Simple SQL question

Arc

$Sql="SELECT * FROM tblReg WHERE OnlineID IS NOT LIKE '$SESSION_UNAME' AND DateAccepted IS NULL";

What exactly is wrong with the syntax of that statement? I'm getting a "supplied argument is not a valid MySQL result resource" error when i run it.

The Field and Table names are correct so it has to be something to do with the DateAccepted IS NULL part i beleive.

Thanks!😃

vallette

I think the problem is with 'IS NOT LIKE'. Try:

$Sql="SELECT * FROM tblReg WHERE OnlineID IS NOT '$SESSION_UNAME' AND DateAccepted IS NULL";

$Sql="SELECT * FROM tblReg WHERE OnlineID !='$SESSION_UNAME' AND DateAccepted IS NULL";

amchargue

Instead of IS NOT LIKE, should be NOT LIKE

I don't really get LIKE anyway. Shouldn't LIKE imply a wildcard? It doesn't. " SELECT name FROM table WHERE name LIKE 'ohnso' " will not return Johnson. Sure seems like Johnson is "like" ohnso to me ... but anyway.

I think LIKE is merely a synonym for =

Can anyone verify/deny that? I'm too lazy to look it up on mysql.com .

Of course to include wildcard, use the % symbol .

Arc

Hi thanks that worked.

Actually LIKE is used when you want to return a non case sensitive result. so LIKE 'A' would return 'a' as well, where as ='A' would not return 'a'.

Ofcourse this depends on how you have the field set in MySQL. You can set the field to be case sensitive or not.

amchargue

Thanks for the scoop on LIKE, I didn't know that. It doesn't seem to matter on my system, though - all my queries are case INsensitive whether I use = or LIKE. Are you saying it's a MySQL configuration option of some kind to turn case sensitivity on/off ?