如何将参数作为GET方法传递给脚本？ [重复]

Question

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• Why file_get_contents() doesn't work without protocol? [duplicate] 1 answer

I have this folder-structure:

\out
    \MakeAvatar.php
oot
    \include
        \Calculator.php
    \img
        \avatar

What's MakeAvatar.php? That's a script which gets a parameter (like id) and makes a avatar based on that parameter. Now I need to pass a argument ($id) from Calculator.php to MakeAvatar.php. How can I do that?

Here is my code:

$_GET['id'] = $id; // passing
file_put_contents("../img/avatar/".$id.".jpg", file_get_contents("../out/MakeAvatar.php"));

But it doesn't work. I mean the result is a unknown-image (unclear).

When I open that image by a editor, it is containing the content of MakeAvatar.php (all its codes). So it seems the problem is passing.

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Note1: If I put MakeAvatar.php into root and pass that argument like this then if works:

... file_get_contents("http://example.com/MakeAvatar.php?id=$id")

But as you see MakeAatar.php is out of root and I cannot use http. So how can I pass an argument without http?

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Answer 1

Your file_get_contents is actually reading the file contents, rather than executing it.

You'll either have to include your file from Calculator.php or expose it to the HTTP server and do a local http request (eg. like file_get_contents('http://localhost/path/to/your/MakeAvatar.php') )

I'd recommend wrapping your important logic in the MakeAvatar file into a function, then includeing it and executing the function.