Re: php/mySQL help

Addressed to: "Stinsman, Scott" <scott.stinsman@uphs.upenn.edu> php-db@lists.php.net ** Reply to note from "Stinsman, Scott" <scott.stinsman@uphs.upenn.edu> Wed, 7 Feb 2001 16:00:29 -0500 > > Hi everyone..been away and off the list for awhile.i am back to work and > need some help. I am not sure if I am having a problem with my PHP or > mySQL. > I have a table called "Restaurant" with a field called "Cuisine". There > are > 5 records in this table. When I run my code, it only returns one cuisine > when there are 5. I am trying to get all the Cuisines in the array called > CuisineMenuArray with no duplicates so I can generate a drop-down menu of > cuisines "on the fly". > Any help would be greatly appreciated! Thanx. > > Here is the code: > > <?php > $db=mysql_connect("localhost", "*******", "********"); > > mysql_select_db("quickcit", $db) OR DIE ("died at connect"); > > $query = "SELECT Restaurant.Cuisine "; > $query .= "FROM Restaurant "; > $query .= "ORDER BY Restaurant.Cuisine ASC "; > $mysql_result=mysql_query($query, $db) OR DIE ("died at query"); > Try changing you Query to: SELECT DISTINCT Cuisine FROM Restaurant ORDER BY Cousine Then rely on the DISTINCT to insure you only get one copy of the value in your array. while( list( $Couisine ) = mysql_fetch_row( $Result )) { $CouisineList[] = $Couisine; } And you are done... Rick Widmer Internet Marketing Specialists http://www.developersdesk.com