管道中的变量范围 [英] scope of variable in pipe

问题描述

以下外壳脚本将检查磁盘空间，如果使用率超过10％，则将变量diskfull更改为1 最后的回声总是显示0 我在if子句中尝试了global diskfull=1，但是它不起作用. 如果磁盘使用量超过10％，如何将变量更改为1?

The following shell scrip will check the disk space and change the variable diskfull to 1 if the usage is more than 10% The last echo always shows 0 I tried the global diskfull=1 in the if clause but it did not work. How do I change the variable to 1 if the disk consumed is more than 10%?

#!/bin/sh
diskfull=0

ALERT=10
df -HP | grep -vE '^Filesystem|tmpfs|cdrom' | awk '{ print $5 " " $1 }' | while read output;
do
  #echo $output
  usep=$(echo $output | awk '{ print $1}' | cut -d'%' -f1  )
  partition=$(echo $output | awk '{ print $2 }' )
  if [ $usep -ge $ALERT ]; then
     diskfull=1
     exit
  fi
done

echo $diskfull

推荐答案

使用管道时，shell接缝将使用子shell进行工作.由于这些子外壳程序不知道$diskfull，因此该值永远不会更改.

When using pipes the shell seams to use sub-shells to do the work. As $diskfull is not known to these sub-shells the value is never changed.

请参阅: http://www.nucleardonkey.net/blog/2007/08/variable_scope_in_bash. html

我修改了您的脚本，如下所示.它对我有用，也应该在您的系统上工作.

I modified your script as follows. It works for me and should work on your system too.

#!/bin/sh
diskfull=0

ALERT=10
stats=`df -HP | grep -vE '^Filesystem|tmpfs|cdrom|none|udev' | awk '{ print $5 "_" $1 }'`
for output in $stats
do
  usep=$(echo $output | awk '{ print $1}' | cut -d'%' -f1  )
  partition=$(echo $output | sed s/.*_// )
  #echo $partition -  $usep
  if [ $usep -le $ALERT ]; then
     diskfull=1
     break
  fi
done
echo $diskfull

这篇关于管道中的变量范围的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！