从页面运行带有变量的PHP脚本

Question

So, I am trying to run a .php script with javascript and some html without the user being able to see any code and passing variables. I tried both CURL and command line:

//curl_setopt($req, CURLOPT_URL,"http://projectvintous.com/Ginav363534.php?activate=10297&to=$Itemname&Shirt=$filename&totype=shop");
//curl_exec($req);

$command = "../../Ginav363534.php 10297 $Itemname $filename shirt shop";
exec( "$ php -f $command");

But I couldn't get either to work. Not getting any error or anything, it is just not running the script because none of the mysqli or file management things are happening. I am trying to get it to render a webgl model, save as an image using the image the user uploaded as a texture, then upload it to the server and save it in a folder and mysqli table.

This is how I am getting the variables on the other side:

$varz = var_dump($argv);

$activate = $varz[1];
$itemname = $varz[2];
$filename = $varz[3];
$filetype = $varz[4];
$type = $varz[5];

I also tried

parse_str(implode('&', array_slice($argv, 1)), $_GET);

before with no luck as well. Is there another way to do this, am I missing something? It works when I run the url directly on the page, so I know that it works.

Answer 1

You're trying to curl a page and have the Javascript on the page executed? I don't think curl bothers to execute any JS on the pages, it just gets the raw file contents. I think you'll need to do something like use Node.js and PhantomJS to open the page in a headless browser, which will execute the JS on the page.