使用php在mysql数据库中保存复选框值

Question

i make a for in which i put multiple check boxes and now i want to save each selected checkbox value in database but i get only array text in database... this is my form code...

<?php session_start();
include '../includes/layouts/header.php';


// Check, if username session is NOT set then this page will jump to login page
if (!isset($_SESSION['admin']))
{
    header('Location:signin.php');
}

?>


<div id="main">
    <div id="navigation" >
        <ul class="nav">
            <li><a href="selected_user.php">Users Statistics</a></li>
            <li><a href="message.php">Post New Anousment</a></li>
            <li><a href="#">Messages</a>
                <ul>
                    <li><a href="#">Inbox</a></li>
                    <li><a href="#">Outbox</a></li>

                </ul>
            </li>
            <li><a href="#">Setting</a>
                <ul>
                    <li><a href="subadmin.php">Add Sub Admin</a></li>
                    <li><a href="#">Update Records</a></li>
                    <li><a href="del.php">Delete Records</a></li>
                </ul>

            </li>
            <li><a href="logout.php">Logout</a></li>
        </ul>
        &nbsp;
    </div>
    <div id="page">

        <h2>Admin Menu</h2>
        <p>Welcome to Admin Area</p>

        <form action="subad.php" method="post">
            <label for="">Admin Name</label><br>
            <input type="text" name="admin_name" value=""/><br>
            <label for="">Admin Password</label><br>
            <input type="password" name="admin_password" value=""/><br><br>
            <label for="">Select Rights For Admin</label><br><br>
            <input type="checkbox" value="track user" name="chk[]"/>
            <label for="">Manage/Track User's</label><br>
            <input type="checkbox" name="chk[]" value=""/>
            <label for="">Delete User's</label><br>
            <input type="checkbox" name="anoucments" value=""/>
            <label for="">Post New Anousment's</label><br>
            <input type="checkbox" name="areject" value=""/>
            <label for="">Accept/Reject User Application</label><br><br>
            <input type="submit" name="add" value="Add Admin"/>
        </form>
        <?php
        if(isset($_SESSION['success']))
        {
            echo $_SESSION['success'];
            unset($_SESSION['success']);
        }
        ?>
    </div>
</div>
<?php include '../includes/layouts/footer.php' ?>

and this is my database.php code

<?php session_start();
include 'conn.php';
$admin=$_POST['admin_name'];
$password=$_POST['admin_password'];
$track=$_POST['chk'];
    for ($i=0; $i<sizeof($track);$i++) {
        $sql="INSERT INTO sub_admin(Admin_Name,Password,Track_User)VALUES('$admin','$password','$track')";
        $res=mysql_query($sql);
        if($res)
        {
            $_SESSION['success']="Sub-Admin is successfuly added!";
            header("location:subadmin.php");
        }
        else
        {
            echo mysql_error();
        }
    }


?>

can any one tell me that what is the reason? i want to store the check box text in the form of true or false, because in my database i made individual column for each checkbox.

and also can any one tell me that how to make a subadmin by super admin and assign them a limited rights? any idea or web link?

Answer 1

Try doing

$track = json_encode($_POST["chk"]);

This is the quickest solution because your database does not match the checkboxes. However, if the database were designed with each checkbox as a column, you could save true or false in each of them.

To fetch the array of data from the database

$data = mysql_result($ret, 0, "Track_User");
$track = json_decode($data, true);

Answer 2

You are currently setting $track equal to the entire array $_POST['chk'] .

As jetawe suggested in the comments you need to run $track[$i] as the value for the Track_User column.