2007/2/10, Dimitar Vasilev <[EMAIL PROTECTED]>:
Hi all, First I would like to thank you for the invaluable pieces of advice. I've made a good progress thanks to you. Now I'm trying to display the output of people who have shareholders at more than one place select COUNT (*),IME,LAND,CNT,ID, (select round(bulstat,0)) from part GROUP BY IME,LAND,CNT,ID HAVING COUNT(*)>'1' AND CNT > '0'. Currently it gives me the number of times a person is a shareholder If I make it like: select IME,LAND,Company,CNT,(select round(bulstat,0)) from part WHERE IME LIKE 'Pooh' AND CNT > '0'; output is like this Pooh, Forest, Big Honeypot, 50 Can someone shed some light how to make it like the second one for all people. I can make a dump to a text file of the names I want, then assign a variable to the names list and then use select IME,LAND,Company,CNT,(select round(bulstat,0)) from part WHERE IME LIKE '@ownerlist' AND CNT > '0'; but I am not sure that text arrays (1 row per line ) will work. Thank you and happy week-end. -- Димитър Василев Dimitar Vassilev GnuPG key ID: 0x4B8DB525 Keyserver: pgp.mit.edu Key fingerprint: D88A 3B92 DED5 917E 341E D62F 8C51 5FC4 4B8D B525
select *,(select round(bulstat,0)) from part GROUP BY IME,LAND,CNT,bulstat HAVING COUNT(*) >'1' and CNT > '0'; Thanks. -- Димитър Василев Dimitar Vassilev GnuPG key ID: 0x4B8DB525 Keyserver: pgp.mit.edu Key fingerprint: D88A 3B92 DED5 917E 341E D62F 8C51 5FC4 4B8D B525
