我有一个功能，如果您在下拉菜单中选择一个值，我的代码会使用所选值更新数据库。现在我尝试添加一个检查来检查数据库中是否已经存在该值。但它不起作用。我究竟做错了什么？
测试.php
<html>
<head>
<script type="text/javascript" src="js/jquery.js"></script>
<link href="style.css" rel="stylesheet" type="text/css">
<script>
function updateDb() {
$.post("buh.php", $("#form").serialize());
}
</script>
</head>
<body>
<form id="form">
<?php
include 'Includes/database_connection.php';
$sql = "select * FROM sims ORDER BY phonenr asc" ;
$result = mysql_query($sql,$con);
$id1 = $_GET["id1"];
echo "<input type='hidden' id='hidden' name='hidden' value='$id1'>";
echo "<select id='select' name='select' onChange='updateDb()'>";
while ($row = mysql_fetch_array($result)) {
echo "<option value='" . $row['phonenr'] . "'>" . $row['phonenr'] . "</option>";
}
echo "</select>";
?>
</form>
<?php
echo '<a href="javascript:window.close()">Sluit venster</a>';
?>
<body>
Buh.php
<?php
include 'Includes/database_connection.php';
$sql = "select * FROM sims WHERE phonenr='".mysql_real_escape_string($_POST["select"])."'" ;
$result = mysql_query($sql,$con);
while ($row = mysql_fetch_array($result)) {
$id = $row['id'];
$id1 = mysql_real_escape_string($_POST["hidden"]);
$checkID = mysql_query("SELECT sim_id from pairings WHERE sim_id = '$id'");
if (mysql_num_rows($checkID) > 0) {
if(confirm("Simkaart is al gekoppeld weet u zeker dat u deze wilt vervangen?")){
mysql_query("UPDATE pairings SET sim_id='$id' WHERE unit_id='$id1'")
or die(mysql_error());
}
}
else {
mysql_query("UPDATE pairings SET sim_id='$id' WHERE unit_id='$id1'")
or die(mysql_error());
}
}