Re: [PHP-DB] Gathering data from a database

2003-03-22 Thread Adam -
Hi, $select = "mysql_select_db ($database)"; << like Mike said you don't need to have this in " and no need for a var to be assigned... Then when you want to print out the data print_r( $row ); you can use $row['table_row_name']; :) Adam At 08:26 PM 3/22/2003 -0500, you wrote: I have been

Re: [PHP-DB] Gathering data from a database

2003-03-22 Thread Paul Burney
on 3/22/03 8:26 PM, Mike Delorme at [EMAIL PROTECTED] appended the following bits to my mbox: > $select = "mysql_select_db ($database)"; Remove the quotes around this. Actually, there is no reason to do the variable assignment either. Sincerely, Paul Burney -- PH

[PHP-DB] Gathering data from a database

2003-03-22 Thread Mike Delorme
I have been having this problem for a while now, and I cant figure it out. the code is: mysql has an error on line 19. Line 19 is while ($row = mysql_fetch_array($result)). What am I doing wrong? Thanks for your help, Mike