’
(i) SELECT
Name, RIGHT(UPPER(Stream),5) FROM Stipend
(j) SELECT
Name, SUBSTR(Stream, 2,5) FROM Stipend WHERE Grade= ‘A’;
8.
Write a SQL commands for the
queries (a) to (f) and output (g) to (k) of the following relation: RELATION: BOOKHOUSE
No |
Title |
Author |
Subject |
Publisher |
Quantity |
Price |
1
|
Data structure |
Lipschute |
DS |
McGraw |
4 |
217.00 |
2
|
DOS Guide |
NORTRON |
OS |
PHI |
3 |
175.00 |
3
|
Turboc C++ |
Robot Lafore |
Prog |
Galgotia |
5 |
270.00 |
4
|
JAVA |
Norton |
Prog |
NULL |
NULL |
500.00 |
5
|
Dbase Dummies |
Palmer |
DBMS |
PustakM |
7 |
130.00 |
6
|
Mastering Windows |
Cowart |
OS |
BPB |
1 |
225.00 |
7
|
Computer Studies |
French |
FND |
Calgotia |
2 |
75.00 |
8
|
COBOL |
Stern |
Prog |
JohnW |
4 |
1000.00 |
9
|
Guide Network |
Freed |
NET |
Zpress |
3 |
200.00 |
10
|
Basic for Beginners |
Norton |
Prog |
BPB |
3 |
40.00 |
11
|
Advanced Pascal |
Schidt |
Prog |
McGraw |
4 |
350.00 |
(a) To
display the title of all books with Price between 100 and 300.
(b) To
Display Title and Author of all the books having type Prog and published by
BPB.
(c) To
display list of all the books with price more than 130 in ascending order Qty.
(d) To
Count the Records.
(e) To
display a report with Title, Price in descending order of price.
(f) To
insert new book in the table LIBRARY with the following data:
11, ‘Java Black Book’,
‘Steven Holzner’, ‘Prog’, DreamTech’, 5, 450.00
(g) SELECT
CHAR(65,86,73,82,65);
(h) SELECT
CHAR(73,110,102,111);
(i) SELECT
COUNT(Quantity) from BookHouse;
(j) SELECT
LEFT(LCASE(Title),4) FROM Bookhouse WHERE Quantity<=3;
(k) SELECT
Title, Length(Subject) as ‘Length’ from BookHouse WHERE Quantity<=3;
9. Write a SQL commands for the queries (a) to
(d) & (e) to ….. of the following
Table:
TABLE
: CONSIGNER
CnorD |
CnorName |
CnorAddress |
City |
ND01 |
R.
Singhal |
24,
ABC Enclave |
New
Delhi |
ND02 |
Amit
Kumar |
123,
Palm Avenue |
New
Delhi |
MU15 |
R.
Kohli |
5/A
South Street |
Mumbai |
MU50 |
S.
Kaur |
27-K,
Westend |
Mumbai |
(a) To
display the names of all Consigners from Mumbai.
(b) Display
all unique cities from the table.
(c) Display
all consigner names who is started with ‘S’.
(d) Arrange
the rows on ascending order of City.
(e) SELECT
CnorName, INSTR(CnorAddress, ‘A’) FROM Consigner;
(f) SELECT
CONCAT(‘My Name is ’, CnorNAme) FROM
Consigner;
10. Create
database TOYSHOP and table as TOY using SQL command of the following structure:
Field Name |
Field Type |
Constraint |
Sno |
Integer |
NOT NULL |
Toy_Name |
Character(20) |
|
Category |
Character(20) |
|
Price |
SMALLINT |
|
Quantity |
Integer |
|
Starting_Age |
Integer |
|
Ending_Age |
Integer |
|
(i) Insert at least 5 records into the TOY table.
(ii) Write a command to describe the structure of
the above table.
(iii) show all records.
(iv) Drop the constraint in the field SNO.
(v) Modify the column Price as FLOAT.
(vi) Drop the column Starting_Age.
(vii) Rename Field Name Quantity to Qty.
(viii) To insert NULL record. Column Price &
Quantity should be NULL.
(ix) To display the record those price is not null.
(x) To display Toy Name, price, quantity with add
new alias Column ‘TotalAmount’ which are calculate Price*Quantity.
(xi) To add Primary Key of column SNo.
11. Write
SQL commands to create the table HOSPITAL with the following specification:
Field Name |
Field Type |
Constraint |
PNo |
Number(4) |
Primary Key |
Name |
Char(20) |
|
Age |
Integer |
|
Department |
Char(15) |
Default ‘Comp’ |
Dateofadm |
Date |
|
Charges |
Float |
<1000 |
Sex |
Char(1) |
|
(i) Insert the 5 records
(ii) Write a command to describe the structure of
the above table.
(iii) Add one more column in the above table as
Address of type char(20).
(iv) Modify the column Address as varchar(30).
(v) Drop the column Address.
(vi) Drop primary key.
12. Write
SQL commands to create the table STUDENT with the following specification:
Field
Name |
Field
Type |
Width |
Constraint |
Roll_No |
Integer |
|
NOT NULL Primary Key |
Name |
Varchar |
25 |
|
Class |
Character |
4 |
|
Section |
Character |
1 |
|
DOB |
Date |
|
|
Total |
Smallint |
|
|
Percentage |
Float |
|
|
Grade |
Character |
1 |
|
Mr. Manoj creates a table Student but he is missing some
requirements. Please help to solve his troubles.
(a) Show
all tables.
(b) To
display the structure.
(c) He
was lost a column Address. Please help to add column address after column name.
(d) He
does not want that blank Name
column. Please help which constraint to add.
(e) He
wants to change data type column Total SMALLINT to MEDIUMINT.
(f) He
wants to change the column Name to Stud_Name.
(g) To
change size address 30.
(h) To
remove the column grade.
(i) Create
a new table Teachers with
appropriate data types and constraint using SQL command of the following
columns:
Tno,Tname,tAddress,salary,dept_no,doj
(j) To
add constraints FOREIGN KEY & references Table Teacher
column tno.
(k) To
remove foreign key.
(l) To
Rename table STUDENT to STUDENTS.
(m)
To insert values only column rollno,
stud_name, address.
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