AWK 输出到 bash 数组
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AWK output to bash Array
提问by Brian Clements
Im trying to put the contents of a simple command in to a bash array however im having a bit of trouble.
我试图将一个简单命令的内容放入一个 bash 数组中,但是我遇到了一些麻烦。
df -h | awk '{ print " " }'
gives percentage used in the file systems on my system output looks like this:
在我的系统输出中给出文件系统中使用的百分比如下所示:
1% /dev
1% /dev/shm
1% /var/run
0% /var/lock
22% /boot
22% /home
22% /home/steve
I would then like to put each of these lines into a bash array array=$(df -h| awk '{ print $5 $6 }')
然后我想将这些行中的每一行放入一个 bash 数组 array=$(df -h| awk '{ print $5 $6 }')
However when I print out the array I get the following:
但是,当我打印出数组时,我得到以下信息:
5%
/
1%
/dev
1%
/dev/shm
1%
/var/run
0%
/var/lock
22%
/boot
22%
/home
22%
/home/steve
Bash is forming the array based on white spaces and not line breaks how can i fix this?
Bash 是基于空格而不是换行符来形成数组,我该如何解决这个问题?
回答by Brian Clements
You need to reset the IFS variable (the delimeter used for arrays).
您需要重置 IFS 变量(用于数组的分隔符)。
OIFS=$IFS #save original
IFS=','
df -h | awk '{ print " ""," }'
回答by ephemient
You could do this in Bash without awk.
你可以在没有 awk 的情况下在 Bash 中做到这一点。
array=()
while read -a line; do
array+=("${line[4]} ${line[5]}")
done < <(df -h)
回答by Diego Torres Milano
You may use this:
你可以使用这个:
eval array=( $(df -h | awk '{ printf("\"%s %s\" ", , ) }') )

