SQL - Class webSite
SQL
(Questions from CBSE Board Exam- from 1998 to 2010)
YEAR: 1998(Outside Delhi)
Q 5 (a) What is need for normalization? Define First, Second and Third Normal Form.
Ans. Need for Normalization: The basic objective of normalization is to reduce redundancy, which means that data to be stored only once. Because storing data many times lead to wastage of space and increase access time. So relation/table are to be normalized so that we can alter them at any time without doing much changes.
First Normal Form: A table / relation is said to be in First Normal Form (1NF) if and only if all underlying domains of the relation contain atomic (individual) values.
Second Normal Form: A table / relation is said to be in Second Normal Form (2NF) if and only if it is in First Normal Form and every non-key attribute is functionally dependent on the Primary Key.
Third Normal Form: A table / relation is said to be in Third Normal Form (1NF) if and only if it is in Second Normal From and every non-key attribute is non-transitively dependent upon the primary key.
NOTE : Write SQL commands for(b) to (g) and write the output for (h) on the basis of table HOSPITAL.
|TABLE : HOSPITAL |
|No |Name |Age |Department |Datofadm |Charges |Sex |
|1 |Sandeep |65 |Surgery |23/02/98 |300 |M |
|2 |Ravina |24 |Orthopedic |20/01/98 |200 |F |
|3 |Karan |45 |Orthopedic |19/02/98 |200 |M |
|4 |Tarun |12 |Surgery |01/01/98 |300 |M |
|5 |Zubin |36 |ENT |12/02/98 |250 |M |
|6 |Ketaki |16 |ENT |24/02/98 |300 |F |
|7 |Ankita |29 |Cardiology |20/02/98 |800 |F |
|8 |Zareen |45 |Gynecology |22/02/98 |300 |F |
|9 |Kush |19 |Cardiology |13/01/98 |800 |M |
|10 |Shaliya |31 |Nuclear Medicine |19/02/98 |400 |M |
(b) To show all information about the patients of cardiology department.
Ans: SELECT * FROM hospital WHERE department=’Cardiology’;
(c ) To list the names of female patients who are in orthopedic dept.
Ans: SELECT name FROM hospital WHERE sex=’F’ AND department=’Orthopedic’;
(d) To list names of all patients with their date of admission in ascending order.
Ans.: SELECT name, dateofadm FROM hospital ORDER BY dateofadm;
e) To display Patient’s Name, Charges, age for male patients only.
Ans: SELECT name, charges, age FROM hospital WHERE sex=’M’;
f) To count the number of patients with age >20.
Ans.: SELECT COUNT(age) FROM hospital WHERE age>20;
g) To insert a new row in the HOSPITAL table wit the following .
11,”mustafa”,37,”ENT”,(25/02/98},250,”M”
Ans.: INSERT INTO hospital VALUES (11, ‘Mustafa’, 37, ‘ENT’, ‘25/02/98’, 250, ‘M’);
h) Give the output of following SQL statement:
i) Select COUNT(distinct departments) from HOSPITAL;
Ans: COUNT(DISTICNTDEPARTMEN)
--------------------------------------------
6
ii) Select Max (Age) from HOSPITAL where SEX = “M”;
Ans: MAX(AGE)
---------------
65
iii) Select AVG(Charges) from HOSPITAL where SEX = “ F”;
Ans.: AVG(CHARGES)
----------------------
400
iv) Select SUM(Charges) from HOSPITAL where Datofadm10;
(½ mark for correct Select statement)
(½ mark for correct Where clause)
(ii) Display the average salary of all doctors working in “ENT” department using
the tables DOCTOR and SALARY. Salary = BASIC + ALLOWANCE
(ii)
SELECT AVERAGE(S.BASIC + S.ALLOWANCE)
FROM DOCTOR D, SALARY S
WHERE D.DEPT = ‘ENT’ AND D.ID = S.ID;
OR
SELECT AVERAGE(BASIC + ALLOWANCE)
FROM DOCTOR, SALARY
WHERE DEPT = ‘ENT’ AND DOCTOR.ID = SALARY.ID;
(1/2 mark for correct Select statement)
