Mutley Posted March 25, 2007 Share Posted March 25, 2007 I need to get a field from the database and attach it with the loop that is created around the <select> options, however doing a hidden field won't work, as it isn't detected when in a <select> Code: <?php $result = mysql_query("SELECT prod_id FROM loadout WHERE user_id = '$user_id' AND slot = 'car' LIMIT 1"); while($row = mysql_fetch_array( $result )) { $selected = $row['prod_id']; } $sql = "SELECT prod_id, dam FROM own WHERE user_id = '$user_id' AND market_id = '0' AND slot = 'car'"; $result = mysql_query($sql); ?> <td><select name="car"> <? if(mysql_num_rows($result)!=0) { // (1) while(list($prod_id, $dam_car) = mysql_fetch_row($result)) { // (2) $result1 = mysql_query("SELECT name FROM products WHERE prod_id = '$prod_id' LIMIT 1");// (3) while($row1 = mysql_fetch_array( $result1 )) { ?> <option value="<?=$prod_id?>"<? if($selected == $prod_id) { echo " selected"; }else{ echo ""; }?>><?=$row1['name']?></option> <? } // (3) } // (2) } // (1) ?> </select></td> "dam" is the field I want posting but where do I put it? So later on I can do an update query but I need that field associated with the selected option from the drop down (same DB row). Thanks in advance. Link to comment https://forums.phpfreaks.com/topic/44250-hidden-input-field-inside-a-select/ Share on other sites More sharing options...
Mutley Posted March 25, 2007 Author Share Posted March 25, 2007 Or maybe another method? Link to comment https://forums.phpfreaks.com/topic/44250-hidden-input-field-inside-a-select/#findComment-214920 Share on other sites More sharing options...
Mutley Posted March 26, 2007 Author Share Posted March 26, 2007 Anyone? Link to comment https://forums.phpfreaks.com/topic/44250-hidden-input-field-inside-a-select/#findComment-215413 Share on other sites More sharing options...
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