php - HipHop PHP -- Passing Arguments to Compiled PHP Executable

Question

I'm trying to understand the HipHop documentation, but it's not very clear. I have a PHP script I want to compile into an executable with HipHop.

How do I pass arguments to this PHP executable by command line? And how is this received by the PHP script ($_GET['arg']?)

I want to do something like this: my_compiled_script --variablename="This is the value"

Which is then received by the script as: $_GET['variablename']

So how does that work?

Answer 1

通过一些测试弄清楚了。

正如 mario 上面所说，参数的格式与运行 PHP 时完全相同：它们从 $argv[1] 开始（$argv[0] 为空）。