(1/2 mark for correct Where clause)
OR
(1 mark for students who have correctly attempted any two parts of Q5b)
(iii) Display the minimum ALLOWANCE of female doctors,
(iii)
SELECT MIN(S.ALLOWANCE)
FROM DOCTOR D, SALARY S
WHERE D.SEX = ‘F’ AND D.ID = S.ID;
OR
SELECT MIN(ALLOWANCE)
FROM DOCTOR, SALARY
WHERE SEX = ‘F’ AND DOCTOR.ID = SALARY.ID;
(½ mark for correct Select statement)
(½ mark for correct Where clause)
(iv) Display the highest consultation fee among all male doctors,
(iv)
SELECT MAX(S.CONSULTATION)
FROM DOCTOR D, SALARY S
WHERE D.SEX = ‘M’ AND D.ID = S.ID;
OR
SELECT MAX(CONSULTATION)
FROM DOCTOR , SALARY
WHERE SEX = ‘M’ AND DOCTOR.ID = SALARY.ID;
(1/2 mark for correct Select statement)
(1/2 mark for correct Where clause)
(v) SELECT count(*) from DOCTOR where SEX = “F”
(v)
4
(1 mark for correct answer)
(vi) SELECT NAME, DEPT, BASIC from DOCTOR, SALARY
where DEPT = “ENT” and DOCTOR.ID = SALARY.ID
(vi)
NAME DEPT BASIC
John ENT 12000
(1 mark for correct answer)
YEAR 2007 (OUTSIDE DELHI)
5(a) Differentiate between primary key and alternate key. 2
(a) All candidate keys, which are not the primary key of the table are called the
alternate keys.
OR
Primary Key:
An attribute/ column used to identify each record in a table
Alternate Key:
All such attributes/columns, which can act as a primary key
but are not the primary key in a table.
(2 Mark for any valid difference/relation between Primary Key and
Alternate Key)
OR
(1 Mark for correct explanation of Primary Key)
(1 Mark for correct explanation of Alternate Key)
(b) Consider the following tables. Write SQL commands for the statements
(i) to (iv) and give outputs for SQL queries (v) to (viii) 6
TABLE:SENDER
SenderlD SenderName SenderAddress SenderCiry
ND01 R Jain 2, ABC Appts New Delhi
MU02 H Sinha 12, Newtown Mumbai
MU15 S Jha 27/A, Park Street Mumbai
ND50 T Prasad 122-K, SDA New Delhi
TABLE : RECIPIENT
RecID SenderlD RecName RecAddress RecCity
KO05 ND01 R Bajpayee 5, Central Avenue Kolkata
ND08 MU02 S Mahajan 116, A Vihar New Delhi
MU19 ND01 H Singh 2A, Andheri East Mumbai
MU32 MU15 P K Swamy B5, C S Terminus Mumbai
ND48 ND50 S Tripathi 13, B1 D, Mayur Vihar New Delhi
(i) To display the names of all Senders from Mumbai
(b) (i)
SELECT SenderName from Sender WHERE City =
‘Mumbai’;
(ii) To display the RecID), SenderName, SenderAddress, RecName,
RecAddress for every Recipient
(ii)
SELECT R.RecID, S.SenderName, S.SenderAddress,
R.RecName, R.RecAddress FROM Sender S,
Recipient R
WHERE S.SenderID = R.SenderID;
(iii) To display Recipient details in ascending order of RecName
(iii)
SELECT * FROM Recipient ORDER BY RecName;
(iv) To display number of Recipients from each city
(iv)
SELECT COUNT(*) FROM Recipient GROUP BY RecCity
(v) SELECT DISTINCT SenderCity FROM Sender;
(v)
SenderCity
Mumbai
New Delhi
(vi) SELECT A. SenderName, B.RecName
FROM Sender A, Recipient B
WHERE A. SenderlD = B.SenderlD AND B.RecCity = ‘Mumbai’;
(vi)
A.SenderName B.RecName
R Jain H Singh
S Jha P K Swamy
(vii) SELECT RecName, RecAddress
FROM Recipient
WHERE RecCity NOT IN (‘Mumbai’, ‘Kolkata’);
(vii)
RecName RecAddress
S Mahajan 116, A Vihar
S Tripathi 13, Bl D, Mayur Vihar
(viii) SELECT RecID, RecName
FROM Recipient
WHERE SenderID=’MU02' ORSenderID=’ND50';
(viii)
RecID RecName
ND08 S Mahajan
ND48 S Tripathi
(Part (i) to (iv) - 1 Mark for each correct query)
(Part (v) to (viii) - ½ Marks for each correct output)
Note:
• Column headings for the output questions to be ignored.
