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Editorial Team
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Asked: 2026-05-20T16:55:37+00:00

i have a simple form to validate if an user exists in the db,

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i have a simple form to validate if an user exists in the db, and permit the login to another page (yes i remove the password validation to simplify the code, and i test without pass and the script always show Your login is not valid).

Database: emprego, table: users, fields: id, username, email, pass

<script type="text/javascript" language="javascript">
$(document).ready(function () {
    $("#login_form").submit(function () {
        $("#msgbox").removeClass().addClass('messagebox').text('Validating....').fadeIn(1000);
        $.post("ajax_login.php", {
            name: $('#name').val(),
        }, function (data) {
            if (data == 'yes') {
                $("#msgbox").fadeTo(200, 0.1, function () {
                    $(this).html('Logging in.....').addClass('messageboxok').fadeTo(900, 1, function () {
                        document.location = 'secure.php';
                    });
                });
            } else {
                $("#msgbox").fadeTo(200, 0.1, function () {
                    $(this).html('Your login is not valid').addClass('messageboxerror').fadeTo(900, 1);
                });
            }
        });
        return false; 
    });
}); 
</script>
    </head>
    <body>
        <form method="post" action="" id="login_form">
                    User Name : <input name="name" type="text" id="name">
                <div class="buttondiv">
                    <input name="Submit" type="submit" id="submit" value="Login" style="margin-left:-10px; height:23px">
                    <span id="msgbox" style="display:none"></span>
                </div>
        </form>

file: ajax_login.php

<?php 

include("includes/f_banco.php");
conecta ();


$check = mysql_query("SELECT username FROM users WHERE username ='".mysql_real_escape_string($_POST['name'])."'")or die(mysql_error());
$row=mysql_fetch_array($check);

if ($row['username'] == $_POST['name']) {
    echo "yes";
    $_SESSION['name']=$row['username'];
}else{echo "no";}

?>

EDIT: updated code

what is the possible cause?

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1 Answer

  1. Editorial Team

    i see lots of problems

    <?php  
//removed the # from the post
$name = mysql_real_escape_string($_POST['name']);
//whys this here?
$row=mysql_fetch_array($check);
//name cant be used (common mistake)
$check = mysql_query("SELECT name FROM users WHERE name ='".$name."'")or die(mysql_error());
$check2 = mysql_num_rows($check);
if ($check2 == 1) {
    echo "yes";
    $_SESSION['name']=$name;
}

else
echo "no";
?>

    try something like this

    <?php  
session_start();
include("includes/f_banco.php");
conecta ();

if(isset($_SESSION['name'])){echo "yes";}else{
    $check = mysql_query("SELECT username FROM users WHERE username ='".mysql_real_escape_string($_POST['name'])."' LIMIT 1")or die(mysql_error());
    $row=mysql_fetch_array($check);

    if ($row['username'] == $_POST['name']) {
        echo "yes";
        $_SESSION['name']=$row['username'];
    }else{echo "no";}
}


?>
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