Xtremer360 Posted March 17, 2011 Share Posted March 17, 2011 I'm trying to figure out why the options aren't appearing inside the select dropdown. Any ideas why? echo "<label for=" . $row2['fullName'] . ">" . $row2['fullName'] . "</label>"; echo "<select name=" . $row2['fullName'] . " id=" . $row2['fullName'] . " class=dropdown title=" . $row2['fullName'] . " />"; if ($styleID == 1 || $styleID == 2 || $styleID == 6) { $charactersQuery = " SELECT characters.ID, characters.characterName FROM characters WHERE characters.styleID = 3 ORDER BY characters.characterName"; $charactersResult = mysqli_query ( $dbc, $charactersQuery ); // Run The Query while ( $row3 = mysqli_fetch_array ( $charactersResult, MYSQL_ASSOC ) ) { print "<option value=" . $row3['ID'] . ">" . $row3['characterName'] . "</option>\r"; } } else { $charactersQuery = " SELECT characters.ID, characters.characterName FROM characters WHERE characters.styleID IN (1,2,6) ORDER BY characters.characterName"; $charactersResult = mysqli_query ( $dbc, $charactersQuery ); // Run The Query while ( $row3 = mysqli_fetch_array ( $charactersResult, MYSQL_ASSOC ) ) { print "<option value=" . $row3['ID'] . ">" . $row3['characterName'] . "</option>\r"; } } echo "</select>"; } Quote Link to comment Share on other sites More sharing options...
monkeytooth Posted March 17, 2011 Share Posted March 17, 2011 What if any errors are you getting T_Variable? T_STRING? unexpected end? other... Quote Link to comment Share on other sites More sharing options...
Xtremer360 Posted March 17, 2011 Author Share Posted March 17, 2011 The options are showing up as text outside the select dropdown. Quote Link to comment Share on other sites More sharing options...
Xtremer360 Posted March 17, 2011 Author Share Posted March 17, 2011 Any ideas? Quote Link to comment Share on other sites More sharing options...
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