elif的語法錯誤

Question

我寫了一個使用if - elif構造的小腳本。出於某種原因，它不會起作用。 錯誤代碼：

./lab4.sh: line 9: syntax error near unexpected token `elif' 
./lab4.sh: line 9: `elif [ $number -eq 2 ] then func2' 

我的實際代碼：

#!/bin/bash 
#ask the user for a number between 1 and 3 
#create some functions, write out the function number 
echo "Enter a number between 1 and 3: " 
read number 

#which function should be called? 
if [ $number -eq 1 ] then func1 
elif [ $number -eq 2 ] then func2 
elif [ $number -eq 3 ] then func3 fi 

function func1 { 
    echo "This message was displayed from the first function." 
} 

function func2 { 
    echo "This message was displayed from the second function." 
} 

function func3 { 
    echo "This message was displayed from the third function." 
} 

Answer 1

您必須返回到一個新行或shell聲明（if，then，elif，fi ...）之前使用;。

你還必須在使用它們之前聲明你的函數。

#!/bin/bash 
#ask the user for a number between 1 and 3 
#create some functions, write out the function number 
echo "Enter a number between 1 and 3: " 
read number 

function func1 { 
    echo "This message was displayed from the first function." 
} 

function func2 { 
    echo "This message was displayed from the second function." 
} 

function func3 { 
    echo "This message was displayed from the third function." 
} 

#which function should be called? 
if [ $number -eq 1 ]; then func1 
elif [ $number -eq 2 ]; then func2 
elif [ $number -eq 3 ]; then func3; fi 

Answer 2

對於這種多if/elif的IT有時候更容易使用case語句，如：

func1() { echo "func1"; } 
func2() { echo "func2"; } 
func3() { echo "func3"; } 

while read -r -p 'Enter a number between 1 and 3:>' number 
do 
    case "$number" in 
     1) func1 ; break ;; 
     2) func2 ; break ;; 
     3) func3 ; break ;; 
     q) exit;; 
     *) echo "Wrong input" >&2;; #error message 
    esac 
done 
echo "end of loop" 

評論：不要使用function關鍵字。這是一個bashism，定義一個殼功能的便攜式方式是

funcname() { 
} 

，這是一個有點短太... :)