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

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();

      $.ajax({
        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;
    });
  });
</script>

The server side:

<?php 
$user="username";
$password="password";
$database="database";

mysql_connect(localhost,$user,$password);
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."')";
mysql_query($query);
mysql_close();

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

echo json_encode($response_array);
?>

Solution 1:

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

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

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

if(mysql_query($query)){
    $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.

Solution 2:

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);
    }
}

Solution 3:

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 ...
}

Solution 4:

In order to build an AJAX webservice, you need TWO files :

• A calling Javascript that sends data as POST (could be as GET) using JQuery AJAX
• A PHP webservice that returns a JSON object (this is convenient to return arrays or large amount of data)

So, first you call your webservice using this JQuery syntax, in the JavaScript file :

$.ajax({
     url : 'mywebservice.php',
     type : 'POST',
     data : 'records_to_export=' + selected_ids, // On fait passer nos variables, exactement comme en GET, au script more_com.php
     dataType : 'json',
     success: function (data) {
          alert("The file is "+data.fichierZIP);
     },
     error: function(data) {
          //console.log(data);
          var responseText=JSON.parse(data.responseText);
          alert("Error(s) while building the ZIP file:\n"+responseText.messages);
     }
});

Your PHP file (mywebservice.php, as written in the AJAX call) should include something like this in its end, to return a correct Success or Error status:

<?php
    //...
    //I am processing the data that the calling Javascript just ordered (it is in the $_POST). In this example (details not shown), I built a ZIP file and have its filename in variable "$filename"
    //$errors is a string that may contain an error message while preparing the ZIP file
    //In the end, I check if there has been an error, and if so, I return an error object
    //...

    if ($errors==''){
        //if there is no error, the header is normal, and you return your JSON object to the calling JavaScript
        header('Content-Type: application/json; charset=UTF-8');
        $result=array();
        $result['ZIPFILENAME'] = basename($filename); 
        print json_encode($result);
    } else {
        //if there is an error, you should return a special header, followed by another JSON object
        header('HTTP/1.1 500 Internal Server Booboo');
        header('Content-Type: application/json; charset=UTF-8');
        $result=array();
        $result['messages'] = $errors;
        //feel free to add other information like $result['errorcode']
        die(json_encode($result));
    }
?>