Re: Forms Question: Options

* Steve Marquez <smarquez@marquez-design.com>: > I created a photo upload utility with individual galleries that images can > be uploaded into. In the MySQL database, there are multiple names of > galleries, some are the same. I want to have a select menu to show just the > unique names of the galleries. > > I have used: > > Select DISTINCT gallery_names from images_upload; > > I have about 15 records, ten are "GALLERY1" and 5 are "GALLERY2." When I > use > the DISTINCT, two are output. It works perfectly in the MySQL terminal. > > However, when I use the same in PHP as a web page, it only outputs one, the > one with only 5 records. > > Here is the code: > > <?php > > // log into our local server using the MySQL root user. > $dbh = mysql_connect( "hostname", "username", "password" ); > > // select the database. > mysql_select_db( "db" ) or die ( mysql_error() . "\n" ); > > //and read it back for printing purposes. > $get_table_data = "SELECT DISTINCT gallery_name FROM images_upload ORDER > BY gallery_name DESC"; > > $response = mysql_query( $get_table_data, $dbh ); > > //now print it out for the user. > if ( $one_line_of_data = mysql_fetch_array( $response ) ) { It's only printing out one record because you're only fetching from the results set once. Change that to : while ($one_line_of_data = mysql_fetch_array($response)) { and you should be set. -- Matthew Weier O'Phinney | mailto:matthew@garden.org Webmaster and IT Specialist | http://www.garden.org National Gardening Association | http://www.kidsgardening.com 802-863-5251 x156 | http://nationalgardenmonth.org