How do you define a variable in php that can be used with .done in jquery

Question

If I have some php logic that results in either :

$staus = "pass";

or:

$status = "fail";

Then in my jQuery/Ajax I have:

$(function() {
$( "#send" ).click(function(e){
e.preventDefault();
$( "#send" ).prop( "disabled", true );
$( ".loader" ).show();
$( "#send" ).html( "Sending <img src='img/ajax-loader.gif'>" );
var form_data = $( "#contact-form" ).serialize();
 $.ajax({
  type: 'POST',
  url: 'send.php',
  data: form_data
}).done(function(response) {
 if(response.status == "pass")
// console log something here

But I am getting "undefined" in the Google Chrome Console when I try to console.log the data.

if($_POST) {


$message = "";

if(empty($_POST['first_name'])) {

    $message .= "First name required<br/>";
}

if(empty($_POST['last_name'])) {

    $message .= "Last name required";
}

if($message) {

    echo "<div class='alert alert-danger'>" . $message . "</div>";
    $pass = "error";

} else {

    echo "<div class='alert alert-success'>Form submitted successfully</div>";
    $pass = "succeed";
}
}

Please add more information like what format response is in and what is the contentType in Ajax request. — Satish Saini, Apr 20 '17 at 06:00

Zaman · Accepted Answer · 2017-04-20T07:44:13.097

1

Try this:

$(function() {
$( "#send" ).click(function(e){
e.preventDefault();
$( "#send" ).prop( "disabled", true );
$( ".loader" ).show();
$( "#send" ).html( "Sending <img src='img/ajax-loader.gif'>" );
var form_data = $( "#contact-form" ).serialize();
 $.ajax({
  type: 'POST',
  url: 'send.php',
  dataType: 'json',
  data: form_data,
  success: function(json){
 if(json['error'])
{
$('#your_notification_div').html("<div class='alert alert-danger'>" +json['message']+ "</div>")
}
else
{
$('#your_notification_div').html("<div class='alert alert-success'>"+json['message']+"</div>")

}
}
});
});
});

Edit the validation PHP file:

    if($_POST) {
    $json = array();
    $message = '';
    if(empty($_POST['first_name'])) {
        $json['error'] = 'true';
        $message .=  "First name required<br/>";
    }
    if(empty($_POST['last_name'])) {
        $json['error'] = 'true';
        $message .=  "Last name required";
    }
    if(!isset($json['error'])) {
        $json['success'] = 'true';
        $message = 'Form submitted successfully';
    }
    $json['message'] = $message;
    echo json_encode($json);
}

edited Apr 20 '17 at 07:44

answered Apr 20 '17 at 06:24

Zaman

551
4
8

I don't really know what you mean? Would I have to change other code or could I just replace the success with .done? – Iggy's Pop Apr 20 '17 at 06:27
please replace my code snippet and let me know if it works. – Zaman Apr 20 '17 at 06:28
Nothing happens when I use this code. There is nothing displayed on the page and there is no console output – Iggy's Pop Apr 20 '17 at 06:37
can you check for js errors, as I have not tested it – Zaman Apr 20 '17 at 06:49
In console? Or how? – Iggy's Pop Apr 20 '17 at 06:51
on Chrome: Right click and "Inspect Element" and checkout the "Console" tab – Zaman Apr 20 '17 at 07:11
Yeah, I did that. There was a simple error about the ) or } but I fixed it and now there are no console errors. – Iggy's Pop Apr 20 '17 at 07:12
Thank you. Still nothing happens and there is no error in console. There was an extra } in the php which I fixed but still nothing. – Iggy's Pop Apr 20 '17 at 07:22
Let us [continue this discussion in chat](http://chat.stackoverflow.com/rooms/142136/discussion-between-zaman-and-jonathan). – Zaman Apr 20 '17 at 07:32

score 0 · Answer 2 · answered Apr 20 '17 at 06:03

0

I guess something like:

<?php
  $staus = "pass";
  ?>
  <script>
    const staus = <?php echo $staus?>
    console.log(staus);
  </script>
  <?php
  // PHP Code...
?>

I do something like that in EJS(Node)

answered Apr 20 '17 at 06:03

Felix Fong

969
1
8
21

score 0 · Answer 3 · edited May 23 '17 at 12:25

0

in ajax response you will get all echo value from PHP. if you want your ajax to give you result in variable then you can use JSON data type to retrieve data in array. refer this link for more details

edited May 23 '17 at 12:25

Community

1
1

answered Apr 20 '17 at 06:23

Bhupat Bheda

1,968
1
8
13

How do you define a variable in php that can be used with .done in jquery

3 Answers3