PHP jQuery AJAX - php

Question

Trying to pass data to the server but it keeps returning a "Parameter Missing"
So either the data is not being passed to the PHP script or I am doing something wrong.
Here is the jQuery:
function quickJob(obj) {
var quickJobNumber = $(obj).text();
//alert(quickJobNumber)
$.ajax({
type: "GET",
url: "quickJobCB.php",
data: quickJobNumber,
success: function(server_response)
{
$("#message").removeClass().html(server_response);
}
});
}
Ok....when tracing the issue I created an alert as seen below. The alert is producing the expected results.
Here is the PHP script:
<?php
require_once("models/config.php");
// Make the connection:
$dbc = #mysqli_connect($db_host, $db_user, $db_pass, $db_name);
if (!$dbc) {
trigger_error('Could not connect to MySQL: ' . mysqli_connect_error());
}
if (isset($_GET['quickJobNumber'])) {
$quickJobNumber = trim($_GET['quickJobNumber']);
$quickJobNumber = mysqli_real_escape_string($dbc, $quickJobNumber);
$query = "SELECT * FROM projects WHERE projectNumber = '" . $quickJobNumber . "'";
$result = mysqli_query($dbc, $query);
if ($result) {
if (mysqli_affected_rows($dbc) != 0) {
while ($row = mysqli_fetch_array($result, MYSQLI_ASSOC)) {
echo $row['projectName'];
}
} else {
echo 'No Results for :"' . $_GET['quickJobNumber'] . '"';
}
}
} else {
echo 'Parameter Missing';
}
?>
<?php include("models/clean_up.php"); ?>

Answer 1

data: quickJobNumber,
should be
data: { 'quickJobNumber': quickJobNumber },

Answer 2

You'll need to pass the data either as a query string like so
data: "quickJobNumber="+quickJobNumber,
or a map like so
data: data { quickJobNumber: quickJobNumber },

Answer 3

If you want to use the GET request, use $.get
$.get("/get_request.php", { quickJobNumber: "myAjaxTestMessage"},
function(data){
console.log("WOW! Server was answer: " + data);
});
In php
<?php
if(isset($_GET['quickJobNumber'])){
header('Content-Type: application/json; charset=utf-8');
echo json_encode(array('answer'=>'Hello user!'));
}
?>
If you want to use the POST request, use $.post
$.post("/post_request.php", { quickJobNumber: "myAjaxTestMessage"},
function(data){
console.log("WOW! Server was answer: " + data);
});
In php
<?php
if(isset($_POST['quickJobNumber'])){
header('Content-Type: application/json; charset=utf-8');
echo json_encode(array('answer'=>'Hello user!'));
}
?>
P.S. or you can use $_REQUEST in php.