Structure Query Language

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Question
CBSEENCO12011543

Write the definition of a method OddSum(NUMBERS) to add those values in the list of NUMBERS, which are odd. 

Solution

def OddSum(NUMBERS):
	n=len(NUMBERS)
	s=0
	for i in range(n):
		if (i%2!=0):
			s=s+NUMBERS[i]
	print(s)

Sponsor Area

Question
CBSEENCO12011551

Write SQL queries for (i) to (iv) and find outputs for SQL queries (v) to (viii), which are based on the tables

TABLE: BOOK

Code

BNAME

TYPE

F101

The priest

Fiction

L102

German easy

Literature

C101

Tarzan in the lost world

Comic

F102

Untold Story

Fiction

C102

War Heroes

Comic


TABLE: MEMBER


MNO

MNAME

CODE

ISSUEDATE

M101

RAGHAV SINHA

L102

2016-10-13

M103

SARTHAK JOHN

F102

2017-02-23

M102

ANISHA KHAN

C101

2016-06-12


(i) To display all details from table MEMBER in descending order of ISSUEDATE.

(ii) To display the BNO and BNAME of all Fiction Type books from the table BOOK

(iii) To display the TYPE and number of books in each TYPE from the table BOOK

(iv) To display all MNAME and ISSUEDATE of those members from table MEMBER who have books issued (i.e ISSUEDATE) in the year 2017.

(v)SELECT MAX(ISSUEDATE) FROM MEMBER;

(vi) SELECT DISTINCT TYPE FROM BOOK;

(vii) SELECT A.CODE, BNAME, MNO, MNAME FROM BOOK A, MEMBER B WHERE A.CODE=B.CODE ;
(viii) SELECT BNAME FROM BOOK
WHERE TYPE NOT IN ('FICTION', 'COMIC');

Solution

(i)SELECT * FROM MEMBER ORDER BY ISSUEDATE DESC;
(ii) SELECT Code, BNAME FROM BOOK WHERE TYPE=’Fiction’;
(iii) SELECT COUNT(*), TYPE FROM BOOK GROUP BY TYPE;
(iv) SELECT MNAME, ISSUEDATE FROM MEMBER WHERE ISSUEDATE>=’2017-01-01’ AND ISSUEDATE<=’2017-12-31’;
(v)MAX(ISSUEDATE)
2017-02-23
(vi) DISTINCT TYPE
      Fiction
      Literature
      Comic
(vii) 

CODE

BNAME

MNO

MNAME

L102

The priest

M101

RAGHAV SINHA

F102

Untold Story

M103

SARTHAK JOHN

C101

Tarzan in the lost world

M102

ANISHA KHAN


(viii) BNAME
       German Easy

Question
CBSEENCO12011646

Observe the following STUDENTS and EVENTS tables carefully and write the name of the RDBMS operation which will be used to produce the output as shown in LIST? Also, find the Degree and Cardinality of the LIST.

STUDENTS

No  Name 
1 Tara Mani 
2 Jaya Sarkar 
3 Tarini Trikha

EVENTS

EVENTCODE EVENTNAME
1001 Programming 
1002 IT Quiz

LIST

No  Name  EVENTCODE EVENTNAME 
1 Tara Mani 1001 Programming 
1 Tara Mani 1002 IT Quiz
2 Jaya Sarkar 1001 Programming
2 Jaya Sarkar 1002 IT Quiz
3 Tarini Trikha 1001 Programming
3 Tarini Trikha 1002 IT Quiz

 

Solution

Cartesian Product
Degree = 4
Cardinality = 6

Question
CBSEENCO12011647

Write SQL queries for (i) to (iv) and find outputs for SQL queries (v) to (viii), which are based on the tables

Table: VEHICLE

CODE VTYPE PERKM
101 VOLVO BUS 160
102 AC DELUXE BUS 150
103 ORDINARY BUS 90
105 SUV 40
104 CAR 20

Note:

    • PERKM is Freight Charges per kilometre
    • VTYPE is Vehicle Type

Table: TRAVEL

NO NAME TDATE KM CODE NOP 
101 Janish  2015-11-13 200 101 32
103 Vedika Sahai 2016-04-21 100 103 45
105 Tarun Ram 2016-03-23 350 102 42
102 John Fen 2016-02-13 90 102 40
107 Ahmed Khan 2015-01-10 75 104 2
104 Raveena  2016-05-28 80 105 4
106 Kripal Anya 2016-02-06 200 101 25

Note:
• NO is Traveller Number
• KM is Kilometer travelled
• NOP is number of travellers travelled in vehicle
• TDATE is Travel Date

  1. To display NO, NAME, TDATE from the table TRAVEL in descending order of NO.
  2. To display the NAME of all the travellers from the table TRAVEL who are travelling by vehicle with code 101 or 102.
  3. To display the NO and NAME of those travellers from the table TRAVEL who travelled between ‘2015‐12‐31’ and ‘2015‐04‐01’.
  4. To display all the details from table TRAVEL for the travellers, who have travelled distance more than 100 KM in ascending order of NOP.
  5. SELECT COUNT(*), CODE FROM TRAVEL
    GROUP BY CODE HAVING COUNT(*)>1;
  6. SELECT DISTINCT CODE FROM TRAVEL;
  7. SELECT A.CODE,NAME,VTYPE
    FROM TRAVEL A,VEHICLE B
    WHERE A.CODE=B.CODE AND KM<90;
  8. SELECT NAME,KM*PERKM
    FROM TRAVEL A,VEHICLE B
    WHERE A.CODE=B.CODE AND A.CODE=’105’;

Solution

  1. SELECT NO, NAME, TDATE FROM TRAVEL
    ORDER BY NO DESC;
  2. SELECT NAME FROM TRAVEL
    WHERE CODE IN (‘101’,’102’)
  3. SELECT NO, NAME from TRAVEL
    WHERE TDATE <= ‘20151231’AND TDATE >= ‘20150401’;
  4. SELECT * FROM TRAVEL
    WHERE KM > 100 ORDER BY NOP;
  5. COUNT(*)   CODE
    2                101
    2                102
  6. DISTINCT CODE
    101
    102
    103
    104
    105
  7. CODE NAME VTYPE
    104 Ahmed Khan CAR
    105 Raveena SUV
  8. NAME KM*PERKM
    Raveena 3200