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I'm a total noob in php and json programming so this is very difficult to me. I'm trying to do a simple login page for my app, that's the only thing that I don't know how to do. On login form click this jquery happens.

function signin(){
$login_form = $('#login-form');
var fields = $login_form.serialize();
$.ajax({
    type: "POST",
    url: "192.168.1.2/_modules_/login.php",
    data: fields,
    dataType: 'json',
    success: function(error){
        if(!error.status){
            {$.ui.loadContent("main", null, null, "fade");}
        }else{
            $('#error').empty().html(error.html);
        }
    }
});

And post to the php

<?php

function conndb(){
    $servername = "localhost";
    $username = "admin";
    $password = "test";
    $dbname = "appsys";
    $conn = mysql_connect($servername,$username,$password,$dbname);
    /*try {
    $conn = new PDO("mysql:host=$servername;dbname=appsys", $username, $password);
    // set the PDO error mode to exception
    $conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
    }
    catch(PDOException $e)
    {
    $this->error['status'] = false;
    $this->error['html'] = "Connection failed: " . $e->getMessage();
    echo json_encode($error);
    }*/
}   

function login(){
    conndb();
    $error = array();
    $error['status'] = $this->error_status; 
    $error['html'] = $this->error_html;
    $exist = mysql_fetch_array(mysql_query("SELECT * FROM eq_tecnicos WHERE codigo='".secure($_POST['user'])."' AND pass='".secure($_POST['pass'])."'"));
    if(!$exist) 
    { unset($_POST['user'],$_POST['pass']); $error['status'] = true; echo json_encode($error);}
        else { 
            $error['status'] = false; 
            $error['html'] = '<p>Error</p>';
            echo json_encode($error);
        }
}

?>

If this works, it's secure?? Thanks in advance!

0 Answers0