$argv — 传递给脚本的参数数组
包含当运行于命令行下时传递给当前脚本的参数的数组。
Note: 第一个参数总是当前脚本的文件名，因此 $argv[0] 就是脚本文件名。
Note: 这个变量仅在 register_argc_argv 打开时可用。
Example #1 $argv 范例
<?php
var_dump($argv);
?>
当使用这个命令执行：php script.php arg1 arg2 arg3
以上例程的输出类似于：
array(4) {
[0]=>
string(10) "script.php"
[1]=>
string(4) "arg1"
[2]=>
string(4) "arg2"
[3]=>
string(4) "arg3"
}
Jesse (2013-01-30 23:55:00)
If your script is read from standard input or with the -r option, $argv[0] will be "-".
If you use the "--" option to separate PHP's arguments from your script's arguments, $argv[1] will be "--" if your script is read from a file. But if your script is read from standard input or with the -r option, the "--" will be removed.
tufan dot oezduman at googlemail dot com (2011-08-22 09:06:45)
Please note that, $argv and $argc need to be declared global, while trying to access within a class method.
<?php
class A
{
public static function b()
{
var_dump($argv);
var_dump(isset($argv));
}
}
A::b();
?>
will output NULL bool(false) with a notice of "Undefined variable ..."
whereas global $argv fixes that.
Steve Schmitt (2009-09-14 16:57:25)
If you come from a shell scripting background, you might expect to find this topic under the heading "positional parameters".
karsten at typo3 dot org (2009-02-18 00:48:48)
Note: when using CLI $argv (as well as $argc) is always available, regardless of register_argc_argv, as explained at http://docs.php.net/manual/en/features.commandline.php