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

Question

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); ?> 

Answer 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.

Answer 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);     } }