PHP HTML Form not posting data to database?

Question

I have this form on my site to change activated statuses of users in the users list, so the admin has the ability to ban people from the admin area. The form is meant to show what the activation status is, then show either 0, or 1 as an option. If you drop down and change to the opposite choice, it will change the activation status number in the database under the specific username in which you've changed the activation number. I am having a problem because when you change the number, the page refreshes as something is happening. But it seems to not send the information or change the database at all? Has anyone got any help here? Thanks.

$outputList .= '<tr><td>'. $id .'</td><td>'. $firstname .'</td>
    <td>'. $lastname .'</td>
    <td><form name="activationsend" action="admin.php?page=2" method="POST">
      <input type="hidden"/>
      <select name="act" onchange="this.form.submit();">
        <option style="display:none">'. $activated .'</option>
        <option name="0" value="0" >0</option>
        <option name="1" value="1" >1</option>
      </select>
    </td>
    <td><a href="'. $username .'">'. $username .'</a></td>
    </form>
  </tr>';
}

$outputList .= '</tbody>
            </table>';

if (isset($_POST['activationsend'])){
$activated = $_POST['act'];
$username = $_POST['$username'];

$exists = mysql_query ("SELECT * FROM users WHERE username='$username'") or die ("not found");
    if (mysql_num_rows($exists) != 0){
        //update the info in database
        mysql_query ("UPDATE users SET activated='$activated' WHERE username='$username'") or die ("update didn't work");
        echo "worked";
} else echo "did not work";
}

$username = $_POST['$username']; also username is never posted, there's no such field — Ivan Yonkov
– Ivan Yonkov, Commented Mar 18, 2014 at 10:32
So how can I make username a field depending on which username is being clicked on? Bare in mind that its a list of usernames with the form on each separate one? — AidanPT
– AidanPT, Commented Mar 18, 2014 at 10:34

Kirill Arutyunov · Accepted Answer · 2014-03-18 10:46:12Z

1

Try this condition

if (isset($_POST['act'])){
// ...
}

I'm not sure, that activationsend send to server. So, you can also try var_dump($_POST) function to show your POST data after request:

$outputList .= '</tbody>
            </table>';

var_dump($_POST);

It will return POST-data or NULL.

UPD.

You also need to pass username data. Try to change

<a href="'. $username .'">'. $username .'</a>

into

<a href="'. $username .'">'. $username .'</a>
<input type="hidden" name="username" value="'.$username.'" />

edited Mar 18, 2014 at 10:46

answered Mar 18, 2014 at 10:29

Kirill Arutyunov

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8 Comments

AidanPT Over a year ago

The form is NOW being initiated but it's echoing "did not work" no idea why?

AidanPT Over a year ago

With the Var_dump on send its saying "array(1) { ["act"]=> string(1) "1" } did not work "

Kirill Arutyunov Over a year ago

oh.you have an error: $username = $_POST['$username']; change to $username = $_POST['username'];

Kirill Arutyunov Over a year ago

You don't need to write $ as array index: $username = $_POST['username']; // 'username' is a name of your form field, but not '$username'

AidanPT Over a year ago

array(1) { ["act"]=> string(1) "1" } did not work, still giving me that.

|

Nishu Tayal · Accepted Answer · 2014-03-18 10:39:11Z

First of all, username is not posted to the server and over that you are using $username variable instead of username in $_POST.

It should be like this :

if (isset($_POST['activationsend'])){
  $activated = $_POST['act'];
  $username = isset($_POST['username'])? $_POST['username'] : '';
  // Check if $username is empty or not
  if($username != ""){
    $exists = mysql_query ("SELECT * FROM users WHERE username='$username'") or die ("not found");
    if (mysql_num_rows($exists) != 0){
        //update the info in database
        mysql_query ("UPDATE users SET activated='$activated' WHERE username='$username'") or die ("update didn't work");
        echo "worked";
    } else echo "did not work";
  }
}

If there is no record in database for that username, it'll show "did not work"

Hope,it'll solve your problem.

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