将参数传递给Bash函数

我正在尝试搜索如何在Bash函数中传递参数，但是出现的总是总是如何从命令行传递参数。

我想在脚本中传递参数。我试过了：

myBackupFunction("..", "...", "xx")

function myBackupFunction($directory, $options, $rootPassword) {
     ...
}

但是语法不正确，如何将参数传递给我的函数？

声明函数有两种典型的方法。我更喜欢第二种方法。

function function_name {
   command...
} 

要么

function_name () {
   command...
} 

要使用参数调用函数：

function_name "$arg1" "$arg2"

该函数通过其位置（而不是名称）引用传递的参数，即$ 1，$ 2，依此类推。$ 0是脚本本身的名称。

例：

function_name () {
   echo "Parameter #1 is $1"
}

另外，您需要在函数声明后调用它。

#!/usr/bin/env sh

foo 1  # this will fail because foo has not been declared yet.

foo() {
    echo "Parameter #1 is $1"
}

foo 2 # this will work.

输出：

./myScript.sh: line 2: foo: command not found
Parameter #1 is 2

参考：高级Bash脚本指南。

熟悉高级编程语言（C / C ++ / Java / PHP / Python / Perl ...）会向外行暗示，bash函数应该像在其他语言中一样工作。相反，bash函数的工作方式类似于shell命令，并且期望将参数传递给它们的方式与可能将选项传递给shell命令的方式相同（例如ls -l）。实际上，bash中的函数参数被视为位置参数（$1, $2..$9, ${10}, ${11}，依此类推）。考虑到getopts工作原理，这不足为奇。不要使用括号在bash中调用函数。

（注意：目前，我碰巧正在使用Open Solaris。）

# bash style declaration for all you PHP/JavaScript junkies. :-)
# $1 is the directory to archive
# $2 is the name of the tar and zipped file when all is done.
function backupWebRoot ()
{
    tar -cvf - $1 | zip -n .jpg:.gif:.png $2 - 2>> $errorlog &&
        echo -e "\nTarball created!\n"
}


# sh style declaration for the purist in you. ;-)
# $1 is the directory to archive
# $2 is the name of the tar and zipped file when all is done.
backupWebRoot ()
{
    tar -cvf - $1 | zip -n .jpg:.gif:.png $2 - 2>> $errorlog &&
        echo -e "\nTarball created!\n"
}


# In the actual shell script
# $0               $1            $2

backupWebRoot ~/public/www/ webSite.tar.zip

想要使用变量名称。就是这样

declare filename=$1 # declare gives you more options and limits variable scope

想要将数组传递给函数吗？

callingSomeFunction "${someArray[@]}" # Expands to all array elements.

在函数内部，像这样处理参数。

function callingSomeFunction ()
{
    for value in "$@" # You want to use "$@" here, not "$*" !!!!!
    do
        :
    done
}

需要传递值和数组，但仍在函数内部使用“ $ @”吗？

function linearSearch ()
{
    declare myVar="$1"

    shift 1 # removes $1 from the parameter list

    for value in "$@" # Represents the remaining parameters.
    do
        if [[ $value == $myVar ]]
        then
            echo -e "Found it!\t... after a while."
            return 0
        fi
    done

    return 1
}

linearSearch $someStringValue "${someArray[@]}"

如果您更喜欢命名参数，则可以（通过一些技巧）将命名参数实际传递给函数（也可以传递数组和引用）。

我开发的方法允许您定义传递给函数的命名参数，如下所示：

function example { args : string firstName , string lastName , integer age } {
  echo "My name is ${firstName} ${lastName} and I am ${age} years old."
}

您还可以将参数注释为@required或@readonly，创建... rest参数，根据顺序参数创建数组（使用eg string[4]），还可以选择在多行中列出参数：

function example {
  args
    : @required string firstName
    : string lastName
    : integer age
    : string[] ...favoriteHobbies

  echo "My name is ${firstName} ${lastName} and I am ${age} years old."
  echo "My favorite hobbies include: ${favoriteHobbies[*]}"
}

换句话说，不仅可以通过名称来调用参数（这构成了更易读的内核），而且实际上可以传递数组（以及对变量的引用-该功能仅在bash 4.3中有效）！另外，映射变量都在本地范围内，就像$ 1（及其他）一样。

