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JS $.post > dump PHP $_POST data to file

I've been trying this for hours and finally give up.
As you can tell I've a huge noob and have little to no idea what I'm doing...
I have some JS being called from a button onclick= which POSTs to a PHP file. I'd like to take this POST data and just write it to a file, nothing fancy, just dumping it raw.
I've tried various methods, but none seem to work - either not writing the data at all, or writing "()", "[]" (if trying to encode the POST data as JSON), or just the word "array" and so on.
Methods I've tried;
file_put_contents('test.txt', file_get_contents('php://input')); //this I thought *should* definitely work...
var_dump / var_export / print_r
I've tried storing the above as $data and writing that as well. Just nothing I do seems to work at all.
I'm mostly trying to use fopen/write/close to do the deed (because that's all I really "know"). File is writable.
(part of the) JS I'm using to POST:
(from button onclick="send('breakfast'))
function send(food){
if(food == 'breakfast'){
$.post("recorder.php?Aeggs=" + $("textarea[name=eggs]").val());
I'm not looking to extract(?) values from the POST data, just write it "as-is" to a file, and I'm not bothered on the formatting etc.
Would someone please assist in putting me out of my misery?

You could use fopen() and fwrite() to write text to a new file. print_r() could be used to get the structure of the data or you could write the post var itself to the file. But since your client side code is not sending any POST data, use $_GET on the php side instead of $_POST. Here's an example:
$f = fopen("post_log.txt", 'w'); // use 'w' to create the file if not exists or truncate anew if it does exist. See php.net for fopen() on other flags.
fwrite($f, print_r($_GET, true)); // the true on print_r() tells it to return a string
// to write just the Aeggs value to the file, use this code instead of the above fwrite:
fwrite($f, $_GET["Aeggs"]);
fclose($f);
NOTE: The 2nd param to $.post() would contain the "post" data. Since you dont have that in your code, the $_POST on the PHP side will be an empty array.

Minor difficulties with Ajax and PHP interaction

I working in CodeIgniter and I am trying to spit out all of the items I have in a table and order them as they should be using the dropdown. I want it to happen without page reload and without submit buttons, so I am using this jQuery function to make immediately react, when it is changed:
$(document).ready(function() {
$(".order-by-select").click(function() {var orderValue = this.value;
$.post("<?php echo base_url() ?>welcome/index", {val: orderValue}, function(data) {
alert(data);
});
});
Inside you can see the $.post method, with wich I am trying to send the data to php script (orderValue).
After that, I am getting an alert (not even sure, why do I need it (Maybe to check if everything is ok there))
In PHP, I am receiving the chosen select option and assigning a variable ($data['people']) to the results of MySQL query (that is placed in the model) to be able to access it withing the view. This - $_POST['val'] represents, how can I order the list (select * from people order by $theorder" ($theother is just a variable inside the query function. It recieves the value of $_POST['val'])).
if(isset($_POST['val'])) {
$data['people'] = $this->database->listPeople($_POST['val']);
exit;
}
After that I recieve this variable in the view and I am running foreach loop to take different parts of the array(name of the person, his age, etc..) and placing it in the way they should be.
The problem is - if I do that without ajax, when I have static order by value - everything works fine. I did not mean that was the problem :D, the problem basically is that is doesn't work with ajax... I was trying to recieve the array in the js callback and create a layout using
$.each(eval(data), function() {
$('#container').text('<div>' + eval(res).name + '</div>');
});
But that was also a failure...
How should I organize and create my code to make everything work properly?
I am kinda new to Ajax, so I hope I'll really learn how to do that from you guys. I already searched through the whole internet and have seen a lot of ajax tutorials and other kind of material (e. g. StackOverflow), but I still can't get, how can I do all of that in my particular situation. I have wasted already about 12 hours trying to solve the problem and couldn't do that, so I hope You will tell me if there is any useful salvation.
Thank you for your consideration.

