Bash的新手.简单的脚本 [英] new to Bash. Simple script

问题描述

我是bash脚本的新手，并且正在练习一些代码.

I am new to bash scripting and I am practicing some code.

我正在尝试创建一个脚本，该脚本显示以下输出并循环播放，直到用户键入x或X

I am trying to create a script that displays the following output and loops until the user types x or X

菜单1

C)计算

X)退出

C

菜单2

输入整数或按X退出:

22

菜单3

+)添加

-)减去

+

菜单2

输入整数或按X退出:

33

22和33的总和是55

The sum of 22 and 33 is 55

菜单1

C)计算

X)退出

c

菜单2

输入整数或按X退出:

50

菜单3

+)添加

-)减去

-

菜单2 输入一个整数或按X退出:

Menu 2 Enter an integer or press X to exit:

23

50和23的差是27

The difference of 50 and 23 is 27

菜单1

C)计算

X)退出

X

这是我的代码:

add () 
{

((sum=n1 + n2))
echo  "The sum of $n1 and $n2 is  $sum"
exit
}

subtract () 
{

((difference=n1 - n2))
echo "The difference of $n1 and  $n2 is $difference"
exit
}


 while true
 do
 echo "Menu 1"
 echo "C) Calculation"
 echo "X) Exit"

 read opr


 if [ ${opr} = 'x' ] || [ ${opr} = 'X' ]
 then
  break

 elif [ ${opr} = 'c' ] || [ ${opr} = 'C' ]
 then
 echo "Menu 2" 
 echo "Enter an integer or press X to exit:"

 fi

 read n1

 if [ $n1 = 'x' ] || [ $n1 = 'X' ]
 then
     break

 else

 echo "Menu3"
 echo "+) Add"
 echo "-) Subtract"

 fi

 read opr

if [ $opr = '+' ]
then

echo "Please enter another integer to perform addition"
read n2
add

elif [ $opr = '-' ]
echo "Please enter another integer to perform subtraction"
read n2
subtract

fi

done

我收到此错误消息:

./myscript.sh:第72行:意外令牌fi' ./myscript.sh: line 72: fi'

./myscript.sh: line 72: syntax error near unexpected token fi' ./myscript.sh: line 72:fi'

我相信，如果将menu1，menu2和menu3设置为函数，则可以实现我希望输出的内容而不是此版本的输出.

I believe if I make menu1, menu2, and menu3 into functions I could achieve what I desire my output to be instead of this version of it.

但是我知道我仍然会对那些文件有问题……任何想法我应该把它们放在哪里，或者我需要做什么才能使我的代码正常工作而不给出错误?

But I know that I will still have a problem with those fi ... any idea where I should put them or what do I need to do for my code to work and not give an error?

谢谢

药丸

推荐答案

您只是在elif

在 http://www.thegeekstuff中查找一般示例. com/2010/06/bash-if-statement-examples/

If [ conditional expression1 ]
then
    statement1
    statement2
.
elif [ conditional expression2 ]
then
    statement3
    statement4
    .
.
.
else
    statement5 
fi

在您的代码中:

...
elif [ $opr = '-' ]
then  <<<<<<< missing in your code!    
    echo "Please enter another integer to perform subtraction"
    read n2
    subtract

fi

为什么程序没有循环

如果在add和subtract方法中编写exit，则脚本将退出.为什么不应该呢?提示:删除exit和循环都有机会:-)

If you write exit in your add and subtract method, your script will exit. Why it should not? Tip: remove both exit and the loop have a chance :-)

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