29.shell脚本正整数加减乘除计算

Posted 2020-12-21

思路：函数先计算数值加减乘除，case赋值输出对应的加减乘除
#!/bin/bash
add() { 
        add=$[$1+$2]
        echo "outcome:$1+$2=$add"
}
minus() {
        min=$[$1-$2]
        echo "outcome:$1-$2=$min"
}
multiply() {
        mult=$[$1*$2]
        echo "outcome:$1x$2=$mult"
}
divide() {
        div=$[$1/$2]
        echo "outcome:$1/$2=$div"
}
positive_integer() {
        if echo $1| grep -q [^0-9];then
        echo "please input a positive number."
        exit 2
        fi

}
if [ $# -ne 3 ];then  #判断输入参数为3
#第二个参数不能为*，shell脚本中*代表是特殊字符，因此乘应使用x代替
        echo "please input three parameter,format like: number1 (+ - x /) number2"
        exit 1
else
        positive_integer $1#调用函数判断正整数
        positive_integer $3#调用函数判断正整数

fi

case $2 in
        x)
        multiply $1 $3
        ;;
        +)
        add $1 $3
        ;;
        -)
        minus $1 $3
        ;;
        /)
        if [ $1 -lt $3 ];then
        echo "input first number $1 must great second number $3,else outcome number is zero."
else
        divide $1 $3
        fi
        ;;
        *)
        echo "please input three parameter,format like: number1 (+ - x /) number2"
        ;;
esac
用法 ： sh  jjcc.sh 3 x 4 
输出：outcome:3 x 4=12
注：除法如果想保留两位小数可以使用 echo "scale=2;1/2"| bc实现