How do I return a proper success/error message for JQuery .ajax()?

05/01/2020 10:30:02

I keep getting the error alert. There is nothing wrong with the MYSQL part, the query gets executed and I can see the email addresses in the db.

The client side:

<script type="text/javascript">
  $(function() {
    $("form#subsribe_form").submit(function() {
      var email = $("#email").val();

        url: "subscribe.php",
        type: "POST",
        data: {email: email},
        dataType: "json",
        success: function() {
          alert("Thank you for subscribing!");
        error: function() {
          alert("There was an error. Try again please!");
      return false;

The server side:


mysql_select_db($database) or die( "Unable to select database");

$senderEmail = isset( $_POST['email'] ) ? preg_replace( "/[^\.\-\_\@a-zA-Z0-9]/", "", $_POST['email'] ) : "";

if($senderEmail != "")
    $query = "INSERT INTO participants(col1 , col2) VALUES (CURDATE(),'".$senderEmail."')";

$response_array['status'] = 'success';    

echo json_encode($response_array);

Verified Answer (123 Votes)

03/13/2012 04:22:02

You need to provide the right content type if you're using JSON dataType. Before echo-ing the json, put the correct header.

    header('Content-type: application/json');
    echo json_encode($response_array);

Additional fix, you should check whether the query succeed or not.

    $response_array['status'] = 'success';  
}else {
    $response_array['status'] = 'error';  

On the client side:

success: function(data) {
    if(data.status == 'success'){
        alert("Thank you for subscribing!");
    }else if(data.status == 'error'){
        alert("Error on query!");

Hope it helps.


Answer #2 (31 Votes)

03/13/2012 04:23:33

Just so you know, you can use this for debugging. It helped me a lot, and still does

error:function(x,e) {
    if (x.status==0) {
        alert('You are offline!!\n Please Check Your Network.');
    } else if(x.status==404) {
        alert('Requested URL not found.');
    } else if(x.status==500) {
        alert('Internel Server Error.');
    } else if(e=='parsererror') {
        alert('Error.\nParsing JSON Request failed.');
    } else if(e=='timeout'){
        alert('Request Time out.');
    } else {
        alert('Unknow Error.\n'+x.responseText);

Answer #3 (23 Votes)

03/13/2012 04:18:54

Some people recommend using HTTP status codes, but I rather despise that practice. e.g. If you're doing a search engine and the provided keywords have no results, the suggestion would be to return a 404 error.

However, I consider that wrong. HTTP status codes apply to the actual browser<->server connection. Everything about the connect went perfectly. The browser made a request, the server invoked your handler script. The script returned 'no rows'. Nothing in that signifies "404 page not found" - the page WAS found.

Instead, I favor divorcing the HTTP layer from the status of your server-side operations. Instead of simply returning some text in a json string, I always return a JSON data structure which encapsulates request status and request results.

e.g. in PHP you'd have

$results = array(
   'error' => false,
   'error_msg' => 'Everything A-OK',
   'data' => array(....results of request here ...)
echo json_encode($results);

Then in your client-side code you'd have

if (!data.error) {
   ... got data, do something with it ...
} else {
   ... invoke error handler ...
Hack Hex uses Stack Exchance API by the Stack Exchange Inc. to scrape questions/answers under Creative Commons license.