[javascript] Pass Javascript variable to PHP via ajax

I am trying to pass a variable from my javascript code over to the server side PHP code. I know this must be done via an ajax call which i believe i have done correctly, however accessing the variable i pass from my ajax into my php is when i run into trouble as i am new to php. Here is my code i have thus far:

$(document).ready(function() {

            $(".clickable").click(function() {
                var userID = $(this).attr('id');
                //alert($(this).attr('id'));
                $.ajax({
                    type: "POST",
                    url: 'logtime.php',
                    data: "userID=" + userID,
                    success: function(data)
                    {
                        alert("success!");
                    }
                });
            });
        });

<?php //logtime.php
$uid = isset($_POST['userID']);
//rest of code that uses $uid
?>

I'm trying to pass my javascript variable "userID" to php ($userID), however i've gone wrong somewhere along the road. Thanks for the help!

$(document).ready(function() {
            $(".clickable").click(function() {
                var userID = $(this).attr('id'); // you can add here your personal ID
                //alert($(this).attr('id'));
                $.ajax({
                    type: "POST",
                    url: 'logtime.php',
                    data : {
                        action : 'my_action',
                        userID : userID 
                    },
                    success: function(data)
                    {
                        alert("success!");
                        console.log(data);
                    }
                });
            });
        });


 $uid = (isset($_POST['userID'])) ? $_POST['userID'] : 'ID not found';
 echo $uid;

$uid add in your functions

note: if $ is not supperted than add jQuery where $ defined

Since you're not using JSON as the data type no your AJAX call, I would assume that you can't access the value because the PHP you gave will only ever be true or false. isset is a function to check if something exists and has a value, not to get access to the value.

Change your PHP to be:

$uid = (isset($_POST['userID'])) ? $_POST['userID'] : 0;

The above line will check to see if the post variable exists. If it does exist it will set $uid to equal the posted value. If it does not exist then it will set $uid equal to 0.

Later in your code you can check the value of $uid and react accordingly

if($uid==0) {
    echo 'User ID not found';
}

This will make your code more readable and also follow what I consider to be best practices for handling data in PHP.

Alternatively, try removing "data" and making the URL "logtime.php?userID="+userId

I like Brian's answer better, this answer is just because you're trying to use URL parameter syntax in "data" and I wanted to demonstrate where you can use that syntax correctly.

To test if the POST variable has an element called 'userID' you would be better off using array_key_exists .. which actually tests for the existence of the array key not whether its value has been set .. a subtle and probably only semantic difference, but it does improve readability.

and right now your $uid is being set to a boolean value depending whether $__POST['userID'] is set or not ... If I recall from memory you might want to try ...

$uid = (array_key_exists('userID', $_POST)?$_POST['userID']:'guest';

Then you can use an identifiable 'guest' user and render your code that much more readable :)

Another point re isset() even though it is unlikely to apply in this scenario, it's worth remembering if you don't want to get caught out later ... an array element can be legitimately set to NULL ... i.e. it can exist, but be as yet unpopulated, and this could be a valid, acceptable, and testable condition. but :

a = array('one'=>1, 'two'=>null, 'three'=>3);
isset(a['one']) == true
isset(a['two']) == false

array_key_exists(a['one']) == true
array_key_exists(a['two']) == true

Bw sure you know which function you want to use for which purpose.