04232005 1728 GMT-6

What about this one.
 $sql = "SELECT * FROM application WHERE appid = '".$prevApp."' AND != 
rejected.appid";
I get the same error so I dont know what is up but, as long as its 
written correctly, Ill keep working.

Wade



Wade wrote:

> 04232005 1718 GMT-6
>
> I get an error: mysql_fetch_array(): supplied argument is not a valid
> MySQL result resouce. I thought I wrote it wrong.
>
> But if not, then I have something else off.
>
> Wade
>
>
> Patrick Bierans wrote:
>
> > > $sql = "SELECT * FROM application WHERE appid != rejected.appid ORDER
> > > BY appid DESC";
> >
> > looks fine so far. so what is your question?
> >
> >
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