$ _SERVER未使用命令行-r选项设置？

问题描述：

When I try to print the $_SERVER variable from a command line php, it thinks $_SERVER is not set.

$ php -r "print_r($_SERVER);"
Warning: print_r() expects at least 1 parameter, 0 given in Command line code on line 1

However, when it's in a file, running it from the command line has it set

$ cat test.php
<?
print_r($_SERVER);

$ php test.php
Array
(
    [TERM] => xterm
    [SHELL] => /bin/bash
    [SSH_CLIENT] => 192.168.1.101 49319 22
    [SSH_TTY] => /dev/pts/0
...

Why?

当我尝试从命令行php打印$ _SERVER变量时，它认为$ _SERVER未设置。

  $ php -r“print_r（$ _ SERVER）;”
警告：print_r（）需要至少1个参数，0在第1行的命令行代码中给出
  代码>  pre> 
 
 但是，当它在文件中时，从命令行运行它已设置 p> 
 
 
  $ cat test.php \  n＆lt;？
print_r（$ _ SERVER）; 
 
 $ php test.php 
Array 
（
 [TERM] =＆gt; xterm 
 [SHELL] =＆gt; / bin / bash 
 [SSH_CLIENT]  =＆gt; 192.168.1.101 49319 22 
 [SSH_TTY] =＆gt; / dev / pts / 0 
 ... 
  code>  pre> 
 
 为什么？ p>  
  div>

答

You need to escape the $ character on the command line.

php -r "print_r(\$_SERVER);"

Otherwise the shell will think it's a shell variable called _SERVER (which you don't have set to anything) and so what's actually been run is php -r "print_r();", which is why you get the error "print_r() expects at least 1 parameter, 0 given".

答

From the official documentation

You may or may not find any of the following elements in $_SERVER. Note that few, if any, of these will be available (or indeed have any meaning) if running PHP on the command line.

$ _SERVER未使用命令行-r选项设置？

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