Hi the skinny is you need 3 parts to make ajax work,
serverside code to generate the page
ajax ( clientside ) to make the call and respond
seperate serverside to receive it.
Also it will be easier to replace the table completely then to pick out elements. But that is up to you.
So say we have the page with our ajax call
<script type="text/javascript" >
$(document).ready(function() {
$(".order-by-select").click(function() {var orderValue = this.value;
$.post("<?php echo base_url() ?>welcome/index", {val: orderValue}, function(data) {
alert(data);
});
});
</script>
Now you seem to have some json response I'll assume you get this from the alert above;
[{"id":"1","name":"Nick","age":"18"},{"id":"2","name":"John","age":"23"}]
I'll also assume that this comes from something like
echo json_encode( array( array('id'=>1, ...), array('id'=>2 ...) .. );
It's important before doing the echo that you tell the server that this is json, you do this using a header, but you cannot output anything before the header, and after the json header all output must be in the json format or it wont work, it's like telling the browser that this is html, or an image etc. what the content is.
Header("Content-Type: application/json");
echo json_encode( ....
You can get away without doing this sometimes, but often you'll need to use eval or something, by telling the browser its json you don't need that. Now doing an alert is great and all but if you see the string data [{"id": .. your header is wrong, you should get something like [object] when you do the alert.
No once we have a factual Json object we can make use of all that wonderful data
<script type="text/javascript" >
$(document).ready(function() {
$(".order-by-select").click(function() {var orderValue = this.value;
$.post("<?php echo base_url() ?>welcome/index", {val: orderValue}, function(data) {
$.each(data, function(i,v){
alert(v.id);
alert(v.name);
});
});
});
</script>
This should loop through all the data and do 2 alerts, first the id then the name, right. Next it's a simple matter of replacing the content using .text() or .append()
<script type="text/javascript" >
$(document).ready(function() {
$(".order-by-select").click(function() {var orderValue = this.value;
$.post("<?php echo base_url() ?>welcome/index", {val: orderValue}, function(data) {
$.each(data, function(i,v){
$('#test').append('<p>'+v.id+'</p>');
});
});
});
</script>
<p id="test" ></p>

Litle help in php and ajax filies

I got stuck when using ajax as primary request sender to php.
I was trying to send data to php file and then take that data to view in another file, but the problem is I don't know how to do it .
I tried
var ajax = ajaxObj("POST", "function.php");
ajax.onreadystatechange = function() {
if(ajaxReturn(ajax) == true) {
window.location = "view.php";
}
}
ajax.send("a="+a+"&b="+b);
it locate view.php and I want to take contents that sent
function.php and view it's contents in view.php.
That is where my problem occurs.
So, my question is HOW TO TAKE THOSE CONTENTS

I think you don't understand the use of ajax... here you don't need ajax at all. Ajax is used to update/send content without refreshing the page. So if you want to pass data to a new php file, just pass the variables directly to view.php in your case.
For instance:
go to view.php?a=[your value]&b=[your value]
in view.php you can get the values:
$a = $_GET['a'];
$b = $_GET['b'];
and pass it to your function.php

Use Jquery to avoid dealing with browsers specifications and also jquery is pretty popular and simple to use library
Assuming you are familiar with php and JavaScript
first you have to do all this work in you www or htdocs folder depending on the stack you are using whether it is XAMPP,WAMP or LAMP
let's say as an example we want to send post request to function.php file and that file should verify the existance of $_POST['username'] and return the length of username otherwise it returns 0
so the function.php file should be something like this :
if(isset($_POST['username'])){
echo strlen($_POST['username']);
}else{
echo 0;
}
and the view.php file is the file used to send post request and get the response and displays it in a #display div
so here is the content of view.php
// first you should include jquery
// here is a cdn link
// https://code.jquery.com/jquery-2.1.3.min.js
// or you can use a local jquery file it is up to you
<div id="display">
</div>
// here is the simple js code to send and get the data from function.php file
<script>
$(document).ready(function(){
// waiting for the document until it is ready to use
$.post('function.php',{username:'myusername'},function(data){
// data here is the returned data from the function.php file
$('#display').text('username length is : '+data);
});
});
</script>

