初学！，请帮忙看看哪里不对，为什么计算结果不对。

biangbiang_cat

刚刚学shell，做练习写了下面的脚本：
#!/bin/bash

echo -e "please enter the first number:\c"
read anum

echo -e "please enter the second number:\c"
read bnum

echo -e "please select a meta between (+,-,*,/)\c"
read yourmeta

var1=`expr $anum + $bnum`
var2=`expr $anum - $bnum`
var3=`expr $anum \* $bnum`
var4=`expr $anum / $bnum`

case $yourmeta in
   +) echo -e "$var1"
   ;;
   -) echo -e "$var2"
   ;;
   *) echo -e "$var3"
   ;;
   /) echo -e "$var4"
   ;;
   *) echo -e "your select is not between (+,-,*,/)"
   ;;
esac

计算结果如下：

marmot@sles11xen:~> sh t10.sh
please enter the first number:30
please enter the second number:15
please select a meta between (+,-,*,/)+
45
marmot@sles11xen:~> sh t10.sh
please enter the first number:30
please enter the second number:15
please select a meta between (+,-,*,/)-
15
marmot@sles11xen:~> sh t10.sh
please enter the first number:30
please enter the second number:15
please select a meta between (+,-,*,/)*
450
marmot@sles11xen:~> sh t10.sh
please enter the first number:30
please enter the second number:15
please select a meta between (+,-,*,/)/
450
marmot@sles11xen:~> sh t10.sh
please enter the first number:30
please enter the second number:15
please select a meta between (+,-,*,/)sdfsdf
450
marmot@sles11xen:~>

这究竟是怎么回子事儿啊。。。为啥只有加减乘对， 而除和其它无效字符都按乘法算。。。。。

Herowinter

回复 1# biangbiang_cat

1. 
   
2. *) echo -e "$var3"  => \*) echo -e "$var3"
   

复制代码

biangbiang_cat

回复 2# Herowinter

哈，知道啦，谢谢，我真是傻呀😄
   

ll104567

占个楼、、、、、