如果命令在Mac终端中运行，如何让shell_exec起作用?

问题描述：

当我在php脚本中运行命令时

When I am running command inside a php script

echo shell_exec("which php");

我得到以下输出:

/usr/bin/php

但是在Mac终端中运行相同命令时

However when running the same command inside mac terminal

 which php

我得到以下输出

 php: aliased to /Applications/MAMP/bin/php/php5.5.10/bin/php

我的问题是，如果命令在Mac终端中运行，如何让shell_exec起作用?

my question is how to let shell_exec act like if commands are running inside mac terminal?

注意:我已经安装了ZSH

note: that i have ZSH installed

答

简短答案:

这是行不通的(可靠地).

This will not work (reliably).

长答案:

问题有三重:

PHP不使用您的登录Shell，但使用/bin/sh
别名必须在使用上下文中设置
which的输出取决于$PATH

PHP does not use your login shell but /bin/sh
Aliases have to be set in the context they are used
The output of which depends on $PATH

要查看 1.您可以通过回显$0

To see 1. You can print the name of the running shell by echoing $0

% echo $0
zsh
% php -r 'echo shell_exec("echo \$0");'
sh

如您所见，PHP启动sh而不是zsh.这意味着它还使用sh的内置函数，或者在没有内置函数的情况下查找命令:

As you can see, PHP starts sh instead of zsh. That meens it also uses the builtins of sh or looks for a command if there is no builtin:

% php -r 'echo shell_exec("which which");'
/usr/bin/which
% zsh -c 'which which'
which: shell built-in command

除非sh链接到zsh，除非您想使用zsh的内置函数，否则必须使用zsh运行命令:

Unless sh links to zsh, that meens, if you want to make use of zsh's builtins you have to run your command with zsh:

echo shell_exec("zsh -c 'which php'");

这将启动/bin/sh，然后反过来启动zsh，然后运行命令.

This starts /bin/sh, which in turn starts zsh, which then runs the command.

虽然您可以使用sh解决PHP，但是第二个问题更加严重:别名仅在定义它们的实例内设置.大多数情况下，这是在某些配置文件(例如~/.zshrc)中发生的.但是，当非交互使用zsh时，不会加载这些配置文件，别名也不会传递给子进程:

While you can work around PHP using sh, the second problem is more grave: Aliases are only set within the instance in which they are defined. Most of the time this happens in some configuration file (e.g. ~/.zshrc). But these configuration files are not loaded when using zsh non-interactively and neither are aliases passed to child processes:

% grep foo ~/.zshrc
alias foo=bar
% which foo
foo: aliased to bar
% zsh -c 'which foo'
foo not found
% php -r 'echo shell_exec("zsh -c which\ foo");'
foo not found

总而言之，这意味着从PHP脚本内部使用which是找出php起源/位置的完全不可靠的方法.更为重要的是，其输出取决于$PATH，对于交互式和非交互式shell来说，它也可能有所不同.

In conclusion that means, that using which from inside a PHP script is an entirely unreliable way to find out the origin/location of php. Even more so as the output of which depends on $PATH, which may also be different for interactive and non-interactive shells.

如果命令在Mac终端中运行，如何让shell_exec起作用?

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