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up vote 1 down vote favorite

I am passing a variable from javascript to another PHP page via ajax. I am not able to do it. This is the code I have so far. 

<script>
function loadXMLDoc()
        {
        $.ajax({
                    type: "POST",
                    url: "update.php",
                    data: { value : masterdata },
                    success: function(data)
                    {
                        alert("success!");
                    }
                }); 

}
$("button").on('click',function(){ loadXMLDoc(); });
</script>

The masterdata variable comes from another javascript function in which I have declared the variable globally, as in this below function. 

<script>

    var masterdata;      
    $("[data-slider]")

       .each(function () {
           var range;
            var input = $(this);
            $("<span>").addClass("output")
                .insertAfter(input);
            range = input.data("slider-range").split(",");
            $("<span>").addClass("range")
                .html(range[0])
                .insertBefore(input);
            $("<span>").addClass("range")
                .html(range[1])
                .insertAfter(input);
        })
        .bind("slider:ready slider:changed", function (event, data) {
            $(this).nextAll(".output:first")
                .html(data.value.toFixed(2));
                masterdata = data.value;

        });


</script>

In my update.php file, I am trying to access the variable using $_REQUEST. The code is as below. 

<?php

    $uid = $_REQUEST['value'];
    echo "Am I getting printed";
    echo $uid;
    // Do whatever you want with the $uid
?>

However, if I click on Update button nothing happens. I just get an alert saying Success for my ajax call. Can someone please help me out?

share|improve this question
 
And what do you expect? –  u_mulder Oct 12 at 19:24
 
Try: "echo $HTTP_RAW_POST_DATA;" in your PHP to see what POST data you get. This may help you debug. –  user2759991 Oct 12 at 19:25
 
I am expecting to print the $uid value in my update.php file, which am calling on the click of Update button. –  ram esh Oct 12 at 19:25
 
Do you understand how ajax request works? –  u_mulder Oct 12 at 19:26
2  
But the only thing that happens when your AJAX comes back is that you alert "success". Try "alert(data);" as an example, or otherwise use the data variable in your success function. This is the response from the server. –  user2759991 Oct 12 at 19:27
show 3 more comments

2 Answers

up vote 0 down vote

in your success function, you're not asking it to display the returned data that's why you are only seeing the alert, change it to:

success: function(data)
   {
     alert(data);
   }

because data here is the values returned from your request

and for the php part, keep in mind one thing, just adjust your php page to display the data you need, and remember it doesn't matter how you make this data, with echo or just plain html, the result of the php code is what you will get in your variable in ajax. and if you go JSON, you can do in the php page:

$uid = $_REQUEST['value'];
$x.uid = $uid;
$x.something = "something";
return $x;

this way you will have all the data you need on client side as an object and then you can manipulate it the way you like, in a much more compact way, just search for RESTful api if you want to read more about the topic

share|improve this answer
 
Yeah. I am getting the response in the alert. But, how can I make it display in a separate PHP file? For example, I am calling the update.php file through ajax request. I am sending the javascript variable through ajax request to the update.php file. I need to display the update.php file rather than displaying the alert(data). Is it possible? –  ram esh Oct 12 at 19:40
 
see the update please –  Labib Ismaiel Oct 12 at 19:42
 
success: function(data) { alert(data); } In the above function, instead of alert(data) I need to display the contents of PHP file. I do not want an alert window but rather I need to display the tables that I have in my update.php file. Is it possible using ajax? –  ram esh Oct 12 at 19:45
 
I would suggest that you return the values from your update.php file as a JSON-encoded array. After that, you can pass the values array back into your AJAX success function, and add the necessary markup to display your results visually. –  Terry Oct 12 at 19:47
 
check the update again –  Labib Ismaiel Oct 12 at 19:51
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up vote 0 down vote

I would recommend JSON-encoding the output you have obtained from your update.php script, and pass it back to the AJAX function. In your update.php file, you can simply use the PHP function json_encode($var), and echo it:

<?php
    $uid = $_REQUEST['value'];
    echo json_encode($uid);
?>

From your AJAX function, you should specify the dataType (usually JS will intelligently guess the type of data returned, but we should always specify it just in case):

function loadXMLDoc() {
    $.ajax({
        type: "POST",
        url: "update.php",
        data: { value : masterdata },
        dataType: "json",
        success: function(data) {
           // Log data into console for reference (and checking, perhaps)
           console.log(data);

           // Parse data here
        }
    }); 
}
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