login form php with jquery validation

Question

hello I am new here and in programming as well. Today I have been made a login form in php / mysql and jquery form validation and was surprised that my form do not work bacause of js script? withou work well.

<div class="top_login_error" id="top_error">All fields must be field</div>
    <div id="logincontainer">
        <div id="loginbox">
            <div id="loginheader">
                <div>Control Panel Login</div>
            </div>
            <div id="innerlogin">
                <form method="POST" action="obr.php" id="login_form">
                    <p>Enter your username:</p>
                    <input type="text" class="logininput" name="login" id="login" />
                    <p>Enter your password:</p>
                    <input type="password" class="logininput" name="password" id="password" />
                    <input type="submit" class="loginbtn" value="Submit" name="enter" id="admin_login" /><br />
                    <p><a href="#" title="Forgoteen Password?">Forgotten Password?</a></p>
                </form>
            </div>
        </div>
        <img SRC="../images/administration/login_fade.png" alt="Fade" />
    </div>
</body>

jquery ccode for validating form

$('#admin_login').click(function(q){

q.preventDefault();

    var login = $('#login').val();
    var password = $('#password').val();

    if (login.length == 0){
        $('#top_error').fadeIn(500);
        $('#login').addClass("logininput_error");
        var error = true;
    } else {
        $('#top_error').fadeOut(500);
        $('#login').removeClass("logininput_error");
        }
    if (password.length == 0){
        $('#top_error').fadeIn(500);
        $('#password').addClass("logininput_error");
        var error = true;
    } else {
        $('#top_error').fadeOut(500);
        $('#password').removeClass("logininput_error");
        }

});

and php

<?php 
session_start(); 



    if (isset($_POST['login'])){$login = $_POST['login'];} if ($login == ''){unset($login); exit('please enter name');}
    if (isset($_POST['password'])){$password = $_POST['password'];} if ($password == ''){unset($password); exit('please enter password');}

    $login = mysql_escape_string($_POST['login']);
    $password = mysql_escape_string($_POST['password']);

    //$password = md5($password); I have been disabled while testing in simle way, login : aaa , password : aaa;

    include("../db.php");

    $result = mysql_query("SELECT * FROM users WHERE login='".$login."' and password='".$password."'") or die(mysql_error());

    $line = mysql_fetch_array($result);

    if (empty($line['id'])){
        exit("Your name or login is incorect <a href='login.php'>back</a>");
    } else {

    $_SESSION['autorized'] = true;
    $_SESSION['login'] = $login;
    $_SESSION['password'] = $password;
    }
    echo "<html><head><meta http-equiv='Refresh' content='0; URL=index.php'></head></html>";

?>

Can someone help me with code, tell what to change that everything work fine. Thank you

Prusse · Accepted Answer · 2011-07-15 21:14:34Z

5

The culprit is q.preventDefault();, only do that when the form shouldn't be submitted.

answered Jul 15, 2011 at 21:14

Prusse

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1 Comment

Viktors Over a year ago

If it is not to hard maybe can someone tell me what code add to my jquery to send data with ajax

Alfred · Accepted Answer · 2011-07-15 23:54:55Z

1

Probably Prusse is right.

the Jquery documentation is very good => You should read http://api.jquery.com/jQuery.post/ to do post request(AJAX) and http://api.jquery.com/submit/ to handle submit.

I would also advice you to use PDO instead of "normal" MySQL to also prevent against SQL-injections. You can then use PHPunit to properly test your code following Uncle Bob's three rules of TDD. I like to use in-memory SQLite database to test my code(new PDO('sqlite::memory:');). You could see my simple authentication library to see how I did that => https://github.com/alfredwesterveld/php-auth

edited Jul 15, 2011 at 23:54

answered Jul 15, 2011 at 23:38

Alfred

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