How to return a ajax error? - php

Question

I want to show a ajax error after submitting a form. It ends now with 'die'
but what is the best way to handle this? Just write something in this php file in 'script' tags?
if($_POST['postForm'] == 'newsletter'){
$newsletterSubscriber = new NewsletterSubscriber();
$newsletterSubscriber->set('CMS_newsletters_id', 2);
$newsletterSubscriber->set('created', date('Y-m-d H:i:s'));
$newsletterSubscriber->set('firstName', $_POST['voornaam']);
$newsletterSubscriber->set('lastName', $_POST['achternaam']);
$newsletterSubscriber->set('companyName', $_POST['beddrijfsnaam']);
$newsletterSubscriber->set('emailAddress', $_POST['email']);
$newsletterSubscriber->set('subscribed', 1);
$saved = $newsletterSubscriber->save();
die('subscriber added');
}
I tried several solutions I found but I can't get it to work.
Thanks!

Answer 1

All you need to do is create a array and place any parameters you want to pass back into that array, then use json_encode() to turn it into a json string that can be easily processed by javascript
if($_POST['postForm'] == 'newsletter'){
$newsletterSubscriber = new NewsletterSubscriber();
$newsletterSubscriber->set('CMS_newsletters_id', 2);
$newsletterSubscriber->set('created', date('Y-m-d H:i:s'));
$newsletterSubscriber->set('firstName', $_POST['voornaam']);
$newsletterSubscriber->set('lastName', $_POST['achternaam']);
$newsletterSubscriber->set('companyName', $_POST['beddrijfsnaam']);
$newsletterSubscriber->set('emailAddress', $_POST['email']);
$newsletterSubscriber->set('subscribed', 1);
$saved = $newsletterSubscriber->save();
$response = array('error_code'=>0,
'message'=>'subscriber added'
);
echo json_encode($response);
exit;
}
The javascript woudl be something like
$.ajax({
type: "POST",
url: "connection.php",
data: {param1: 'aaa'},
dataType: JSON
})
.done( function(data){
if(data.error_code == 0) {
alert(data.message);
}
});
Note when you use dataType:JSON the browser automatically converts the json string returned to a javascript object so you can address data.error_code and data.message in simple javascript object notation

Answer 2

You can do like:
if($saved) {
die('subscriber added');
} else {
echo "error";
}
and In ajax you can check:
$.ajax({
type: "POST",
url: "savedata.php",
data: form,
cache: false,
success: function(data){
if(data == "error") {
alert("Data has not been saved successfully. Please try again.");
window.location.reload(true);
}
}
});

Answer 3

Did you check jQuery Ajax API? this comes directly from their example.
It says that you can use the .done() .fail and .always() functions
var jqxhr = $.ajax( "example.php" )
.done(function() {
alert( "success" );
})
.fail(function() {
alert( "error" );
})
.always(function() {
alert( "complete" );
});

Answer 4

the best solution is to make you custom json and send it to ajax:
instead of die try:
$message = array('error'=>'subscriber added');
echo json_encode($message);
and in you ajax callback do:
function(success) {
if(success.error) {
//do stuff
}
//do stff
}

Answer 5

Use a json message followed by a error number:
if($saved) {
echo json_encode(array('message'=>'Successfully saved','erno'=>0));
} else {
echo json_encode(array('message'=>'Error on save','erno'=>1));
}
js:
success:function(data) {
if(data.erno == 1) {
alert(data.message)
//do other stuf here
} else {
alert(data.message)//if save was successful
}
}