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

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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 ...
}

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最佳答案

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.

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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);
}
}

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adding to the top answer: here is some sample code from PHP and Jquery:

$("#button").click(function () {
$.ajax({
type: "POST",
url: "handler.php",
data: dataString,


success: function(data) {


if(data.status == "success"){


/* alert("Thank you for subscribing!");*/


$(".title").html("");
$(".message").html(data.message)
.hide().fadeIn(1000, function() {
$(".message").append("");
}).delay(1000).fadeOut("fast");


/*    setTimeout(function() {
window.location.href = "myhome.php";
}, 2500);*/




}
else if(data.status == "error"){
alert("Error on query!");
}








}




});


return false;
}
});

PHP - send custom message / status:

    $response_array['status'] = 'success'; /* match error string in jquery if/else */
$response_array['message'] = 'RFQ Sent!';   /* add custom message */
header('Content-type: application/json');
echo json_encode($response_array);

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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));
}
?>

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I had the same issue. My problem was that my header type wasn't set properly.

I just added this before my json echo

header('Content-type: application/json');

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Server side:

if (mysql_query($query)) {
// ...
}
else {
ajaxError();
}

Client side:

error: function() {
alert("There was an error. Try again please!");
},
success: function(){
alert("Thank you for subscribing!");
}

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...you may also want to check for cross site scripting issues...if your html pages comes from a different domain/port combi then your rest service, your browser may block the call.

Typically, right mouse->inspect on your html page. Then look in the error console for errors like

Access to XMLHttpRequest at '...:8080' from origin '...:8383' has been blocked by CORS policy: No 'Access-Control-Allow-Origin' header is present on the requested resource.