SELECT problem

[Anon]

I can´t get this working!
I need to sort out artist whos names start with B and have the field show, containing the value yes.
Please help me!

$resultat =
mysql_query("SELECT * FROM artist WHERE namnet LIKE 'B%' AND show ='yes') ORDER BY namnet",$db);

[crusty_php]

Try this

$resultat =
mysql_query("SELECT * FROM artist WHERE namnet LIKE 'B%' AND show ='yes' ORDER BY namnet",$db);

[Anon]

Thank you for your help, but it didn´t seem to help.

[crusty_php]

Have you tried running the query directly to the db command line.
Do you get any error messages?

[Anon]

This is what I get
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource

[Anon]

Can you give us a desc of your table !
In Mysql command line : desc artist

[nielskg]

The query that Martin have is definately correct. Your error may have something to do with the database setup or settings. We indeed need more info to really know what's up...