As far as I know, that error means that the query that was supposed to produce $result did not run because of errors in the SQL (or the database). Suggest you check your query in myPHPAdmin or MySQLfront or whatever to be sure that it has no errors. Terry ----------Original Message--------- > I am getting the following error when attempting to pull data out of a > mysql > DB > > Warning: mysql_fetch_array(): supplied argument is not a valid MySQL > result > resource in /home/pffl/public_html/pffl/webpage/html/dataentry.php on > line > 125 > > I can take this same code to a different web server and it is pulling > correctly! > > I am currently running Apache 2.0.46 with PHP 4.3.2 I was running Apache > 1.3.x with PHP 4.3.1 and getting the same thing, any ideas? here is the > chunk of code where I use the function. > > while ($myrow=mysql_fetch_array($result)) > { > ?> > <option value="<?print $myrow['name'];?>, <?print > $myrow['team'];?>"> <? echo $myrow["name"]; ?></option> > <? > } > > Any Ideas. > -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php