在Bash循环内设置变量 [英] Setting variables inside a Bash loop

问题描述

刚刚开始学习Linux Bash Shell编程，而我只是不知道我是否正确理解它.查看下面的示例程序:

Just started learning Linux Bash Shell Programming, and I just don't know if I understood it correctly. Looking at the sample program below:

#!/bin/bash
n=1
sumRSS=1000
sumSZ=2000

echo Before sumRSS=$sumRSS sumSZ=$sumSZ
ps -ly | while
read c1 c2 c3 c4 c5 c6 c7 c8 c9 c10
do
    if (( n>1 ))
    then
        echo n=$n rss=$sumRSS sz=$sumSZ
        ((sumRSS = sumRSS +  c8))
        ((sumSZ = sumSZ + c9))

    fi
    ((n++))
done

echo Sum of RSS = $sumRSS
echo Sum of SZ = $sumSZ

输出:

Before sumRSS=1000 sumSZ=2000
n=2 rss=1000 sz=2000
n=3 rss=2368 sz=29118
n=4 rss=3792 sz=55644
n=5 rss=4780 sz=82679
Sum of RSS = 1000
Sum of SZ = 2000

我不知道为什么总和仍然回到RSS = 1000和SZ = 2000.我实际上期望的是RSS = 4780和SZ = 82679.

I don't know why the sum still goes back to RSS=1000 and SZ=2000. I actually was expecting RSS=4780 and SZ=82679.

我知道我缺少一些基本知识.我正在通过编写简单的脚本来学习bash.

I know I am missing something basic. I am learning bash by writing simple scripts.

推荐答案

您应该避免使用@linuxfan建议的管道.您可以将代码更改为:

you should avoid the pipe as @linuxfan proposes. You can change your code to:

while read c1 c2 c3 c4 c5 c6 c7 c8 c9 c10
do
    ...
done < <(ps -ly)

这样，您的变量将保持在同一范围内.

this way your variables stay in the same scope.

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