我如何创建具有多种选择的菜单变量的shell脚本？ [英] how do i create a shell script with multiple choice menu variables?

问题描述

下面就是我要做的。

Please pick from the following  
1) area1  
2) area2  
3) area3

Now pick a place based on the area you chose  
1) place1  
2) place2  
3) place3

Based on $area and $place here is what you need...

我的猜测是，我需要if语句的情况下和组合，但我不能环绕这个问题我的头。

My guess is i need a combination of case and if statements but i am unable to wrap my head around the problem.

推荐答案

使用选择语句@Barmar建议。

Use the select statement as @Barmar suggested.

根据您的例子：

echo 'Please pick from the following'

select area in area1 area2 area3
do
        case $area in 
        area1|area2|area3)   
                break
                ;;
        *)
                echo "Invalid area" 
                ;;
        esac
done

echo 'Now pick a place based on the area you chose'
select place in place1 place2 place3
do
        case $place in
        place1|place2|place3)
                break
                ;;
        *)
                echo "Invalid place"
                ;;
        esac
done

echo "Based on $area and $place here is what you need..."

我用 A | B | C）将多个选择，但你可以为每一个单独的code路径。 *）可以作为一个其他条款。使用破打出来的菜单循环。使用 ;; 在情况下，子句之间语句和; 一个子句中语句之间。

I used a|b|c) to group multiple options, but you could provide separate code paths for each one. *) can be used as an "else" clause. Use break to break out of the menu loop. Use ;; between clauses in the case statement and ; between statements in a single clause.

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