Re: pg_result
От | Sharmad Naik |
---|---|
Тема | Re: pg_result |
Дата | |
Msg-id | 20030124223834.377cd8fb.sharmad@goatelecom.com обсуждение исходный текст |
Ответ на | pg_result (ryanne cruz <ryanne.cruz@up.edu.ph>) |
Список | pgsql-php |
Hi, PLease look at the query.The syntax is "SELECT attribute_name[,attr2,..] FROM tables WHERE attribute =[LIKE,IN,..] ETC.. It looks in ur query that province is an attribute as well as a query. Reagrds Sharmad On Wed, 22 Jan 2003 14:18:32 +0800 ryanne cruz <ryanne.cruz@up.edu.ph> wrote: > > hi list. > > this is my script: > > $query7="select provinceid from province where province like '$tablename1';"; > $result7=pg_exec($conn,$query7); > if(!$result7){ > . > . > .} > else { > . > . > . > } > > my problem is when the value of the variable $tablename1 is not on the table > province, it doesn't go to the if statements. what am i doing wrong? > > thanks! > > > ---------------------------(end of broadcast)--------------------------- > TIP 2: you can get off all lists at once with the unregister command > (send "unregister YourEmailAddressHere" to majordomo@postgresql.org)
В списке pgsql-php по дате отправления: