Question

我必须提取提交消息以及从git log更改的相应文件。基本上，我在提交消息中给出了BUC ID。因此，对于某些文件集，我想获得BUC ID。

输出我的命令：git log --dirstat

  commit bcc9d8be62b6ac9b8dc02c0bf56d0f433df59466
  Author: 
  Date:   Tue Sep 5 12:08:04 2017 +0530

  BUC:BUC3565-EPIC14 | Review Page Manage permission and junit

 Change-Id: Ice73d25f77d9f6c4afe647e35bdf9ec280dd7dcf

 31.3% manager/src/main/java/com/ericsson/cm/manager/web/listner/
 68.6% manager/src/test/java/com/ericsson/cm/manager/web/listners/

 commit ab70d068b2bcea4060028a3457551cd1cc35a1f1
 Author: 
 Date:   Tue Sep 5 11:30:29 2017 +0530

 BUC:BUC12345 MADE some changes

Change-Id: I7e6733afaf8064c1279e5b217c8d4fba469fd061

  59.0% webui/servermanagement/src/servermanagement/regions/left/
  40.9% webui/servermanagement/src/servermanagement/regions/main/

现在我想BUC ID仅对webui/进行更改。{ 我怎样才能做到这一点？

Answer 1

$ awk -F'[: ]' '/BUC/{buc=$3;next}/webui/{print buc;exit}' logfile
BUC12345

# OR
$ awk  '/BUC/{$1=$1;gsub(/BUC:| .*/,"");buc=$0}/webui/{print buc;exit}' logfile
BUC12345

输入：

$ cat logfile
 commit bcc9d8be62b6ac9b8dc02c0bf56d0f433df59466
  Author: 
  Date:   Tue Sep 5 12:08:04 2017 +0530

  BUC:BUC3565-EPIC14 | Review Page Manage permission and junit

 Change-Id: Ice73d25f77d9f6c4afe647e35bdf9ec280dd7dcf

 31.3% manager/src/main/java/com/ericsson/cm/manager/web/listner/
 68.6% manager/src/test/java/com/ericsson/cm/manager/web/listners/

 commit ab70d068b2bcea4060028a3457551cd1cc35a1f1
 Author: 
 Date:   Tue Sep 5 11:30:29 2017 +0530

 BUC:BUC12345 MADE some changes

Change-Id: I7e6733afaf8064c1279e5b217c8d4fba469fd061

  59.0% webui/servermanagement/src/servermanagement/regions/left/
  40.9% webui/servermanagement/src/servermanagement/regions/main/

Answer 2

该命令将从输出中删除空白行，在webui匹配之前按2过滤并打印webui行。

git log --dirstat | grep -ve '^$' | grep webui -B 2

如何使用awk或grep从git log中提取数据？

2 个答案: