shell脚本实例之Charpter8-8

要求:如果一个整数各位数字之和可以被9整除，那么该数就能被9整除。编写一个脚本，提示用户输入一
个整数，然后输出该整数，并告知能够被9整数。

这里只实现了四位数的判断:

#!/bin/bash
#chapter_8-8
#trap 'echo "Before executing the line:$LINENO,num=$num,i=$i,sum=$sum" ' DEBUG
echo "Please input a integer number:"
read num

for ((i=3;i>=0;i--))
do
let divisor=10**i
let temp=num/divisor
if [ $temp -ne 0 ]
then
break
fi
done
echo this is a $((i+1)) bit number.
sum=0
case "$i" in
3)
let sum=num/10**3+num%10**3/10**2+num%10**2/10+num%10 ;;
2)
let sum=num/10**2+num%10**2/10+num%10 ;;
1)
let sum=num/10+num%10 ;;
0)
let sum=num ;;

esac

echo  sum=$sum

if [ $((sum%9)) -eq 0 ]
then
echo "$num can be divided by 9!"
else
echo  "$num can not be divided by 9!"

fi

执行脚本效果:

[root@localhost charpter8]# sh 8-8
Please input a integer number:
4321
this is a 4 bit number.
sum=10
4321 can not be divided by 9!
[root@localhost charpter8]# sh 8-8
Please input a integer number:
432
this is a 3 bit number.
sum=9
432 can be divided by 9!
[root@localhost charpter8]# sh 8-8
Please input a integer number:
9333
this is a 4 bit number.
sum=18
9333 can be divided by 9!
[root@localhost charpter8]#

这能实现不大于4位数的数值。
谁有好的脚本思路可以共享下。

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