brown2005 Posted May 3, 2006 Share Posted May 3, 2006 hi i have:-$select_type_sql = "SELECT * FROM affiliates_types ORDER BY types_name ASC;";$select_type_result = mysql_query($select_type_sql) or die(mysql_error());echo" <select name='types' class='text'>";while ($select_type_array = mysql_fetch_array($select_type_result)){echo '<option value="'.$select_type_array['types_id '].'">'.$select_type_array['types_name'].'</option>';}echo"</select>";it shows up in the combo box, but dosnt post the value...any ideas why? Link to comment https://forums.phpfreaks.com/topic/8998-select/ Share on other sites More sharing options...
.josh Posted May 3, 2006 Share Posted May 3, 2006 hmm.. well your code looks fine. so when you click submit and in the new page if you try to echo $types or $_POST['types'] you get nothing?also, you are populating your dropdown with type_name but you are passing types_id with it, is this really what you are trying to do?example: dropdown box might show Betty, Sue, George but you are actually passing 23, 19, 56 not Betty, Sue or George if so, is types_id a valid column name in your db? Link to comment https://forums.phpfreaks.com/topic/8998-select/#findComment-33077 Share on other sites More sharing options...
brown2005 Posted May 3, 2006 Author Share Posted May 3, 2006 sorry for bothering you...ive solved it myself..i left a bloody gap after the types_id... silly clown..thanks for the help Link to comment https://forums.phpfreaks.com/topic/8998-select/#findComment-33084 Share on other sites More sharing options...
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