• Since in part (iv) the fieldname RecCity is not mentioned
specifically, so full 1 mark to be given if any part of 5 (b) is
answered correctly.
YEAR 2007 (DELHI)
5(a) What is the importance of a Primary Key in a table ? Explain with a suitable
example. 2
(a) The Primary Key is an attribute/set of attributes that identifies a tuple/ row/
record uniquely.
Example:
Rollnumber in the table STUDENT
OR
AccessionNumber in the table LIBRARY
OR
EmpNumber in the table EMPLOYEE
OR
PanNumber in the table INCOMETAX
OR
MemberNumber in the table MEMBER
OR
AccNumber in the table BANK
OR
Any other suitable example
(1 Mark for correct definition/explanation of Primary Key)
(1 Mark for suitable example)
(b) Consider the following tables Consignor and Consignee. Write SQL
commands for the statements (i) to (iv) and give outputs for SQL queries (v)
to (viii). 6
TABLE : CONSIGNOR
CnorlD 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
TABLE : CONSIGNEE
CneelD CnorlD CneeName CneeAddress CneeCity
MU05 ND01 Rahul Kishore 5, Park Avenue Mumbai
ND08 ND02 P Dhingra 16/J, Moore Enclave New Delhi
KO19 MU15 A P Roy 2A, Central Avenue Kolkata
MU32 ND02 S Mittal P 245, AB Colony Mumbai
ND48 MU50 B P Jain 13, Block D, A Vihar New Delhi
(i) To display the names of all Consignors from Mumbai.’
(b) (i)
SELECT CnorName FROM CONSIGNOR WHERE
City=’Mumbai’;
(½ Mark for correct use of SELECT and FROM)
(½ Mark for correct use of WHERE clause)
(ii) To display the CneelD, CnorName, CnorAddress, CneeName,
CneeAddress for every Consignee.
(ii)
SELECT eeID, orName, orAddress,
eeName , eeAddress
FROM Consignor A, Consignee B
WHERE orID=orID;
OR
SELECT eeID, CnorName, CnorAddress,
CneeName, neeAddress
FROM Consignor, Consignee
WHERE orID= orID;
(½ Mark for correct use of SELECT and FROM)
(½ Mark for correct use of WHERE clause)
(iii) To display consignee details in ascending order of CneeName.
(iii)
SELECT * FROM CONSIGNEE ORDER BY CneeName;
(½ Mark for correct use of SELECT and FROM)
(½ Mark for correct use of ORDER BY clause)
(iv) To display number of consignors from each city,
(iv)
SELECT City,Count(CnorID) FROM CONSIGNOR Group
By City;
OR
SELECT City,Count(*) FROM CONSIGNOR Group By
City;
(½ Mark for correct use of SELECT and FROM)
(½ Mark for correct use of GROUP BY clause)
(v) SELECT DISTINCT City FROM CONSIGNEE;
(v)
DISTINCT City
Mumbai
New Delhi
Kolkata
(½ Mark for correct output)
OR
(½ Mark for mentioning Error as CITY not present in the table
CONSIGNEE)
(vi) SELECT orName, eeName
FROM Consignor A, Consignee B
WHERE orID = orlD AND eeCity = ‘Mumbai’;
(vi)
orName eeName
R Singhal Rahul Kishore
Amit Kumar S Mittal
(½ Mark for correct output)
OR
(½ Mark for any 2 correct output out of (v),(vii)and (viii) even if
part (vi) not attempted)
(vii) SELECT CneeName, CneeAddress
FROM Consignee
WHERE CneeCity NOT IN (‘Mumbai’, ‘Kolkata’);
(vii)
CneeName CneeAddress
P Dhingra 16/J,Moore Enclave
B P Jain 13,Block D,A Vihar
(½ Mark for correct output)
(viii) SELECT CneelD, CneeName
FROM Consignee
WHERE CnorID=’MU15' OR CnorID=’ND01';
(viii)
CneeID CneeName
MU05 Rahul Kishore
KO19 A P Roy
(½ Mark for correct output)
Note: Column Headings for all Outputs may be ignored
YEAR 2008 (OUTSIDE DELHI)
5 a) Differentiate between Candidate Key and Alternate Key in context of
RDBMS. 2
Ans: Candidate Key: It is the one that is capable of becoming primary key i.e., a
column or set of columns that identifies a row uniquely in the relation.
