PHP 5.6.0 released

$argv

$argv传递给脚本的参数数组

说明

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。

范例

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"
}

参见

  • getopt() - 从命令行参数列表中获取选项

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User Contributed Notes 4 notes

up
16
tufan dot oezduman at googlemail dot com
3 years ago
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.
up
9
hamboy75 at example dot com
9 months ago
To use $_GET so you dont need to support both if it could be used from command line and from web browser.

foreach ($argv as $arg) {
    $e=explode("=",$arg);
    if(count($e)==2)
        $_GET[$e[0]]=$e[1];
    else   
        $_GET[$e[0]]=0;
}
up
4
Steve Schmitt
4 years ago
If you come from a shell scripting background, you might expect to find this topic under the heading "positional parameters".
up
-9
Jesse
1 year ago
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.
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