Posted 03 May 2006 - 07:14 PM
$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>';
it shows up in the combo box, but dosnt post the value...
any ideas why?
Posted 03 May 2006 - 07:22 PM
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?
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Posted 03 May 2006 - 07:34 PM
ive solved it myself..
i left a bloody gap after the types_id...
thanks for the help
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