Alternate Key: A candidate key that is not selected as a primary key is called
an Alternate Key.
(1 Mark each for correct definition/explanation of Candidate Key and
Alternate Key)
OR
(Full 2 Marks for illustrating the concept of Candidate and Alternate key with
appropriate example)
.
(b) Consider the following tables Item and Customer. Write SQL commands
for the statements (i) to (iv) and give outputs for SQL queries (v) to (viii) 6
TABLE : ITEM
[pic]
TABLE : CUSTOMER
[pic]
(i) To display the details of those Customers whose City is Delhi
Ans:
SELECT * FROM CUSTOMER WHERE City=’Delhi’ ;
(½ Mark for correct use of SELECT and FROM)
(½ Mark for correct use of WHERE clause)
(ii) To display the details of Items whose Price is in the range of 35000 to
55000 (Both values included)
Ans:
SELECT * FROM ITEM WHERE PRICE BETWEEN 35000 AND 55000;
OR
SELECT * FROM ITEM WHERE PRICE>=35000 AND PRICE=50 AND Price = ‘08-DEC-07’
AND LAUNCHDATE =
1260 ;
Ans DESCRIPTION TYPE
INFORMAL SHIRT COTTON
INFORMAL PANT COTTON
FORMAL PANT TERELENE
(½ Mark for correct output)
(vii) SELECT MAX (FCODE) FROM FABRIC;
Ans MAX (FCODE)
F04
(½ Mark for correct output)
(viii) SELECT COUNT (DISTINCT PRICE) FROM GARMENT ;
Ans COUNT(DISTINCT PRICE)
7
(½ Mark for correct output)
Year 2009 (Outside Delhi)
5 (a) What is the purpose of a key in a table? Give an example of a key in a Table. (2)
Ans An attribute/group of attributes in a table that identifies each tuple uniquely is
known as a Key.
OR
Any correct definition of Key / Primary Key / Candidate Key / Alternate Key
Table:Item
|Ino |Item |Qty |
|I01 |Pen |560 |
|I02 |Pencil |780 |
|I04 |CD |450 |
|I09 |Floppy |700 |
|I05 |Eraser |300 |
|I03 |Duster |200 |
(1 Mark for writing correct definition/purpose of any valid Key)
(1 Mark for giving suitable example)
OR
(2 Marks for illustrating the purpose of Key with/without showing it as a part
of a Table)
(b) Consider the following tables DRESS and MATERIAL. Write SQL commands for the statements (i) to (iv) and give outputs for SQL queries (v) to (viii). 6
|DCODE |DESCRIPTION |PRICE |MCODE |LAUNCHDATE |
|10001 |FORMAL SHIRT |1250 |M001 |12-JAN-08 |
|10020 |FROCK |750 |M004 |09-SEP-07 |
|10012 |INFORMAL SHIRT |1450 |M002 |06-JUN-08 |
|10019 |EVENING GOWN |850 |M003 |06-JUN-08 |
|10090 |TULIP SKIRT |850 |M002 |31-MAR-07 |
|10023 |PENCIL SKIRT |1250 |M003 |19-DEC-08 |
|10089 |SLACKS |850 |M002 |20-OCT-08 |
|10007 |FORMAL PANT |1450 |M001 |09-MAR-08 |
|10009 |INFORMAL PANT |1400 |M002 |20-OCT-08 |
|10024 |BABY TOP |650 |M003 |07-APR-07 |
Table: MATERIAL
|MCODE |TYPE |
|MOO1 |TERELENE |
|MOO2 |COTTON |
|MOO4 |POLYESTER |
|MOO3 |SILK |
(i) To display DCODE and DESCRIPTION of each dress in ascending order of DCODE.
(ii) To display the details of all the dresses which have LAUNCHDATE in between 05-DEC-07 and 20-JUN-08 (inclusive of both the dates).
(iii) To display the average PRICE of all the dresses which are made of material with MCODE as M003
(iv) To display material-wise highest and lowest price of dresses from DRESS table. (Display MCODE of each dress along with highest and lowest price).
