Ryan Because you are actually getting a result, a result with no records... Might be better to use http://au2.php.net/manual/en/function.mysql-num-rows.php instead
Martin > -----Original Message----- > From: Ryan A [mailto:[EMAIL PROTECTED] > Sent: Tuesday, 23 March 2004 11:37 AM > To: [EMAIL PROTECTED] > Subject: [PHP] Passing by conditional IF statement...why? > > > Hi, > I have this simple code in my php script: > > * * * * * > $res = mysql_query("SELECT product_id, now()-1 FROM > ".$tc."_prods where > cno=$cno AND product_id='$product_id' LIMIT 1"); > > if($res) > { > $r = mysql_fetch_row($res); > $product_id2 = $r[0]; > $th_pres= $r[1]; > echo "debug echo"; > }else {echo "No results, sorry";} > * * * * * > > its working great when the data actually exists but when there are no > matches it still executes the "if($res)" part instead of > displaying "No results, sorry". > Why is that? or am I using the syntax wrong? > > Thanks, > -Ryan -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