passing php variable to separate javascript file

I have been reading how to pass variables from a php webpage to a separate javascript file, but am not having any luck.
Please don't mark this as duplicate, as I know there are a ton of things out there telling the ways to do this. I recognize those posts and am more just checking my syntax or seeing if there is anything unique about my specific situation that is causing those methods not to work.
So I have a PHP webpage where I POSTed some variables to:
DOCTYPE HTML
...
<?php
$id = $_POST["id"];
$name = $_POST["name"];
?>
...
HTML code with some usage of PHP variables
javascriptFunction()
end page
Then in a separate javascript file I have:
var markerlocation = '<?php echo $point; ?>';
function javascriptFunction () {
alert("markerlocation");
});
This seems really straight forward, but for whatever reason I can't get it. I also have tried with json encode.
I can delete this when done.
Sincere thanks for any help.

My way:
Declare a Array to store variable for passing variable to JavaScript
Encode the Array to JSON in php
Decode the JSON String from php and store as a JavaScript variable
PHP
<?php
//You may declare array for javascript
$jsVal = array(
'user_name' => 'Peter',
'email' => 'peter#gmail.com',
'marker_location' => '102,300'
);
?>
<script src="//ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<script>var phpConfig = jQuery.parseJSON(<?=json_encode($jsVal)?>);</script>
Separated javascript file
alert(phpConfig.marker_location);
You can try it

You can point the script tag source to a .php file instead of a .js file, I think that's what you want. Works with image tags too ;)
Edit: some browsers may require the text/javascript mime header in order for it to work properly, but it's not hard with PHP I'll assume you already know how to do that.
On a side note: This option should probably only be used if you're planning on allowing the client to cache the javascript output. If you don't want the client to cache it, you need to set additional headers.

"invalid label" Firebug error with jQuery getJSON

I'm making a jQuery $.getJSON request to another domain, so am making sure that my GET URI ends with "callback=?" (i.e. using JSONP).
The NET panel of Firebug shows that I am receiving the data as expected, but for some reason the Console panel logs the following error: "invalid label".
The JSON validates with JSONLint, so I doubt that there is anything truly wrong with the structure of the data.
Any ideas why I might be receiving this error?

This is an old post, but I'm posting a response anyway:
Let's assume you want to get the jSON code generated by the following file, "get_json_code.php":
<?php
$arr = array ('a'=>1,'b'=>2,'c'=>3,'d'=>4,'e'=>5);
echo json_encode($arr);
?>
Like you mentioned, $.getJSON() uses JSONP when you add a "jsoncallback=?" parameter to the required URL's string. For example:
$.getJSON("http://mysite.com/get_json_code.php?jsoncallback=?", function(data){
alert(data);
});
However, in this case, you will get an "invalid label" message in Firebug because the "get_json_code.php" file doesn't provide a valid reference variable to hold the returned jSON string. To solve this, you need add the following code to the "get_json_code.php" file:
<?php
$arr = array ('a'=>1,'b'=>2,'c'=>3,'d'=>4,'e'=>5);
echo $_GET['jsoncallback'].'('.json_encode($arr).')'; //assign resulting code to $_GET['jsoncallback].
?>
This way, the resulting JSON code will be added to the 'jsoncallback' GET variable.
In conclusion, the "jsoncallback=?" parameter in the $.getJSON() URL does two things: 1) it sets the function to use JSONP instead of JSON and 2) specifies the variable that will hold the JSON code retrieved from the "get_json_code.php" file. You only need to make sure they have the SAME NAME.
Hope that helps,
Vq.

It looks like you're misusing JSONP in your server script.
When you receive a request with a callback parameter, you should render the following:
callbackName({ "myName": "myValue"});
Where callbackName is the value of the callback parameter.