为什么要采购具有“ set var = value”的脚本？打破$ @？ [英] Why does sourcing a script with "set var = value" break $@?

问题描述

我正在尝试在Centos VM上配置自动注销。我已经注意到，如果我在/etc/profile.d/autologout.sh中创建一个文件，并且该文件中仅 set autologout = 30 ，那么它将破坏任何源于/ etc / profile的脚本。

I am trying to configure auto logout on my Centos VM. I have noticed that if I create a file at /etc/profile.d/autologout.sh with only set autologout = 30 in the file, then it breaks passing arguments for any script that sources /etc/profile.

示例脚本显示如下：

#!/bin/bash 
source /etc/profile 
echo ${@}

当我运行它时，脚本只看到参数 autologout 30，并且没有任何参数，我试图在运行时传递它。

When I run it, the script only sees the arguments "autologout 30" and it doesn't get any parameter I try to pass it when running it.

无论自动注销脚本的名称，属性的名称如何，或者是否我设置了 set autologout 30 ，都会发生这种情况。

This occurs regardless of the name of the autologout script, the name of the property, or if I have set autologout 30 instead.

有人可以解释发生了什么吗？好像autologout.sh劫持了参数。我对正在发生的事情不知所措，研究profile.d和set命令什么也没发现。

Can someone please explain what is happening? It is as if autologout.sh is hijacking the arguments. I am at a loss for what is happening, and researching profile.d and the set command has turned up nothing.

推荐答案

set 不能用于修改POSIX兼容shell中的shell变量。而是在给定位置参数时，它会修改命令行参数列表。

set is not used to modify shell variables in POSIX-compliant shells. Rather, when given positional arguments, it modifies the command-line argument list.

如果只想为变量赋值，请不要使用 set 。相反，您的文件应该只包含：

If you just want to assign a value to a variable, don't use set. Instead, your file should just contain:

autologout=30

...和 $ @ 将保持其原始状态。

...and "$@" will remain in its original state.

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