删除awk命令中的引号

Question

我有一个需要使用awk处理的文本文件.

"5","1211274723","0","D","2"
"1","1211292921","0","A","2"
"5","1211295793","0","A","2"
"5","1211310146","0","A","2"
"5","1211310310","0","A","2"
"4","1211315271","0","A","2"
"5","1211318203","0","D","2"
"2","1211323658","0","A","2"
"5","1211329224","0","A","2"
"5","1211330064","0","A","2"


# cat testme.csv | awk -F',' '{print "set", $2, $3}'
set "1211274723" "0"
set "1211292921" "0"
set "1211295793" "0"
set "1211310146" "0"
set "1211310310" "0"
set "1211315271" "0"
set "1211318203" "0"
set "1211323658" "0"
set "1211329224" "0"
set "1211330064" "0"

唯一的问题是我不知道如何删除电话号码周围的报价.所以我的最终输出看起来像这样......

set 1211274723 "0"
set 1211292921 "0"
set 1211295793 "0"
set 1211310146 "0"
set 1211310310 "0"
set 1211315271 "0"
set 1211318203 "0"
set 1211323658 "0"
set 1211329224 "0"
set 1211330064 "0"

Answer 1

你可以使用gsub函数:

awk -F',' '{gsub(/"/, "", $2); print "set", $2, $3}' testme.csv 
set 1211274723 "0"
set 1211292921 "0"
set 1211295793 "0"
set 1211310146 "0"
set 1211310310 "0"
set 1211315271 "0"
set 1211318203 "0"
set 1211323658 "0"
set 1211329224 "0"
set 1211330064 "0"

Answer 2

使用双引号作为字段分隔符;

 awk -F'"' '{print "set", $4, "\""$6"\""}' testme.csv