(v) SELECT SUM(PRICE) FROM DRESS WHERE MCODE='M001';
(vi) SELECT DESCRIPTION, TYPE FROM DRESS, MATERIAL WHERE DRESS.MCODE=MATERIAL.MCODE AND DRESS.PRICE>=1250;
(vii) SELECT MAX(MCODE) FROM MATERIAL;
(viii) SELECT COUNT(DISTINCT PRICE) FROM DRESS;
(i) To display DCODE and DESCRIPTION of each dress in ascending order of DECODE.
SQL> SELECT DCODE, DESCRIPTION FROM DRESS ORDER BY DCODE ASC;
(1 Mark for correct query)
(½ Mark for partially correct answer) :
DCODE DESCRIPTION
----------- ---------------
10001 FORMAL SHIRT
10007 FORMAL PANT
10009 INFORMAL PANT
10012 INFORMAL SHIRT
10019 EVENING GOWN
10020 FROCK
10023 PENCIL SKIRT
10024 BABY TOP
10089 SLACKS
10090 TULIP SKIRT
10 rows selected.
(ii) To display the details of al the dresses which have LAUNCHDATE in between 05-DEC-07 and 20-JUN-08 (inclusive of both the dates).
SQL> SELECT * FROM DRESS WHERE LAUNCHDATE BETWEEN '05-DEC-07' AND '20-JUN-08';
OR
SELECT * FROM DRESS WHERE LAUNCHDATE >= ‘05-DEC-07’
AND LAUNCHDATE SELECT AVG(PRICE) FROM DRESS WHERE MCODE='M003';
(1 Mark for correct query)
(½ Mark for partially correct answer)
AVG(PRICE)
----------
900
(iv) To display material-wise highest and lowest price of dresses from DRESS table. (Display MCODE of each dress along with highest and lowest price).
Ans SELECT MCODE, MAX(PRICE), MIN (PRICE) FROM DRESS GROUP BY MCODE;
(1 Mark for correct query)
(½ Mark for partially correct answer)
MCOD MAX(PRICE) MIN(PRICE)
---- ---------- ----------
M001 1450 1250
M002 1450 850
M003 1250 650
M004 750 750
SQL> SELECT SUM(PRICE) FROM DRESS WHERE MCODE='M001';
SUM(PRICE)
----------------
2700
(½ Mark for correct output)
SQL> SELECT DESCRIPTION, TYPE FROM DRESS, MATERIAL WHERE DRESS.MCODE=MATERIAL.MCODE AND DRESS.PRICE>=1250;
DESCRIPTION TYPE
------------------- ---------
FORMAL SHIRT TERELENE
FORMAL PANT TERELENE
INFORMAL SHIRT COTTON
INFORMAL PANT COTTON
PENCIL SKIRT SILK
(½ Mark for correct output)
SQL> SELECT MAX(MCODE) FROM MATERIAL;
MAX(MCODE)
-------------------
M004
(½ Mark for correct output)
SQL> SELECT COUNT(DISTINCT PRICE) FROM DRESS;
COUNT(DISTINCTPRICE)
----------------------------------
6
(½ Mark for correct output)
SQL> SELECT DISTINCT PRICE FROM DRESS;
PRICE
----------
650
750
850
1250
1400
1450
6 rows selected.
(½ Mark for correct output)
(vi) SELECT DESCRIPTION, TYPE FROM DRESS, MATERIAL WHERE
DRESS.DCODE = MATERIAL.MCODE AND DRESS.PRICE>=l250;
Ans DESCRIPTION TYPE
(NO OUTPUT)
(½ Mark for the above)
OR
(½ Mark for attempting the question)
Year 2010 (Outside Delhi)
5.(a) What do you understand by Primary Key ? Give a suitable example of Primary Key from a table containing some meaningful data. 2
Ans An attribute/group of attributes in a table that identifies each tuple uniquely is known as a Primary Key.
Table:Item
|Ino |Item |Qty |
|I01 |Pen |560 |
|I02 |Pencil |780 |
|I04 |CD |450 |
|I09 |Floppy |700 |
|I05 |Eraser |300 |
|I03 |Duster |200 |
In above table Ino is the Primary Key.
