Cannot pass values into a php function

Question

I am having trouble passing in a value or array of values into my php script. When the submit button on the page is clicked, the following function executes:

$('#rangeSubmit').click( function (e) {
        e.preventDefault();
        $.ajax({
            type: 'POST',
            url: 'http://.../test.php',
            data: {test1:"DONE"},
            success: function(data)
            {
                $("#myDiv").html(data);
            }
    });
});

The PHP script looks like this:

<?php

    echo $argv[1]; //should return the first item in the array of values passed to the php function?
    var_dump($_SERVER['argv']);

?>  

And this script just returns "NULL" instead of the value I sent to the script. Any ideas?

Answer 1

$argv[1] is a reference to parameter passed from command line, like:

php page.php argument

This call page.php with "argument" as parameter than can acessible with $argv[1].

To get post variables use $_POST[$name]. In your case use:

echo $_POST['test1'];

Answer 2

try ;)

<?php
    echo $_POST['test1'];    
    var_dump($_POST);
?> 

Answer 3

use $_POST['test1']; in test.php to receive value passed through jQuery.

Answer 4

Dirty but may help, If your PHP install allows it, you could simply try using exec(). In your receiving script, extract the POST arguments and format them into a command string then pass that to exec() to run.

$filein =$_POST[$arg1];
$fileout =$_POST[$arg2];

$return=exec("php target.php -i $filein -o $fileout ");

Answer 5

USE POST

type: 'POST', --->POST REQUEST

var_dump($_POST['test1']);