如何从 csv 中删除除特定字段之外的所有双引号

Question

我想从 csv 中删除所有双引号，但不是第四个字段（因为这四个字段代表文件的路径）

请建议如何通过sed或awk或perl oneliner等来实现这一点

我现在所知道的是使用简单的 sed 命令：

   sed s"/\"//g"  file.csv  | sed 's/ //g'

但这个命令不那么优雅，也适用于第四个字段（第四个字段不应该编辑）

备注 - 还需要删除引号之间的空格以接近字符

示例（之前的 csv 文件）

"24  ","COsc   ","LINUX","/VP/Ame/AR/Celts/COf","  fbsutamante ",fbu2012,"kkk","&^#$@J  ",,,,,
25,COsc,LINUX,"/VP/Ame/AR/Celts/COf","fbsutamante ",fbu2012,"iiii ","   *****",,,,,

示例（之后的 csv 文件）

24,COsc,LINUX,"/VP/Ame/AR HR/Ce   lts/COf",fbsutamante,fbu2012,kkk,&^#$@J,,,,,
25,COsc,LINUX,"/VP/Ame/AR HR/Ce   lts/COf",fbsutamante,fbu2012,iiii,*****,,,,,

Answer 1

这可以是一种方式：

awk 'BEGIN{FS=OFS=","}              # set input and output field separator as comma
     {for (i=5; i<=NF; i++) {       # loop from 5th field
            gsub("\"","", $i);      # remove "
            gsub(/^[ \t]+/,"", $i); # remove leading spaces
            gsub(/[ \t]+$/,"",$i)}  # remove trailing spaces
     }1' file

删除前导和尾随是基于 BMW 的答案：Removeleading and Trailing space in field in awk。

测试

$ awk 'BEGIN{FS=OFS=","} {for (i=5; i<=NF; i++) {gsub("\"","", $i); gsub(/^[ \t]+/,"", $i); gsub(/[ \t]+$/,"",$i)}}1' file
24,COsc,LINUX,"/VP/Ame/AR/Celts/COf",fbsutamante,fbu2012,kkk,&^#$@J,,,,,
25,COsc,LINUX,"/VP/Ame/AR/Celts/COf",fbsutamante,fbu2012,iiii,*****,,,,,

如果还必须清理第一个到第三个字段，只需添加if (i!=4)并循环所有字段：

$ awk 'BEGIN{FS=OFS=","} {for (i=1; i<=NF; i++) {if (i!=4) {gsub("\"","", $i); gsub(/^[ \t]+/,"", $i); gsub(/[ \t]+$/,"",$i)}}}1' a
24,COsc,LINUX,"/VP/Ame/AR/Celts/COf",fbsutamante,fbu2012,kkk,&^#$@J,,,,,
25,COsc,LINUX,"/VP/Ame/AR/Celts/COf",fbsutamante,fbu2012,iiii,*****,,,,,