How to echo PHP to Ajax? Login System

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I’m working on a login system that warns the user if the username already exists in the database. I basically have a popup window on the same page where the input fields are for both login and user registration. The idea was to submit the fields and return a paragraph that would inform the user whether the account had been added successfully or not.

Here is the code:

 if(isset($_POST['submitlogin'])) {

        global $connection;
        $username = mysqli_real_escape_string($connection, trim($_POST['username']));
        $password = mysqli_real_escape_string($connection, trim($_POST['password']));

        if(!empty($username) && !empty($password)) {

            /*Register*/
                if(isset($_POST['classmember'])){
                    $class_member= mysqli_real_escape_string($connection, trim($_POST['classmember']));

                    $query = "SELECT member FROM members WHERE member='$username';";
                    $result = mysqli_query($connection, $query);
                    $num_rows = mysqli_num_rows($result);

                    if($password == GUILDPASS) {
                         if($num_rows == 0) {

                         $query2 = "INSERT INTO members(member,pass,class_id,status_id) VALUES ('$username',". '\''. GUILDPASS . '\'' . ",$class_member,3);";

                         $insert = mysqli_query($connection, $query2);

                         echo "Added Successfully";

                         } else {
                             echo "player already exists";
                         }
                     } else {
                         echo "Wrong password";
                     }
                }        
            /* LOGIN*/

                else {    
                    $query = "SELECT member,pass FROM members WHERE member='$username' AND pass='$password';";
                    $result = mysqli_query($connection, $query);
                    $num_rows = mysqli_num_rows($result);

                    if($num_rows == 0) {
                         echo "Create Account or Username/Password incorrect..";   
                    } else if ($num_rows == 1){
                         $_SESSION['sessionid'] = session_id();
                    }
                }


        } else {
            echo "Password/Username missing";
        }
     header("Location: ../index.php");
     exit();
}
php jquery ajax

  • Buddy, do you get any errors when you run this code? Be very careful when using "SELECT Member FROM Members WHERE Member='$username';", as this opens the doors to SQL Injection. Never use a variable directly in the sql query, without filtering.

  • It is not possible to perform echo by ajax because the page has already been loaded. An alternative would be to collect the return of the ajax request, and manipulate the page’s DOM to simulate an "Echo". Notes that PHP (Hypertext Preprocessor) means something that manipulates the text before it is rendered.

  • @Stéfano At the moment this in production later I will pass everything to Prepared statements and review the security of the code.

1 answer

2

You can take the returned value and use in ajax, example:

$.ajax({
  type: 'POST',
  url: 'SUA URL',
  dataType: 'html',
  success: function(data){
     alert(data)
  }
});

The date is echo you returned from PHP, the above example is using dataType as html, but we usually use as json and return the json of PHP with json_encode()

  • Thank you for your reply. Unfortunately I’ve already tried this, and what happens in my case is that when I do console.log(data) what appears is the entire HTML of the page. I’ve even tried not to do by ajax, and just submit the POST to the same page but for some reason when I leave the form action="" it returns to the previous Folder and goes to the Dashboard of XAMPP

  • In your file, it only has the PHP code, or you are using PHP and HTML in the same code?

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