Question

我有一个以下形式的csv文件：

$ head purchases.csv 
id,userID,itemID,price,platform,day
1,9132,id_005,3600,2,2014-10-30 17:29:46
2,67894,id_005,3000,1,2015-04-23 21:22:55
3,272780,id_004,1000,1,2014-11-27 16:58:30
4,302396,id_001,100,1,2014-12-11 08:35:07

现在，我想更改csv的最后一栏。目前，它在day形式的2014-10-30 17:29:46列中，即日期和时间之间有空格。但我想将此列拆分为两列day和time，以便在更改后csv文件变为：

$ head purchases.csv 
id,userID,itemID,price,platform,day,time
1,9132,id_005,3600,2,2014-10-30,17:29:46
2,67894,id_005,3000,1,2015-04-23,21:22:55
3,272780,id_004,1000,1,2014-11-27,16:58:30

如何从终端进行操作？

Answer 1

在$ 6上使用split：

$ awk -v OFS=\, -F\, 'NR==1{print $0,"time";next} {split($6,a," "); print $1,$2,$3,$4,$5,a[1],a[2]}' test.in
id,userID,itemID,price,platform,day,time
1,9132,id_005,3600,2,2014-10-30,17:29:46
2,67894,id_005,3000,1,2015-04-23,21:22:55
3,272780,id_004,1000,1,2014-11-27,16:58:30
4,302396,id_001,100,1,2014-12-11,08:35:07

或者您可以使用gsub，只需用逗号代替空格：

$ awk -v OFS=\, -F\, 'NR==1{print $0,"time";next} {gsub(/ /,",",$6); print $0}' test.in

Answer 2

James Brown's answer很有帮助，但是对列进行硬编码修改，同时还假设它是 last 。

一些简单的调整概括解决方案：

awk -v ndx=6 -F, 'NR==1 {sub(/$/, ",time", $ndx); print; next} sub(" ", ",", $ndx)' \
  purchases.csv

拆分字符串列＆＃34;日期时间＆＃34;进入＆＃34; date＆＃34; ＆＃34;时间＆＃34;在CSV文件中

2 个答案: