$.ajax query not posting to PHP - php

Question

I've spent the better part of today trying to send a jQuery variable to my PHP file and use it in a SQL search. My first issue is that can't even get the variable into PHP properly. I am also getting an error about some of my SQL code which I've included as well.
What I want: the variable in jQuery called name sent to the PHP file where it is used in an SQL Query to find the real name of musicians. the real name would then be sent back to the HTML file where it is displayed on the webpage. MyPHPAdmin and my database has already been setup.
jQuery:
$('li').click(function () {
var name = $(this).text();
$('#prf').attr("class",name);
$('#pic').attr("class",name);
$('#info').attr("class",name);
JSON.stringify (name);
$.ajax({
type: "POST",
url: "ajax-name.php",
data: {name: name}
})
PHP:
$rname = $_POST['name'];
var_dump($rname);
try {
$db = new PDO('mysql:dbname=dbname;host=myhost;charset=utf8', 'user', 'pass');
$db->setAttribute(PDO::ATTR_EMULATE_PREPARES, false);
$db->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
$query="SELECT realname FROM artistmaster WHERE stagename = :name";
$stmt = $db->prepare($query);
$stmt->execute(array("name"=>$rname));
var_dump($rname);
echo 'Success';
$result=$stmt->fetchAll();
print "<pre>";
print_r($result);
print "</pre>";
}
catch(PDOExeception $e)
{
echo $e->getMessage();
}
OUTPUT:
NULL NULL
I've read the documentation on $.ajax() and still can't figure out what I'm doing wrong. I appreciate any help on this, it's been giving one hell of time.

Answer 1

To evaluate the output of the PHP script from the AJAX request add a return handler to your request:
$.ajax({
type: "POST",
url: "ajax-name.php",
data: {name: name}
}).always(function( html ) {
console.log(html); // this will show the content of the php script in the developer console of your browser
});
The function defined in .always() is called every time an AJAX request finishes. When you made sure that your code works you can change it to .done(), which is only called when the AJAX call is successfull. You can also add a .failure() function to handle errors gracefully.
Also note this from the documentation:
The $.ajax() function underlies all Ajax requests sent by jQuery. It is often unnecessary to directly call this function, as several higher-level alternatives like $.get() and .load() are available and are easier to use.
I'd suggest to use the $.post function instead:
$.post("ajax-name.php", {name: name}, function( html ) {
$('#outputdiv').html(html);
// this will replace the content of a <div id="outputdiv"></div> with the output of the AJAX request
});

Answer 2

Try this:
$.ajax({
type: "POST",
url: "ajax-name.php",
data: {name: name},
success : function(html) {
// html code, do something with it here
}
});
then
$rname = $_POST['name'];
$query="SELECT realname FROM artistmaster WHERE stagename = :name";
$stmt = $db->prepare($query);
$stmt->execute(array("name"=>$rname));
You get your result with
$stmt->fetchAll();
You don't retrieve PDO driver results with mysqli

Answer 3

alert variable name in javascript and make sure it is not null.you may use exit;
echo $_POST['name'] in the php file.receive the processed data in javascript with the success:function(result) and alert it.
store every data in a single variable in the php file. and finally echo them.