我有下面的代码，我想问一下，如何根据之前的数据将源插入到其他表中的数据库中？
我从数据库中获取数据并将其显示在复选框上，但是当我检查复选框并按下按钮 Tambah 时，数据被插入到表中，但数据错误，只有 0 和数组。
<!DOCTYPE html>
<html>
<head>
<title>kumpulan data command</title>
<link rel="stylesheet" type="text/css" href="style.css">
</head>
<body>
<h1>Evaluation</h1>
<h2>Data command</h2>
<?php
$db = new mysqli("localhost","root","","FWS_online");
echo $db->connect_errno?'Koneksi gagal :'.$db->connect_error:'';
$query = ("SELECT * FROM printer_function_commands") or die(mysql_error());
$result = mysqli_query($db, $query);
$pilihan = '';
$data = array();
if (isset($_POST['submit']) && isset($_POST['commands']))
{
if (count($_POST['commands']) > 0)
{
$pilihan = serialize($_POST['commands']);
echo "<p>Data berhasil disimpan ke database berupa ".$pilihan."</p>";
}
}
if ($pilihan <> '')
{
$data = unserialize($pilihan);
}
?>
<table border="1">
<tr>
<th>ID</th>
<th>No</th>
<th>Commands</th>
</tr>
<?php
include "database.php";
$data1 = mysql_query("select * from result_commands");
$no = 1;
while($d = mysql_fetch_array($data1)){
?>
<tr>
<td><?php echo $no++; ?></td>
<td><?php echo $d['no_commands']; ?></td>
<td><?php echo $d['commands']; ?></td>
</tr>
<?php } ?>
</table>
<br/>
<h2>Input Banyak Data</h2>
<form method="post" action="tambah.php">
<table border="1">
<tr>
<th>ID</th>
<th>No</th>
<th>Commands</th>
<th>Pilih</th>
</tr>
<?php
include "database.php";
$data = mysql_query("select * from printer_function_commands");
$no = 1;
while($d = mysql_fetch_array($data)){
?>