(1 Mark for writing correct definition/purpose of Primary Key)
(1 Mark for giving suitable example)
OR
(2 Marks for illustrating the purpose of Key with/without showing it as a part
of a Table)
c) Consider the following tables STOCK and DEALERS and answer (bi) and (b2) parts of this question
Table: STOCK
|ItemNo |Item |Dcode |Qty |UnitPrice |StockDate |
|5005 |Ball Pen 0.5 |102 |100 |16 |31-Mar-10 |
|5003 |Ball Pen 0.25 |102 |150 |20 |01-Jan-10 |
|5002 |Gel Pen Premium |101 |125 |14 |14-Feb-10 |
|5006 |Gel Pen Classic |101 |200 |22 |01-Jan-09 |
|5001 |Eraser Small |102 |210 |5 |19-Mar-09 |
|5004 |Eraser Big |102 |60 |10 |12-Dec-09 |
|5009 |Sharpener Classic |103 |160 |8 |23-Jan-09 |
Table: DEALERS
|Dcode |Dname |
|101 |Reliable Stationers |
|103 |Classic Plastics |
|102 |Clear Deals |
(bl) Write,SQL commands for the following statements: 4
(i) To display details of all Items in the Stock table in ascending order of StockDate.
Ans:
SQL> SELECT * FROM Stock ORDER BY StockDate ASC;
ITEMNO ITEM DCODE QTY UNITPRICE STOCKDATE
---------- -------------------- ---------- ---------- ---------- ---------
5006 Gel Pen Classic 101 200 22 01-JAN-09
5009 Sharpener Classic 103 160 8 23-JAN-09
5001 Eraser Small 102 210 5 19-MAR-09
5004 Eraser Big 102 60 10 12-DEC-09
5003 Ball Pen 0.25 102 150 20 01-JAN-10
5002 Gel Pen Premium 101 125 14 14-FEB-10
5005 Ball Pen 0.5 102 100 16 31-MAR-10
7 rows selected.
(ii) To display ItemNo and Item name of those items from Stock table whose UnitPrice is more than Rupees 10.
Ans.:
SQL> SELECT ItemNo, Item FROM Stock WHERE UnitPrice>10;
ITEMNO ITEM
---------- --------------------
5005 Ball Pen 0.5
5003 Ball Pen 0.25
5002 Gel Pen Premium
5006 Gel Pen Classic
(iii) To display the details of those items whose dealer code (Dcode) is 102 or Quantity in Stock (Qty) is more than 100 from the table Stock.
Ans.:
SQL> SELECT * FROM Stock WHERE Dcode=102 OR Qty>100 ;
ITEMNO ITEM DCODE QTY UNITPRICE STOCKDATE
---------- -------------------- ---------- ---------- ---------- ---------
5005 Ball Pen 0.5 102 100 16 31-MAR-10
5003 Ball Pen 0.25 102 150 20 01-JAN-10
5002 Gel Pen Premium 101 125 14 14-FEB-10
5006 Gel Pen Classic 101 200 22 01-JAN-09
5001 Eraser Small 102 210 5 19-MAR-09
5004 Eraser Big 102 60 10 12-DEC-09
5009 Sharpener Classic 103 160 8 23-JAN-09
7 rows selected.
(iv) To display Maximum UnitPrice of items for each dealer individually as per Dcode from the table Stock.
Ans.:
SQL> SELECT Decode, MAX(UnitPrice) FROM Stock GROUP BY Decode;
DECODE MAX(UNITPRICE)
---------- --------------
101 22
102 20
103 8
(b2) Give the output of the following SQL queries 2
(i) SELECT COUNT (DISTINCT Dcode) FROM Stock;
Ans.:
COUNT(DISTINCTDCODE)
---------------------
3
(ii) SELECT Qty*UnitPrice FROM Stock WHERE ItemNo=5006;
Ans.:
QTY*UNITPRICE
-------------
4400
(iii) SELECT Item, Dname FROM Stock S, Dealers D WHERE S.Dcode=D.Dcode AND ItemNo=5004;
Ans.:
ITEM DNAME
-------------------- --------------------
Eraser Big Clear Deals
(iv) SELECT MIN(StockDate) FROM Stock;
Ans.:
MIN(STOCKDATE)
---------
01-JAN-09
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