刪除 awk 命令中的引號

Question

我有一個需要使用 awk 處理的文本文件。

"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"

唯一的問題是我不知道如何刪除電話號碼周圍的引號。 所以我的最終輸出看起來像這樣......

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

您可以使用 gsub 功能：

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

使用雙引號作為字段分隔符；

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