将参数传递给 Bash 函数 [英] Passing parameters to a Bash function

问题描述

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

I am trying to search how to pass parameters in a Bash function, but what comes up is always how to pass parameter from the command line.

我想在我的脚本中传递参数.我试过了:

I would like to pass parameters within my script. I tried:

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?

But the syntax is not correct. How can I pass a parameter to my function?

推荐答案

有两种典型的函数声明方式.我更喜欢第二种方法.

There are two typical ways of declaring a function. I prefer the second approach.

function function_name {
   command...
} 

或

function_name () {
   command...
} 

调用带参数的函数:

function_name "$arg1" "$arg2"

该函数通过它们的位置(而不是名称)来引用传递的参数，即 $1、$2 等等.$0 是脚本本身的名称.

The function refers to passed arguments by their position (not by name), that is $1, $2, and so forth. $0 is the name of the script itself.

示例:

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

此外，您需要在声明函数之后调用它.

Also, you need to call your function after it is declared.

#!/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.

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