为什么jquery .fail没有收到故意触发错误的参数

Question

I am trying to use custom json encoded data after firing an error on server (throwing error intentionally) under jquery ajax request.

I don´t understand what is wrong, because the header is right and the data arrives. But it seems that the exception arguments are not passed to the exception callback in javascript.

The php code

<?php

 if ( valid_input() )
 {
    ...      
 } 
 else // The validation goes wrong so, I have to return error
 {
   // Error code
   header('HTTP/1.1 500 Format does not match');
   header('Content-Type: application/json');    
   die(json_encode(array('message' => 'Bad format', 'code' =>1000)));
 }
?>

The javascript code

$("#uploadbutton").on('click',function(e)
{
    $.ajax({
        url:"http://www.oneurl.com/upload.php",
        async:true,
        type:'post',
        data:formdata
    })
    .done(render_data)
    .fail(upload_error);
});

//
function render_data(params) {...}

function upload_error(jqxhr,textstatus,ex)
{
 alert("Error:-> ", ex.message);   
}

Response from server

Status Code: 500 Format does not match
Content-Type: application/json
Response: {"message":"Bad Format","code":1000}

But the alert is undefined instead Bad Format. What is wrong ?¿

Answer 1

The response is stored as the responseText property of the jqXHR parameter. It comes back as a JSON string, so you'll need to JSON.parse it before you can directly access the properties.

function upload_error(jqxhr,textstatus,ex)
{
    var response = JSON.parse(jqxhr.responseText);
    alert("Error:-> ", response.message);   
}