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jquery ajax dont work without firebug break point

I am using following method to call the php:

function validateEmaiAjax(email){
    val = null;
    $("#warning").load("https://localhost/Continental%20Tourism/register_ajax.php",{email: email}, function(rspns, stat, xml){
        val = rspns;
    });

    if(val == ".")
        return true;
    else {
        return false;
    }
}

my php code is:

<?php
    $dbc = mysqli_connect("localhost","root","pass","continental_tourism") OR die(mysqli_connect_error());

    $email = $_REQUEST['email'];

    $query = "SELECT email FROM customer_info WHERE email = '$email' ";

    $r = mysqli_query($dbc, $query) OR die(mysqli_error($dbc));

    if(mysqli_num_rows($r) > 0)
        echo "Email address exists!";
    else
        echo ".";   
?>

Basically this do check the database and if email exists shows "Email address exists!" if not I want to return true(so I echo "." and compare it). The weird thing is if i put a break point using firebug near if(val == ".") program works correctly and returns true. If I remove that break point function always return false. I cant understand why this happens. Please help! Thanks.

like image 242
Jayanga Kaushalya Avatar asked Dec 21 '22 23:12

Jayanga Kaushalya


1 Answers

The reason you have this problem is because you have performed an asynchronous request. This means that the if(rspns == ".") will be reached before the response has been received from the server, and the result will always be false.

In order to wrap this code in a function the returns a boolean and does not require a callback function (a blocking procedure) you will need to use a synchronous request:

function validateEmaiAjax(email) {

  // This is the correct way to initialise a variable with no value in a function
  var val;

  // Make a synchronous HTTP request
  $.ajax({
    url: "https://localhost/Continental%20Tourism/register_ajax.php",
    async: false,
    data: {
      email: email
    },
    success: function(response) {
      // Update the DOM and send response data back to parent function
      $("#warning").html(response);
      val = response;
    }
  });

  // Now this will work
  if(val == ".") {
    return true;
  } else {
    $("#warning").show();
    return false;
  }

}
like image 160
DaveRandom Avatar answered Jan 11 '23 07:01

DaveRandom