如何使用awk或grep从git日志中提取数据？

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

我的命令的输出：

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，我只想获取

BUC ID

，以便在

webui/

中进行更改
我怎样才能做到这一点

$ 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/

---

该命令将从输出中删除空行，按

webui

过滤，并在

webui

匹配之前打印

2

行

---

git-log--dirstat | grep-ve'^$'| grep webui-b2

git-log--pretty=%s--webui/| awk'{print$1}

@ElpieKay谢谢这是我们想要的。非常感谢@DevDio。这正是我想要的。谢谢@Akshay Hegde的回复，但我发现另外两个提供的解决方案更容易使用。