Question

我正在尝试为useradd命令创建简单的函数，并快速提高我糟糕的shell编程技能。

useradd -m -g [initial_group] -G [additional_groups] -s [login_shell] [username]

现在我有点不确定如何使用可选参数进行操作。经过一些谷歌搜索，我认为可能有一个处理，只需要玩代码。

我不确定的一件事是逻辑，我很好奇你们将如何写这篇文章。我相信它会比我可以破解的更好。

以下是我将尝试设置我的函数参数的方法，对于登录shell和初始组，我希望它们具有通用默认值。

arg1 - userName, required
arg2 - loginShell, optional (default: /bin/bash)
arg3 - initGroup, optional (default: users)
arg4 - otherGroups, optional (default: none)

这是关于我如何构建这个问题的一些蹩脚的伪代码。

function addUser( userName, loginShell, initGroup, otherGroups){
// Not how I would go about this but you should get the point
  string bashCmd = "useradd -m -g ";

// Adding the initial user group
  if(initGroup == null){
    bashCmd += "users";
  } else {
    bashCmd += initGrop;
  }

// Adding any additional groups
  if(otherGropus != null){
    bashCmd += " -G " + otherGroups;
  }

  if(loginShell == null){
    bashCmd += " -s /bin/bash " + userName;
  } else {
    bashCmd += " -s " + loginShell + " " + userName;
  }
}

这些是我要去的链接

http://tldp.org/HOWTO/Bash-Prog-Intro-HOWTO-8.html

Passing parameters to a Bash function

How to write a bash script that takes optional input arguments?

Using Functions inside here document

Answer 1

您可能会发现${parameter:+word}扩展很有用。来自Bash Reference Manual：

如果参数为null或未设置，则不替换任何内容，否则将替换 word 的扩展。

所以：

function addUser {
    useradd -m ${2:+-s "$2"} ${3:+-g "$3"} ${4:+-G "$4"} "$1"
}

请注意，如果任何参数包含有趣的字符（如空格，美元符号或其他shell元字符），则此函数会正确处理引用。如果你试图拼凑一个命令字符串，那么正确引用这些部分要困难得多。如果这仅仅是为了您的个人短期使用并且您知道输入是安全的，那可能无关紧要。但是如果要以root身份运行并且不小心处理它的输入，最好不要留下脚本或函数。

Answer 2

@rob mayoff的答案是实现这一目标的最简单方法，但我认为我会把你的伪代码转换成真正的shell语法，为那些习惯于“真正的”编程语言的人指出一些标准问题。首先是三个一般性说明：

不同的shell具有不同的功能，因此如果您需要任何bash扩展，请使用bash（即使用#!/bin/bash启动脚本或使用bash命令运行它）。如果您只使用基本的Bourne shell功能和语法，请使用sh（#!/bin/sh或sh命令）来运行它。如果你不知道，假设你需要bash。
在构建命令以供以后执行时，您可以遇到各种各样的解析奇怪（请参阅BashFAQ#050: I'm trying to put a command in a variable, but the complex cases always fail!）。最好的方法通常是将其构建为数组，而不是字符串。 '当然，数组是bash扩展，而不是基本的shell功能......
在shell语法中，空格重要。例如，在命令if [ -n "$2" ]; then中，分号后面的空格是可选的（分号前面也可能有空格），但是其他空格的所有都是必需的（没有它们）命令会做一些完全不同的事情）。此外，在赋值中，不能是等号周围的空格，或者（再次）它会做一些完全不同的事情。

考虑到这一点，这是我对功能的看法：

addUser() {
# The function keyword is optional and nonstandard, just leave it off. Also,
# shell functions don't declare their arguments, they just parse them later
# as $1, $2, etc

bashCmd=(useradd -m)
# you don't have to declare variable types, just assign to them -- the
# parentheses make this an array. Also, you don't need semicolons at the
# end of a line (only use them if you're putting another command on the
# same line). Also, you don't need quotes around literal strings, because
# everything is a string by default. The only reason you need quotes is to
# prevent/limit unwanted parsing of various shell metacharacters and such.

# Adding the initial user group
if [ -z "$3" ]; then
# [ is actually a command (a synonym for test), so it has some ... parsing
# oddities. The -z operator checks whether a string is empty (zero-length).
# The double-quotes around the string to be tested are required in this case,
# since otherwise if it's zero-length it'll simply vanish. Actually, you
# should almost always have variables in double-quotes to prevent accidental
# extra parsing.
# BTW, since this is a bash script, we could use [[ ]] instead, which has
# somewhat cleaner syntax, but I'm demonstrating the difficult case here.
    bashCmd+=(-g users)
else
    bashCmd+=(-g "$3")
    # Here, double-quotes here are not required, but a good idea in case
    # the third argument happens to contain any shell metacharacters --
    # double-quotes prevent them from being interpreted here. -g doesn't
    # have any shell metacharacters, so putting quotes around it is not
    # necessary (but wouldn't be harmful either).
fi

# Adding any additional groups
if [ -n "$4" ]; then
    bashCmd+=(-G "$4")
fi

# Set the login shell
if [ -z "$2" ]; then
    bashCmd+=(-s /bin/bash "$1")
else
    bashCmd+=(-s "$2" "$1")
fi

# Finally, run the command
"${bashCmd[@]}"
# This is the standard idiom for expanding an array, treating each element
# as a shell word.
}

Answer 3

google for ABS获取许多复杂的样本

function addUser{
userName=$1;
loginShell=$2;
initGroup=$3
otherGroups=$4;

  args=(-m -g);

// Adding the initial user group
  if [[ $initGroup == '' ];then
    args+=(users);
  else 
    args+=("$initGrop");
  fi;

# Adding any additional groups
  if [[ $otherGroups != '' ]];then
    args+=(-G "$otherGroups");
  fi;

  if [[ $loginShell == '' ]];then
    args+=(-s /bin/bash "$userName");
  else
    args+=(-s "$loginShell" "$userName");
  fi;
  useradd "${args[@]}"
}

代码没有检查，但我希望我不会错过任何东西

Bash - 具有可选参数和缺少逻辑的函数

3 个答案: