如何在shell Scripting中调用函数？

Question

我有一个有条件地调用函数的shell脚本.

例如:-

if [ "$choice" = "true" ]
then 
  process_install
elif [ "$choice" = "false" ]
then 
  process_exit
fi

process_install()
{
  commands...
  commands...
}

process_exit()
{
  commands...
  commands...
}

请让我知道如何做到这一点.

Answer 1

你没有指定哪个 shell(有很多),所以我假设Bourne Shell,我认为你的脚本以:

#!/bin/sh

请记住使用shell类型标记未来的问题,因为这将有助于社区回答您的问题.

您需要在调用之前定义函数.使用():

process_install()
{
    echo "Performing process_install() commands, using arguments [${*}]..."
}

process_exit()
{
    echo "Performing process_exit() commands, using arguments [${*}]..."
}

然后你可以调用你的函数,就像你调用任何命令一样:

if [ "$choice" = "true" ]
then
    process_install foo bar
elif [ "$choice" = "false" ]
then
    process_exit baz qux

您可能还希望在此时检查无效选择...

else
    echo "Invalid choice [${choice}]..."
fi

看它运行三个不同的$ {choice}值.

祝好运!

Answer 2

#!/bin/bash

process_install()
{
    commands... 
    commands... 
}

process_exit()
{
    commands... 
    commands... 
}


if [ "$choice" = "true" ] then
    process_install
else
    process_exit
fi

Answer 3

在 bash 中使用 function() 的示例：

#!/bin/bash
# file.sh: a sample shell script to demonstrate the concept of Bash shell functions
# define usage function
usage(){
    echo "Usage: $0 filename"
    exit 1
}

# define is_file_exists function
# $f -> store argument passed to the script
is_file_exists(){
    local f="$1"
    [[ -f "$f" ]] && return 0 || return 1
}
# invoke  usage
# call usage() function if filename not supplied
[[ $# -eq 0 ]] && usage

# Invoke is_file_exits
if ( is_file_exists "$1" )
then
 echo "File found: $1"
else
 echo "File not found: $1"
fi