I want to remove all lines after a line which contains the string “Date:” in it. Specifically, I am writing the results of git show --name-only to a logging file, but I do not want to log the data after the string that starts with “Date:”
This is the output of git show --name-only:
commit <commit-hash> (HEAD -> <branch-name>, origin/<branch-name>)
Author: bitbucket-pipelines <[email protected]>
Date: Mon Sep 18 18:26:04 2023 +0000
[skip ci] Commit message
logs/log.txt
And this is what I want to output:
commit <commit-hash> (HEAD -> <branch-name>, origin/<branch-name>)
Author: bitbucket-pipelines <[email protected]>
Date: Mon Sep 18 18:26:04 2023 +0000
I tried to remove lines using specific line numbers and writing to a file and then parsing it there, but I am looking for a solution where I can get the output directly as I want it without any extra steps
You can do this in Awk. This will print all lines, and if the line starts with Date: the script will exit after printing.
git show --name-only | awk '{print} ; /^Date:/ { exit }'
Or Sed (quit when line matches):
git show --name-only | sed '/^Date:/q'
Or if you want to explicitly specify a number of lines:
git show --name-only | awk '{print} ; NR==3 { exit }'
git show --name-only | head -n 3
You can easily do this with only Git by providing the format with which the commit should be printed.
git show -s --pretty='commit %H%d%nAuthor: %aN <%aE>%nDate: %ad'
%Hthe commit hash%dref names like--decorate%aNauthor name (respecting.mailmap)%aEauthor email (respecting.mailmap)%adauthor date
See the PRETTY FORMATS section of git help log for a detailed explanation.
If you don’t want to use Git to control the output format, head -3 is usually a pretty simple way to get only the first three lines of any output.
git show --name-only --no-patch | grep -v '^ .*' | grep -v '^$'
--no-patch to remove file-change lines (stat). Inverted greps to remove empty and indented lines, which is the message part.
A bit overkill for git show but can also be used for git log.
Don’t forget the good old grep! Here I use regex to match and print lines starting with commit, Author:, and Date:.
git show --name-only | grep '^\(commit\|Author:\|Date:\)'
There’s lots of ways to do this – please edit your question to include what you tried so we can help you with that. Does a solution need to key on the string
Dateor could you just print the first 3 lines or the lines before the first empty line or something else?Do you want to specifically only remove lines after a line contiaining
Date:or do you want to printgit showa commit without the commit message and file changes? Are you asking XY question? Did you readman git show? In particular there isgit show --format=.I think the question explains very clearly what is wanted, with example input and output.