PHP multiple select form to insert multiple rows in MySQL - php

Question

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>

Answer 1

add [] to the name of the select, to get an array of all selected values.
<select name="icons[]" size="10" multiple="multiple">
Then, read from $_POST / $_GET... (assuming form method="post")
$sql = '';
foreach ($_POST['icons'] as $icon) {
// no idea where to get EmpNo from, let's assume it is in $empno.
$sql .= "('$empno','$icon'),";
}
$sql = "INSERT INTO tablename (EmpNo, IconId) VALUES " . trim($sql,',');
// Now, please use mysqli_* or PDO instead of mysql_*