I have the following code used to insert a record using PHP into a MySQL database. The form element is a multiple select that works fine when only one option is selected. When I choose more than 1 option, only the last option is inserted. How do I have the form create a new row for each option selected in the multiple select?
if ((isset($_POST["MM_insert"])) && ($_POST["MM_insert"] == "InsertForm")) { $insertSQL = sprintf("INSERT INTO emplikons (EmpNo, IconId) VALUES (%s, %s)", GetSQLValueString($_POST['insertRecordID'], "text"), GetSQLValueString($_POST['icons'], "int")); mysql_select_db($database_techsterus, $techsterus); $Result1 = mysql_query($insertSQL, $techsterus) or die(mysql_error()); This is the code of the form element. It uses a recordset to dynamically pull values from another table:
<select name="icons" size="10" multiple="multiple"> <?php do { ?> <option value="<?php echo $row_icons['id']?>"><?php echo $row_icons['name']?></option> <?php } while ($row_icons = mysql_fetch_assoc($icons)); $rows = mysql_num_rows($icons); if($rows > 0) { mysql_data_seek($icons, 0); $row_icons = mysql_fetch_assoc($icons); } ?> </select>
mysql_*functions in new code. They are no longer maintained and are officially deprecated. See the red box? Learn about prepared statements instead, and use PDO or MySQLi - this article will help you decide which. If you choose PDO, here is a good tutorial.