30

awk を使用して処理する必要があるテキスト ファイルがあります。

"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"

唯一の問題は、電話番号を囲む引用符を削除する方法がわからないことです。私の最終的な出力は次のようになります...

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"
4

2 に答える 2

60

gsub 関数を使用できます：

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"
于 2013-10-20T07:16:16.453 に答える
20

フィールド区切り文字として二重引用符を使用します。

 awk -F'"' '{print "set", $4, "\""$6"\""}' testme.csv
于 2013-10-20T07:16:44.890 に答える