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Editorial Team
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Asked: 2026-06-07T16:27:10+00:00

I want to show the selected ID data in the form and EDIT it

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I want to show the selected ID data in the form and EDIT it and UPDATE in the database. I selected the data from the database and put it in the input tag but it doesn’t work. Please help!

<html>
<body>
<?
$db = mysql_connect("localhost", "root","");
mysql_select_db("db_ncs",$db);
$id = $_GET['s_id'];
if($id)
{
    $result=mysql_query("SELECT * FROM tbl_student WHERE s_id=$id");
    $row = mysql_fetch_assoc($result);
}
?>

<form method="post" action="update.php">
    Name:<input type="Text" name="name" value="<?php echo $row['s_name'];?>" /><br>
    Contact:<input type="Text" name="contact" value="<?php echo $row['s_contact'];?>" /><br>
    Address:<input type="Text" name="address" value="<?php echo $row['s_address'];?>" /><br>
    E-mail:<input type="Text" name="email" value="<?php echo $row['s_email'];?>" /><br>
    <input type="submit" name="update" value="Update">
</form>

<?
if(isset($_POST['update']))
    {
    $name = $_POST['s_name'];
    $contact = $_POST['s_contact'];
    $address = $_POST['s_address'];
    $email = $_POST['s_email'];
    $sql = "UPDATE tbl_student
            SET (s_name='$name', s_contact='$contact', s_address='$address', s_email='$email')
            WHERE s_id=$id";
    $res = mysql_query($sql);       
    if($res)
        {
            echo "Upadate Successfull!";
        }
        else
        {
            echo "Sorry!";
        }
    }

?>

</body>
</html>
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1 Answer

  1. Editorial Team

    You forgot to pass the id.

    Add this between the <form> tags.

    <input type="hidden" name="s_id" value="<?php echo $id;?>" />

    You also need to make your methods consistent. The form submits the data via method="get" but you ask for it via $_POST. You also need to make the input names consistent with the names you ask for, by either adding or removing the “s_” in the appropriate places.

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