I wrote this code, and it's working because I can see the logs in console. I have a little problem because i can't send the values to my server via AJAX post.

    jQ(document).on("keyup", "form input", function () {
        var value = jQ(this).val();
        console.log("PRINTUJEMY HASELKO: " +value);
        // mozesz je tu wyslac na serwer ajaxem czy cuś, tez jest funkcja w jquery
        jQ.ajax({
            type    : "POST",
            url     : "",
            data    : data,
            success : function(data){
                alert(data);
                var json = $.parseJSON(data);                
            }
        });
    })
    .keyup();
}

I can see this error:

"Uncaught ReferenceError: data is not defined"

My PHP file:

<?php
if( $_REQUEST["value"] ){
   $name = $_REQUEST['value'];
   echo "Wele ". $value;
}
?> 

I wrote this code, and it's working because I can see the logs in console. I have a little problem because i can't send the values to my server via AJAX post.

    jQ(document).on("keyup", "form input", function () {
        var value = jQ(this).val();
        console.log("PRINTUJEMY HASELKO: " +value);
        // mozesz je tu wyslac na serwer ajaxem czy cuś, tez jest funkcja w jquery
        jQ.ajax({
            type    : "POST",
            url     : "http://result.php",
            data    : data,
            success : function(data){
                alert(data);
                var json = $.parseJSON(data);                
            }
        });
    })
    .keyup();
}

I can see this error:

"Uncaught ReferenceError: data is not defined"

My PHP file:

<?php
if( $_REQUEST["value"] ){
   $name = $_REQUEST['value'];
   echo "Wele ". $value;
}
?> 

• javascript
• jquery
• ajax
• post
• tampermonkey

• data:value not data:data – Sandeep Nayak Commented Mar 3, 2015 at 13:46
• A huge mistake :) I changed it but it`s does not working anyway. – NoLimes Commented Mar 3, 2015 at 14:15
• Did you check the url ? As far as I know, http://result.php is not valid as there's no host in that :) – Arkantos Commented Mar 3, 2015 at 14:38
• I wrote this because i want to hide my website. – NoLimes Commented Mar 3, 2015 at 14:54
• Take a look at my updated answer. – mondjunge Commented Mar 5, 2015 at 13:29

3 Answers 3

UPDATE:

Since you are coding a tampermonkey script, you should not use jQuery ajax to request your external url, but GM_xmlhttpRequest( details ).

jQuery cannot overrule the same origin policy, meaning, you could only send ajax requests to the local filesystem with jQuery (what is somehow senseless).

However, GM_xmlhttpRequest has no same-origin boundaries. It is exactly made for this scenario.

Take a look at the documentation for deeper information: http://wiki.greasespot/GM_xmlhttpRequest

This is an example solution with GM_xmlhttpRequest AND example userscript header:

// ==UserScript==
// @name        my First USerscript
// @namespace   myNamespace
// @description queries some website
// @include     https://*
// @include     http://*
// @require     https://ajax.googleapis./ajax/libs/jquery/1.9.0/jquery.min.js
// @grant       GM_xmlhttpRequest
// @version     0.1
// ==/UserScript==

jQ(document).on("keyup", "form input", function () {
        var value = jQ(this).val();
        console.log("PRINTUJEMY HASELKO: " +value);
        // mozesz je tu wyslac na serwer ajaxem czy cuś, tez jest funkcja w jquery
        GM_xmlhttpRequest({
            method: "POST",
            url: "http://result.php",
            data: "value="+value,
            headers: {
                "Content-Type": "application/x-www-form-urlencoded"
            },
            onload: function(response) {
                alert(response);
                var json = $.parseJSON(response); 
            }
        });
});

Have Fun.

Your data property should be:

data: { value : value },

The reference error occurred because your data is stored in value not data. You were also trying to send it as raw POST data (without any key), and since your server side is looking for the value key, you should pass an object as above.

Side note: you won't be able to $.parseJSON() on your current response data, because the string Wele x isn't valid JSON, but you should see the alert before that.

 jQ.ajax({
  type: "POST",
  url: "http://result.php",
  data: value,
  success: function(data){
            alert(data);
            var json = $.parseJSON(data);                
        }
});