我试图搜索如何在Bash函数中传递参数,但出现的总是如何从命令行传递参数。

我想在脚本中传递参数。我试着:

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

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

但是语法不正确。如何将参数传递给函数?


当前回答

一个简单的例子,将在执行脚本或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-separated value
# You can also pass parameters during executing the script
print_param_value "$1" "$2" # Parameter $1 and $2 during execution

# Suppose our script name is "param_example".
# Call it like this:
#
# ./param_example 5 5
#
# Now the parameters will be $1=5 and $2=5

其他回答

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...
} 

or

function_name () {
   command...
} 

调用带参数的函数:

function_name "$arg1" "$arg2"

该函数根据参数的位置(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。

例子:

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

此外,还需要在函数声明之后调用它。

#!/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脚本编写指南。

一个简单的例子,将在执行脚本或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-separated value
# You can also pass parameters during executing the script
print_param_value "$1" "$2" # Parameter $1 and $2 during execution

# Suppose our script name is "param_example".
# Call it like this:
#
# ./param_example 5 5
#
# Now the parameters will be $1=5 and $2=5

如果您更喜欢命名形参,则可以(使用一些技巧)实际将命名形参传递给函数(也可以传递数组和引用)。

我开发的方法允许你像这样定义传递给函数的命名参数:

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

你也可以注解参数为@required或@readonly, create…其余参数,从顺序参数创建数组(例如使用字符串[4]),并可选地在多行中列出参数:

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,您不仅可以通过名称调用参数(这使核心更具可读性),还可以实际传递数组(以及对变量的引用——该特性仅在Bash 4.3中有效)!另外,映射的变量都在局部作用域中,就像$1(和其他变量)一样。

实现这一功能的代码非常简单,并且可以在Bash 3和Bash 4中运行(这是我测试过的唯一版本)。如果您对更多这样的技巧感兴趣,可以让bash开发变得更好更简单,您可以看看我的bash Infinity Framework,下面的代码是它的功能之一。

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 parameter:
      shift
    fi
  done
  echo "${evalString} local -i __paramIndex=${paramIndex};"
}

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

另一种向Bash传递命名参数的方法…是通过引用传递的。Bash 4.0支持这一点

#!/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无缝地取消了引用,因此更加方便,但一些较旧的受支持发行版仍然发布了较旧的版本,因此我暂时不推荐使用它。

去掉括号和逗号:

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

函数应该是这样的:

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