* Steve Marquez [EMAIL PROTECTED]:
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:[EMAIL PROTECTED]
Webmaster and IT Specialist | http://www.garden.org
National Gardening Association| http://www.kidsgardening.com
802-863-5251 x156 | http://nationalgardenmonth.org
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