Two questions:
I know that git use parent to link the commit, it's easy to log the commit from end to start.
like: git log HEAD~10
But i need to query from the start to end, is it possible?
Simply log everything with one line format and tail the output:
git log --pretty=oneline | tail -n 10
git rev-list HEAD | tail -n 10 to get the 10 IDs to list, which you can then show in whatever format you like (not limited to the one-line method). –
Bolt tail seems to lose the nice coloring that I get in my git console. Passing a parameter like git log --oneline -10 as in the other answer works as I expected. –
Invalidism git log --pretty=oneline -10 will show with colors –
Hexone git log -10
Would show 10 latest commits matching the revision spec (a missing spec means "all commits").
See manpage:
git help log
section Commit Limiting
-<number>, -n <number>, --max-count=<number>
Limit the number of commits to output.
tail being available on your platform. –
Overshoot first 10, not the latest. –
Dosimeter Simply log everything with one line format and tail the output:
git log --pretty=oneline | tail -n 10
git rev-list HEAD | tail -n 10 to get the 10 IDs to list, which you can then show in whatever format you like (not limited to the one-line method). –
Bolt tail seems to lose the nice coloring that I get in my git console. Passing a parameter like git log --oneline -10 as in the other answer works as I expected. –
Invalidism git log --pretty=oneline -10 will show with colors –
Hexone Here my approach,
To get first 10 commits:
git log -n 10
-n is number
Additional To get next 10 commit skip first 10 :
git log --skip=10 -n 10
i would use below simple syntax command;
git log -10 --oneline
To get the last 10 commits:
git log HEAD~10..HEAD
To get them in oldest-to-newest order:
git log --reverse HEAD~10..HEAD
Note that if there are merges, this may show more than 10 commits; add --first-parent if you only want to traverse through the first parent of each branch.
For far more detail, see the documentation for git rev-list.
git log --no-walk `git rev-list HEAD | tail -n 10`
and:
git log --no-walk `git rev-list --reverse HEAD | head -n 10`
depending on which order you want the results.
the best result comes with combination of both best answers:
git log --pretty=oneline -10
lkj = log --pretty=oneline --abbrev-commit -10. Now I realize it doesn't answer the OP's question but I came here looking for the simple number switch. –
Clipfed Simply log everything reverse -1 means list one log
git log --reverse -1
Because... more detail :p
By (no branch), you might be asking about the reflog rather than any given ancestry chain. The following has nothing to do with the branch you are on.
git log -g --pretty=oneline | tail -10
<sha> HEAD@{###}: action: summary (old)
<sha> HEAD@{###}: action: summary (older)
...
<sha> HEAD@{###}: action: summary (oldest)
-g is --walk-reflogs Instead of walking the commit ancestry chain, walk reflog entries.q|cut -d ' ' -f 2|tr -d ':' > log to log only the reflog commit index.The following will show the earliest ancestors of the currently checked out branch.
git log --reverse --pretty=oneline | head -10 | cat -n
1 <sha> summary (oldest)
2 <sha> summary (second)
--reverse Output the commits in reverse order.-n 10 or -10 since it breaks --reversecat -n adds line numbers (commit index?)Try this neat output format:
git log --date=short -10 --pretty="%C(Yellow)%h %x09 %C(reset)%ad %x09 %C(Cyan)%an: %C(reset)%s"
It will print short lines with colors, like:
In case someone wants more than just git one-line log:
git log --reverse | awk 'NR>1 {print last} {last=$0}; /^commit/ && ++c==11{exit}'
where the 11 at the end should be set to 1 more than the number of commits you want.
As here points out git log --reverse -n 10 doesn't work as you need it to. (I suppose it would need to be non-commutative to give you the ability to chose the first 10 commits in reverse order or the last 10 commits 🤪)
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