定义在bash中接受参数的子命令

时间:2016-05-16 15:26:55

标签: linux bash

所以我想做一个"程序"这有助于像yum命令和其他命令...当程序完成后我想把它放在/ usr / bin中,名字是" dafs"

我用这个例子测试了文件名为dafs

#!/bin/bash

$1 $2 $3

function yum {
    function maintenance {
        yum -y update
        yum -y upgrade
        yum clean all
    }

    function download {
        yum -y install --downloadonly $3
    }

}

但是当我运行./dafs yum maintenance./dafs yum download http时,我猜它不起作用,因为语法不正确..

那么,我如何将参数传递给函数或子函数,如上例所示?

2 个答案:

答案 0 :(得分:6)

定义子命令的最佳实践方法是使用带前缀的命名空间和“启动器”功能。例如git就是这样做的(对git-foogit-bar使用git foogit bar命令。

这里,我使用双下划线而不是单个短划线作为分隔符,因为下划线(与破折号不同)在POSIX sh标准中定义为函数名称内的有效。

yum__maintenance() {
  command yum -y update
  command yum -y upgrade
  command yum clean all
}

yum__download() {
  command yum -y install --downloadonly "$@"
}

yum() {
  local cmdname=$1; shift
  if type "yum__$cmdname" >/dev/null 2>&1; then
    "yum__$cmdname" "$@"
  else
    command yum "$cmdname" "$@" # call the **real** yum command
  fi
}

# if the functions above are sourced into an interactive interpreter, the user can
# just call "yum download" or "yum maintenance" with no further code needed.

# if invoked as a script rather than sourced, call function named on argv via the below;
# note that this must be the first operation other than a function definition
# for $_ to successfully distinguish between sourcing and invocation:
[[ $_ != $0 ]] && return

# make sure we actually *did* get passed a valid function name
if declare -f "$1" >/dev/null 2>&1; then
  # invoke that function, passing arguments through
  "$@" # same as "$1" "$2" "$3" ... for full argument list
else
  echo "Function $1 not recognized" >&2
  exit 1
fi

注意事项:

  • "$@"扩展为传递给范围中当前项的参数的完整列表,保留参数边界并避免全局扩展(与$*和不带引号的$@不同)。
  • shift弹出列表前面的第一个参数($1),使新值"$@"比旧列表短一些。
  • command内置函数会导致调用真实yum命令,而不是仅在没有子命令时再次递归yum函数。
  • 如果真的传递了函数,
  • declare -f funcname将返回true(并打印该函数的定义)。相反,type如果传递任何类型的runnable命令,则返回true。因此,使用type "yum__$cmdname"允许将yum__foo定义为外部脚本或任何其他类型的命令,而不仅仅是函数,而稍后完成的declare -f "$1"仅允许运行函数。 / LI>

最后要考虑的是,如果你不打算支持被采购,那么将遗漏yum函数,但扩展你的启动器以识别子命令:

if declare -f "${1}__$2" >/dev/null; then
  func="${1}__$2"
  shift; shift    # pop $1 and $2 off the argument list
  "$func" "$@"    # invoke our named function w/ all remaining arguments
elif declare -f "$1" >/dev/null 2>&1; then
  "$@"
else
  echo "Neither function $1 nor subcommand ${1}__$2 recognized" >&2
  exit 1
fi

在这种情况下,始终搜索前两个参数命名的子命令,后跟仅由第一个参数命名的函数。

答案 1 :(得分:2)

您也可以这样做:

#!/bin/sh
yum() {
    if [ "$1" = "maintenance" ]; then
        command yum -y update
        command yum -y upgrade
        command yum clean all
    elif [ "$1" = "download" ]; then
        command yum -y install --downloadonly "$2"
    else
       echo "Invalid arg..."
    fi

}

if [ "$1" = "yum" ];then
    shift
    yum "$@"
fi

现在,您可以使用./dafs yum maintenance./dafs yum download http