使这项工作有效的代码非常轻巧，并且可以在bash 3和bash 4中使用（这是我测试过的唯一版本）。如果您对更多类似的技巧感兴趣，这些技巧使使用bash进行开发变得更加轻松便捷，则可以查看我的Bash Infinity Framework，以下代码可作为其功能之一。

shopt -s expand_aliases

function assignTrap {
  local evalString
  local -i paramIndex=${__paramIndex-0}
  local initialCommand="${1-}"

  if [[ "$initialCommand" != ":" ]]
  then
    echo "trap - DEBUG; eval \"${__previousTrap}\"; unset __previousTrap; unset __paramIndex;"
    return
  fi

  while [[ "${1-}" == "," || "${1-}" == "${initialCommand}" ]] || [[ "${#@}" -gt 0 && "$paramIndex" -eq 0 ]]
  do
    shift # first colon ":" or next parameter's comma ","
    paramIndex+=1
    local -a decorators=()
    while [[ "${1-}" == "@"* ]]
    do
      decorators+=( "$1" )
      shift
    done

    local declaration=
    local wrapLeft='"'
    local wrapRight='"'
    local nextType="$1"
    local length=1

    case ${nextType} in
      string | boolean) declaration="local " ;;
      integer) declaration="local -i" ;;
      reference) declaration="local -n" ;;
      arrayDeclaration) declaration="local -a"; wrapLeft= ; wrapRight= ;;
      assocDeclaration) declaration="local -A"; wrapLeft= ; wrapRight= ;;
      "string["*"]") declaration="local -a"; length="${nextType//[a-z\[\]]}" ;;
      "integer["*"]") declaration="local -ai"; length="${nextType//[a-z\[\]]}" ;;
    esac

    if [[ "${declaration}" != "" ]]
    then
      shift
      local nextName="$1"

      for decorator in "${decorators[@]}"
      do
        case ${decorator} in
          @readonly) declaration+="r" ;;
          @required) evalString+="[[ ! -z \$${paramIndex} ]] || echo \"Parameter '$nextName' ($nextType) is marked as required by '${FUNCNAME[1]}' function.\"; " >&2 ;;
          @global) declaration+="g" ;;
        esac
      done

      local paramRange="$paramIndex"

      if [[ -z "$length" ]]
      then
        # ...rest
        paramRange="{@:$paramIndex}"
        # trim leading ...
        nextName="${nextName//\./}"
        if [[ "${#@}" -gt 1 ]]
        then
          echo "Unexpected arguments after a rest array ($nextName) in '${FUNCNAME[1]}' function." >&2
        fi
      elif [[ "$length" -gt 1 ]]
      then
        paramRange="{@:$paramIndex:$length}"
        paramIndex+=$((length - 1))
      fi

      evalString+="${declaration} ${nextName}=${wrapLeft}\$${paramRange}${wrapRight}; "

      # continue to the next param:
      shift
    fi
  done
  echo "${evalString} local -i __paramIndex=${paramIndex};"
}

alias args='local __previousTrap=$(trap -p DEBUG); trap "eval \"\$(assignTrap \$BASH_COMMAND)\";" DEBUG;'

错过括号和逗号：

 myBackupFunction ".." "..." "xx"

该函数应如下所示：

function myBackupFunction() {
   # here $1 is the first parameter, $2 the second etc.
}

我希望这个例子能对您有所帮助。它从用户那里获得两个数字，并将它们提供给所调用的函数add（在代码的最后一行），然后add将它们求和并打印出来。

#!/bin/bash

read -p "Enter the first  value: " x
read -p "Enter the second value: " y

add(){
    arg1=$1 #arg1 gets to be the first  assigned argument (note there are no spaces)
    arg2=$2 #arg2 gets to be the second assigned argument (note there are no spaces)

    echo $(($arg1 + $arg2))
}

add x y #feeding the arguments

一个简单的示例，它将在执行脚本期间或在调用函数时在脚本内部清除。

#!/bin/bash
echo "parameterized function example"
function print_param_value(){
    value1="${1}" # $1 represent first argument
    value2="${2}" # $2 represent second argument
    echo "param 1 is  ${value1}" #as string
    echo "param 2 is ${value2}"
    sum=$(($value1+$value2)) #process them as number
    echo "The sum of two value is ${sum}"
}
print_param_value "6" "4" #space sparted value
#you can also pass paramter durign executing script
print_param_value "$1" "$2" #parameter $1 and $2 during executing

#suppose our script name is param_example
# call like this 
# ./param_example 5 5
# now the param will be $1=5 and $2=5

以为我会提到另一种将命名参数传递给bash的方法...通过引用传递。从bash 4.0开始支持

#!/bin/bash
function myBackupFunction(){ # directory options destination filename
local directory="$1" options="$2" destination="$3" filename="$4";
  echo "tar cz ${!options} ${!directory} | ssh root@backupserver \"cat > /mnt/${!destination}/${!filename}.tgz\"";
}

declare -A backup=([directory]=".." [options]="..." [destination]="backups" [filename]="backup" );

myBackupFunction backup[directory] backup[options] backup[destination] backup[filename];

bash 4.3的另一种语法是使用nameref

尽管nameref可以无缝地取消引用，因此更加方便，但是某些受支持的较早发行版仍提供较旧的版本，因此我暂时不会推荐它。