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thanks for the fish

Author

Topic

pureclass85
Starting Member

29 Posts

Posted - 2009-03-07 : 15:30:02

thank you for your help guys!

sodeep
Master Smack Fu Yak Hacker

7174 Posts

Posted - 2009-03-07 : 15:52:01

You query is not making any sense. Can you explain what you are looking for?

pureclass85
Starting Member

29 Posts

Posted - 2009-03-07 : 16:03:42

sorry missed a bit

SELECT DISTINCT a.bird_name
FROM expectation a
WHERE (SELECT count(*)
FROM expectation b
WHERE a.bird_name = b.bird_name
AND a.location_name = b.location_name) = 4

basicly it is a compusite query that displays a list of birds spotted 4 times in a location

SwePeso
Patron Saint of Lost Yaks

30421 Posts

Posted - 2009-03-07 : 16:05:43

...
FROM Expectation AS a
INNER JOIN (SELECT Bird_Name, Location_Name FROM Expectation GROUP BY Bird_Name, Location_Name HAVING COUNT(*) = 4) AS b
ON b.Bird_Name = a.Bird_Name AND b.Location_Name = a.Location_Name

E 12°55'05.63"
N 56°04'39.26"

pureclass85
Starting Member

29 Posts

Posted - 2009-03-07 : 16:10:18

thanks peso but it contians a sub query so that means it is composite and i am looking for a way with out using a composite query. thanks

guptam
Posting Yak Master

161 Posts

Posted - 2009-03-07 : 17:58:38

How about ...

SELECT a.bird_name
FROM expectation a
GROUP BY a.bird_name, a.location_name
HAVING COUNT(*) = 4

?

--
Mohit K. Gupta
B.Sc. CS, Minor Japanese
MCITP: Database Administrator
MCTS: SQL Server 2005
http://sqllearnings.blogspot.com/

pureclass85
Starting Member

29 Posts

Posted - 2009-03-08 : 04:26:20

that worked perfect thank you just added distinct
but again thank you

visakh16
Very Important crosS Applying yaK Herder

52326 Posts

Posted - 2009-03-08 : 12:43:54