Re: Table shows even when if () is false

Ah, I see. Changed and working. Thanks Jay! On 8/22/07, Jay Blanchard <jblanchard@pocket.com> wrote: > > [snip] > [snip] > $deferred_comments= "SELECT * FROM comments WHERE credit_card_id = > '$credit_card_id' AND request_type = 'D'"; > $result_deferred_comments = mssql_query($deferred_comments) or > die(mssql_error()); > > if(!empty($result_deferred_comments)) { > [/snip] > > $result_deferred_comments is not empty, a query resource was returned. > To see this; > > echo $result_deferred_comments; > > What you want to do is something like this; > > $num_Deferred_Comments = mysql_num_rows($result_deferred_comments); > > if(0 != $num_Deferred_Comments)) {.... > [/snip] > > My bad, just noticed you were using mssql, so use the proper functions > there, like mssql_num_rows >