Bash替换为CSV多列 [英] Bash replace in CSV multiple columns

问题描述

我具有以下CSV格式:

I have the following CSV format:

data_disk01,"/opt=920MB;4512;4917;0;4855","/=4244MB;5723;6041;0;6359","/tmp=408MB;998;1053;0;1109","/var=789MB;1673;1766;0;1859","/boot=53MB;656;692;0;729"

我想从每一列中获取数据，除了第一列，它是数组中的最后一个值，像这样:

I would like to take from each column, except the first one, the last value from the array, like this:

data_disk01,"/opt=4855","/=6359","/tmp=1109","/var=1859","/boot=729"

我尝试过类似的事情:

awk 'BEGIN {FS=OFS=","} {if(NF==!1);gsub(/\=.*/,",")} 1'

仅仅是字符串，我设法做到了:

Just the string, I managed to do it with:

string="/opt=920MB;4512;4917;0;4855"
echo $string | awk '{split($0,a,";"); print a[1],a[5]}' | sed 's#=.* #=#'
/opt=4855

但是无法使它适用于整个CSV. 任何提示表示赞赏.

But could not make it work for the whole CSV. Any hints are appreciated.

推荐答案

如果输入的引号中从未包含逗号，则简单的sed脚本应该起作用:

If your input never contains commas in the quoted fields, simple sed script should work:

sed 's/=[^"]*;/=/g' file.csv

这篇关于Bash替换为CSV多列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！