How to verify receipt of my Ajax string query by the server? - php

Question

I am using Javascript to send form data to a php via Ajax for validation and mailing. The query string which I named formString looks like this 'name= John Smith'. I have one input field only for testing purpose. The Ajax communication between my client page and the server is fine and I checked it successfully with the scripts shown below.
The problem is that I am not able to capture the formString query at the server-side. I am providing below the method I am used to capture the data unsuccessfully. The echo json_encode($name) is returning nothing to the html server.
I tried the query with several input fields values serialized and did not work. I tried to submit the query string a simple string including only the first name 'John', but it did not work either.
processForm()
var name = document.getElementById("fullName").value;
var formString = name;
var name = document.getElementById("fullName").value;
var formString = name;
var xhr = new XMLHttpRequest();
xhr.open('POST', formfile.php, true);
xhr.setRequestHeader('Content-type', 'application/x-www-form-urlencoded');
xhr.setRequestHeader('X-Requested-With', 'XMLHttpRequest');
xhr.onreadystatechange = function () {
if(xhr.readyState == 4 && xhr.status == 200) {
var result = xhr.responseText;
xhr.send(formString);
button.addEventListener("click", function(event) {
event.preventDefault();
processForm();
PHP snippet:
header('Content-Type: application/json');
function is_ajax_request(){
return isset($_SERVER['HTTP_X_REQUESTED_WITH']) &&
$_SERVER['HTTP_X_REQUESTED_WITH'] == 'XMLHttpRequest';
}
if(is_ajax_request()) {
$Ajax_results = array (
'Testing Text' => 'Hello World',
'Your Ajax submittal succeeded.
);
echo json_encode($Ajax_results);
} else {
$None_Ajax_results = array (
'errors' => 'None Ajax - short'
'Your Ajax submittal failed. Errors.'
);
echo "Form is Non Ajax Submitted";
echo json_encode($None_Ajax_error);
exit;
}
Define and set variables:
global $name;
$errors = null;
if (isset($_POST['name'])) { $name = $_POST['name']; }
else { $name = ''; }
echo '$name';
echo json_encode($name);

Answer 1

If I am reading your question correctly and assuming you have proper heartbeat between Ajax and the server as you state you do, taking a quick look at your code as provided you are not properly formatting your "formString". In order for your formString to properly show up in the $_POST['name'] it should be:
var formString = "name="+name
This is because the the post string being sent ("formString" in your case) should have the format:
field1=val1&field2=val2& ... fieldN=valN
where the name of each field is stated, followed by '=' and the value of the field. Multiple fields are separated by the'&' character. Which in PHP which will translate to
$_POST = {field1=>val1, field2=>val2, ... fieldN=>valnN}
on the server side. This is of course not literal code above but an example of the standard API. Take a closer look at how to format Post strings for HTML GET/POST