Cant get value back from PHP - php

Question

JavaScript
$.ajax({
type : 'POST',
url : 'post.php',
data: dataString,
success:function (data) {
if (data==null) { alert("darnit!!!!");}
//$("#response").append(data);
alert(dataString);
}
});
});
in PHP file just a simple
print_r($_REQUEST);
Also tried
echo "got iT!";
But nothing, been looking of over tried differant things but no luck
first //alert (dataString); works
but after the success:function (data) I don't get any alerts - no response within the page!
What am I doing wrong?

Answer 1

There's a SyntaxError in your snippet. I'm not sure if that's also in your real code.
Be sure to use json_encode in your PHP file and dataType: 'json' in jQuery.ajax. And always use an error callback as well. You don't want your application to become indefinitely frozen if something fails.
Something like this:
$.ajax({
url: 'api.php',
data: {
action: 'greet',
foo: 'bar',
baz: 'quux'
},
type: 'POST',
dataType: 'json',
}).then(function (response) {
console.log(response); // DEBUG
if (response.error) {
alert('Greet Error: ' + response.error);
} else {
alert(response.greet);
}
}).catch(function (jqXHR) {
console.log('AJAX Error', jqXHR); // DEBUG
alert('AJAX Error: Request failed');
});
PHP:
<?php
$input = $_POST;
$response = array();
if (!isset($input['action'])) {
$response['error'] = 'Action parameter required';
} else {
if ($input['action'] === 'greet') {
if (!isset($input['foo']) || !isset($input['bar'])) {
$response['error'] = 'Invalid greet request';
} else {
$response['greet'] = 'Welcome home, David!';
}
} else {
$response['error'] = 'Unknown action';
}
}
header('Content-Type: application/json; charset=utf8');
echo json_encode($response);

Answer 2

First: you missed dataType property from ajax function, next, you need to pass json type data, and you had a syntax error in code, try this:
$.ajax({
type : 'POST',
url : 'post.php',
dataType: 'text',
data: {
data_to_pass: dataString
},
success:function (data) {
if (data==null) {
alert("darnit!!!!");
} else {
//$("#response").append(data);
alert(dataString);
}
});
});
in PHP:
$dataString = $_POST['data_to_pass'];
if($dataString) {
echo "got IT!";
}

Answer 3

JQuery ignoring your data return usually means it doesn't understand the format that's being returned, or doesn't know what to expect. Set the dataType to an acceptable format and also double check that your PHP script is actually sending something back to the page in Firebug's console.
$.ajax({
type : 'POST',
url : 'post.php',
data: dataString,
dataType: 'text',
success:function (data) {
if (data==null) { alert("darnit!!!!");}
//$("#response").append(data);
alert(dataString);
}
});
});

Answer 4

I would like share the following, and I think I've solved the problem.
Besides the fact that I did made a mistake, retrieving the values from the form a bit wrong.
so the first alert gave me name=[Object name], now that was stupid.
The problem of not getting results seemed to be a problem with jquery itself, in my case.
I do not know if it is the same problem other people are having. I replaced the included file of jquery 1.7 with 1.4.
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.4/jquery.min.js"></script>
than with the following code (only from $.ajax).
var dataString = $("#contactForm").serialize();
$.ajax({
type: 'POST',
url: './php/mail.php',
data: dataString,
success: function(msg){
if (msg === 'sent'){
$('#success').text('Message sent!');
} else{
$('#success').text('Mail Error. Please Try Again.!');
}
}
});
Now i have it working - and gonna tweak it a bit to my needs!
Thx all for all help!

Answer 5

Try this:
$.ajax({
type : 'POST',
url : 'post.php',
data: dataString,
success:function (data) {
if (data===undefined) { alert("darnit!!!!");}
//$("#response").append(data);
alert(data);
}
});
});
Also I would look into using json_encode on an array in your php file and returning a json object.