How am I SUPPOSED to check for !$result ??

[mOrloff]

I want some feedback when there aren't any matches found, and I know I've done it in the past, but I'm not getting it right now.

This is where I'm at:

PHP Code:

while($row=sqlsrv_fetch_array($rzStock, SQLSRV_FETCH_ASSOC)){
    $stockPN=$row['pn'];   //  grab the part number
    $select='pn,description';
    $from='table';
    $where="pn=\"$stockPN\"";
    $result=qryFunc($select,$from,$where);
}
if(!$result){
    echo " - No match found for $stockPN.<br/>";  //  for testing
}else{
    $array=mysql_fetch_assoc($result);
    echo '<pre>'; print_r($array); echo '</pre>';  //  for testing
} 


I never get No match found echoed, but when there is a match, I do get the array printed back successfully.

Where am I going off ?

~ Mo

[MattF]

if (!mysql_num_rows($result)){

Edit: Why do you have three different formats for DB query functions? Change the above mysql_ to whichever function name is relevant for your scenario.

[mOrloff]

Ahhh, there we go.

Thanks.

[DJCMBear]

From what it looks like you looking in one table to get the part number and then looking in the other table for the part information, I personaly would use a mysql table join so you only use one query but it looks in two different tables

[Fou-Lu]

Originally Posted by DJCMBear

From what it looks like you looking in one table to get the part number and then looking in the other table for the part information, I personaly would use a mysql table join so you only use one query but it looks in two different tables

We've actually been over this one a little while ago; the problem is one is an SQLServer and the other is a MySQL server.