PHP Login form validation failed

Question

I have following login form pointing to the file '../exe/form-exec.php'.

<form id="loginForm" name="loginForm" method="post" action="../exe/login-exec.php">
  <table width="300" border="0" align="center" cellpadding="2" cellspacing="0">
    <tr>
      <td width="112"><b>Login</b></td>
      <td width="188"><input name="login" type="text" class="textfield" id="login" /></td>
    </tr>
    <tr>
      <td><b>Password</b></td>
      <td><input name="password" type="password" class="textfield" id="password" /></td>
    </tr>
    <tr>
      <td>&nbsp;</td>
      <td><input type="submit" name="Submit" value="Login" /></td>
    </tr>
  </table>
</form>

Where login-exec.php should validate where inputs login & password has been field and if NOT return to the form file with $errflag.

File login-exec.php:

//Start session
    session_start();

    //Include database connection details
    require_once('../inc/config.php');

    //Array to store validation errors
    $errmsg_arr = array();

    //Validation error flag
    $errflag = false;



..... 



//Input Validations
    if($login == '') {
        $errmsg_arr[] = 'Login ID missing';
        $errflag = true;
    }
    if($password == '') {
        $errmsg_arr[] = 'Password missing';
        $errflag = true;
    }

    //If there are input validations, redirect back to the login form
    if($errflag) {
        $_SESSION['ERRMSG_ARR'] = $errmsg_arr;
        session_write_close();
        header("location: ../form/login-form.php");
            exit();
}

And partly is working correctly - it is returning to the form - but no error is displayed. Any suggestion much appreciated.

There isn't any. Sorry. I doing some exercises with PHP - it's new to me. PHP checking the array of $errmsg_arr should be places in login-form.php, should it not? — Iladarsda
– Iladarsda, Commented Aug 5, 2011 at 10:32

JK. · Accepted Answer · 2011-08-05 10:39:26Z

1

On the same page as your html form, do something similar to this:

session_start();
foreach ($_SESSION['ERRMSG_ARR'] as $msg) {
  echo $msg . "<br />";
}
etc...

To actually display the messages.

answered Aug 5, 2011 at 10:39

JK.

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4 Comments

Iladarsda Over a year ago

This is working. But, straight away on login page there is an error displayed - even without entering any data. Just a guess that array is saved from previous login failure. How to change that?

Iladarsda Over a year ago

If session is dead: Notice: Undefined index: ERRMSG_ARR in C:\wamp\www\form\login-form.php on line 3 and Warning: Invalid argument supplied for foreach() in C:\wamp\www\form\login-form.php on line 3 If session is live but loggout - error msg straight away. If ignore the error when the session is dead and press 'submit' correct error msg is displayed without errors.

JK. Over a year ago

Try to surround the code with this: if(isset($_SESSION['ERRMSG_ARR'])) { // Code here }

Iladarsda Over a year ago

Yep. This is working now. Now need to handle incorrect data entry :) Thanks!

VAShhh · Accepted Answer · 2011-08-05 10:48:57Z

1

You are valorizing the $_SESSION['ERRMSG_ARR'] but then when you return to the form you're not checking/using it... try to add this to the form code

<?php 
session_start();
if(!is_null($_SESSION['ERRMSG_ARR'])) {
    echo '<div class="errmsg">Error...</div>';
}
?>
<form id="loginForm" name="loginForm" method="post" action="../exe/login-exec.php">
  <table width="300" border="0" align="center" cellpadding="2" cellspacing="0">
    <tr>
...

Anyway this type of validation handling (using session parameters) is heavily discouraged...

Try to follow some design patterns instead ;)

edited Aug 5, 2011 at 10:48

answered Aug 5, 2011 at 10:37

VAShhh

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2 Comments

Iladarsda Over a year ago

This will return Notice: Undefined variable: _SESSION in C:\wamp\www\form\login-form.php on line 2. Sorry, I'm new to PHP - first login system of my

VAShhh Over a year ago

I would suggest to use any PHP framework in the first place ;) or use an ajax validation. Anyway, if this is for "practice" purpose, you can simply put the validation & the form code in the same file and make the form put to itself. So you will have to check the $_POST parameters and if they are correct forward the page to the desired page (or directly show the error)

gmadd · Accepted Answer · 2012-01-10 16:14:51Z

0

When getting data submitted from a form with POST method, it is better to use $_POST['login'] instead of $login since if register_globals is off it won't work. This might be your problem.

edited Jan 10, 2012 at 16:14

gmadd

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answered Aug 5, 2011 at 10:32

Pierre

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2 Comments

Iladarsda Over a year ago

Yep. register_globals = off, I will try to change it.

plandolt Over a year ago

@Pete: No, let register_globals off. This feature has been DEPRECATED as of PHP 5.3.0. Relying on this feature is highly discouraged.

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