PHP Ajax alert wont work

Question

I am trying to check user is exist from data using PHP and Ajax. Using the following codes:

Ajax

$('#btn_check_pc').click(function() {
  username = $("#username").val();
  $.ajax({
    type: 'POST',
    url: 'process.php',
    data: "username=" + username,
    dataType: 'json',
    beforeSend: function() { //Do Something    
    },
    success: function(response) {
      if (response == 'used') {
        console.log("Username Already In Use");
      }
    }
  });
});

PHP

<?php
    include_once "includes/get_data.php" ;

    $username = $_POST["username"];
    //Validating purchase

    $checkUserRegistered = mysqli_query($db,"SELECT  * FROM  users WHERE username = '$username'") or die(mysqli_error($db));

    if(!mysqli_num_rows($checkUserRegistered)){
        // Do Something
    }else{
         echo 'used';
    }
?>

So When I click the #btn_check_pc button then the PHP code will response used if the username already exists. I want to show it with console.log("Username Already In Use"), but it doesn't show. What I am missing here to show console.log() anyone can help me?

`success` should not be nested in `beforeSend`, it should be at the top level of the ajax options object — Rob M., Aug 23 '17 at 18:40
Did you try to use a debugger to check the value of `response` and whether `success` is even called? — litelite, Aug 23 '17 at 18:41
You are open to SQL injections. log `response`, it might have a new line. — chris85, Aug 23 '17 at 18:42
Also, your code is at risk of [SQL Injections](https://stackoverflow.com/questions/601300/what-is-sql-injection) — litelite, Aug 23 '17 at 18:42
@litelite I know there are SQL Injection and I just test it now. Thanks for your attention. I just want to show `console.log();` after success here. — AlwaysStudent, Aug 23 '17 at 18:44
Returning a string like `used` and then checking for that string isn't always the best thing to do, as any space or unseen character breaks your code, and expecting JSON back, you'd be better of actually returning JSON — adeneo, Aug 23 '17 at 18:45
It's not going in the `success` callback since it's not receiving a JSON response. — Chin Leung, Aug 23 '17 at 18:46
@RiggsFolly No worries, it's understandable that you made that comment :) — Rob M., Aug 23 '17 at 18:48

Sadhon · Answer 1 · 2017-08-23T20:28:46.920

You have not use alert function so you can not see any alert box.

Your code is almost correct. If you want to see the result then use alert function in lieu of console.log() function inside success:function(response){} blcok.

just do comment the datatype: 'json' inside ajax. Otherwise you can use datatype:'text' .Reason is your expected data is not of json type. It is text type.

$('#btn_check_pc').click(function() {
  username = $("#username").val();
  $.ajax({
    type: 'POST',
    url: 'process.php',
    data: "username=" + username,
    // (alternative in case of data) data: {username:username},

    dataType: 'text',

   //(expected data is not json) dataType: 'json',
    beforeSend: function() { //Do Something    
    },
    success: function(response) {
      if (response == 'used') {
        console.log("Username Already In Use");
      }
    }
  });
});

Hope this will work.

of course data: {username: username} should work will work – Sadhon Aug 23 '17 at 19:47 — Sadhon, Aug 23 '17 at 19:47

score -1 · Answer 2 · answered Aug 23 '17 at 20:09

You have to make two changes in your code.

First since you are using dataType json in your ajax call so you have to pass json in response from php

if(!mysqli_num_rows($checkUserRegistered)){
    echo json_encode(["success"=>false]);
}else{
    echo json_encode(["success"=>true]);
}

Second in JS. Since you are calling the ajax request via POST so you have to send data like the following.

$.ajax({
    ...
    data: {username: username}
    ...
});

Then in your success handler you can make a check like this

success: function(response) {
    if (response.success === true) {
        console.log("Username Already In Use");
    }
}

PHP Ajax alert wont work

2 Answers2