linux - 为什么在Linux Shell脚本中使用％不能正常工作？

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想改进这个问题吗？Update the question所以堆栈溢出的值小于aa>。
我用%来确定数字是否可以被5整除。但它似乎总是返回“0”，表示所有内容都可以被5整除。
我做错什么了？

#!/bin/sh

n=20
i=1
# loop while i <= n
while [ $i -lt $n ] || [ $i -eq $n ]
do

# find the numbers that are evenly divisible by '5'

  m=`expr $n % 5`
  if [ $m -eq 0 ]; then
    echo $i" is evenly divisible by 5"
  fi

  # do i = i + 1
  i=`expr $i + 1`
done

我的输出如下：

1 is evenly divisible by 5
2 is evenly divisible by 5
3 is evenly divisible by 5
4 is evenly divisible by 5
5 is evenly divisible by 5
6 is evenly divisible by 5
7 is evenly divisible by 5
8 is evenly divisible by 5
9 is evenly divisible by 5
10 is evenly divisible by 5
11 is evenly divisible by 5
12 is evenly divisible by 5
13 is evenly divisible by 5
14 is evenly divisible by 5
15 is evenly divisible by 5
16 is evenly divisible by 5
17 is evenly divisible by 5
18 is evenly divisible by 5
19 is evenly divisible by 5
20 is evenly divisible by 5

最佳答案

n始终为20，因此“m=`expr$n%5”将始终为0