Database does not update, mysql_error does not show errors

Question

I've seen the related solutions about this and tried it but none of those works. This is my code for Updating Database, the mysql_error shows no sign of error. Don't know what to do. Help!

    <?php 
        mysql_select_db("dbUsers");     
        if(isset($_POST['btnSave'])){

        $firstname=$_POST['firstname'];
        $lastname=$_POST['lastname'];
        $address=$_POST['address'];

        $query1 =("UPDATE `webuser` SET `firstname`='".$firstname."',
                `lastname`='".$lastname."',`address`='".$address."'
                 WHERE empNo = '".$selEmp."' ");

        $result = mysql_query($query1) or die(mysql_error());
        }
        mysql_close($conn);

   ?>

Echo the statement and put exit after the update line and check.. Whether all the particulars that are to be updated are present in the query. — Naresh Kumar P
– Naresh Kumar P, Commented Aug 28, 2016 at 8:25
Danger: You are using an obsolete database API that has been removed from PHP. You should select a modern replacement. You are vulnerable to SQL injection attacks that a modern API would make it easier to defend yourself from. — Quentin
– Quentin, Commented Aug 28, 2016 at 8:36
@NareshKumar.P thank you. I've seen the error. the empNo is not showing, I'm still trying to figure why. I can echo it but I can't use it in query. — eshi
– eshi, Commented Aug 28, 2016 at 8:47
have updated the query @eshi. And you could find how to pass the Emp_id from HTML Form while updating the particular row in the DB. — Naresh Kumar P
– Naresh Kumar P, Commented Aug 28, 2016 at 10:17

Vivek Tiwari · Accepted Answer · 2016-08-28 08:33:33Z

1

Instead of using mysql go for mysqli as it preffered and secured.

use the below code for connection (create variable with its values)

$conn = mysqli_connect($servername, $username, $password, $dbname);

For update quesry... use

$sql = "UPDATE webuser SET firstname='".$firtname."' WHERE empNo='".$selEmp."';

and to Query use...

mysqli_query($conn, $sql)

I have not mentioned your whole code but hope you understand it now.

answered Aug 28, 2016 at 8:33

Vivek Tiwari

275 bronze badges

Sign up to request clarification or add additional context in comments.

3 Comments

Ryan Vincent Over a year ago

fyi, This is vulnerable to SQL injection attacks. Please use prepared queries with parameters.

eshi Over a year ago

Thank you Vivek. I have fix the problem tho :)

Vivek Tiwari Over a year ago

it is just part of it and thanks for your comment .. @RyanVincent

Naresh Kumar P · Accepted Answer · 2016-08-28 10:16:21Z

0

Change the update query like this. Since the braces are not allowed.

Ensure that your empNo is present at the time while updating.

<?php
mysql_select_db("dbUsers");     
if(isset($_POST['btnSave'])){
$selEmp = $_POST['emp_id'];
$firstname=$_POST['firstname'];
$lastname=$_POST['lastname'];
$address=$_POST['address'];
$query1 ="UPDATE `webuser` SET `firstname`='".$firstname."',
`lastname`='".$lastname."',`address`='".$address."'
WHERE empNo = '".$selEmp."'";    
$result = mysql_query($query1) or die(mysql_error());
}
mysql_close($conn);
?>

<form method="POST" action="">
    <input type="hidden" name="emp_id" value="<?php echo $data['id']; ?>" />
    <input type="text" name="firstname" value="<?php echo $data['firstname']; ?>" />
    <input type="text" name="lastname" value="<?php echo $data['lastname']; ?>" />
    <textarea name="address"><?php echo $data['address']; ?></textarea>
    <input type="submit" name="btnSave" value="UPDATE" />
</form>

edited Aug 28, 2016 at 10:16

answered Aug 28, 2016 at 8:31

Naresh Kumar P

4,2802 gold badges18 silver badges34 bronze badges

2 Comments

Naresh Kumar P Over a year ago

Share thoughts if you face any hindrance in my codes.

eshi Over a year ago

Thank you Naresh. I have fix the problem tho. the variable i need is not accesible inside the isset function so what I did is I made a hidden input then pass the value there and get it. Thank you again :)

Kostas Mitsarakis · Accepted Answer · 2016-08-28 09:29:03Z

0

To avoid SQL Injections and the deprecated mysql_* functions use PDO with prepared statements:

define('DB_HOST', 'localhost');
define('DB_NAME', 'dbUsers');
define('DB_USER', 'your_username');
define('DB_PASSWORD', 'your_password');

try {
    //Make your connection handler to your database
    $conn = new PDO("mysql:host=".DB_HOST.";dbname=".DB_NAME, DB_USER, DB_PASSWORD, array(PDO::ATTR_ERRMODE => PDO::ERRMODE_WARNING));

    $firstname = $_POST['firstname'];
    $lastname = $_POST['lastname'];
    $address = $_POST['address'];

    $sql = "UPDATE webuser SET firstname = :firstname, lastname = :lastname, address = :address WHERE empNo = :emp_no";
    $stmt = $conn->prepare($sql);
    $stmt->execute(array(':firstname' => $firstname, ':lastname' => $lastname, ':address' => $address, ':emp_no' => $selEmp));

} catch(PDOException $e) {
    echo $e->getMessage();
    die();
}

More info here.

edited Aug 28, 2016 at 9:29

answered Aug 28, 2016 at 9:04

Kostas Mitsarakis

4,7473 gold badges26 silver badges38 bronze badges

3 Comments

Kostas Mitsarakis Over a year ago

Where is the catch?

eshi Over a year ago

Thank you Kostas, I have fix the problem tho. :)

Kostas Mitsarakis Over a year ago

Nice. Keep in mind to check about the PDO and how to avoid SQL Injection.

Collectives™ on Stack Overflow

Database does not update, mysql_error does not show errors

3 Answers 3

3 Comments

2 Comments

3 Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

3 Comments

2 Comments

3 Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related