I do not have the die statment -- the DB is the same except the name and I have double checked that I just did a back up of the previous db and moved it to the new server. I have done some more digging, my query appears to work, but I almost appears that I have a problem with selecting the database -- if I comment out the mysql_select -- statment I get the same error, perhaps I just can't get to that DB. intresting. I guess I can assume at this point there is something flaky with my DB "Jeffrey N Dyke" <jeffrey_n_Dyke@Keane.com> wrote in message OF34955652.B9ABBCBE-ON85256D5E.005C604A@keane.com">news:OF34955652.B9ABBCBE-ON85256D5E.005C604A@keane.com... > > Do you have an 'or die(mysql_error())' statement following your > mysql_query($result) line. 99% of the time, this error means your query > failed. if it works on another server....are the fields the same, the > dbname, the tablename? > > hth > jeff > > > > "Greg Hetrick" > <ghetrick@avalon. To: php-db@lists.php.net > net> cc: > Subject: mysql_fetch_array issues. > 07/09/2003 12:04 > PM > > > > > > > I am getting the following error when attempting to pull data out of a > mysql > DB > > Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result > resource in /home/pffl/public_html/pffl/webpage/html/dataentry.php on line > 125 > > I can take this same code to a different web server and it is pulling > correctly! > > I am currently running Apache 2.0.46 with PHP 4.3.2 I was running Apache > 1.3.x with PHP 4.3.1 and getting the same thing, any ideas? here is the > chunk of code where I use the function. > > while ($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 > > > > > -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
