Two questions:
the best result comes with combination of both best answers:
git log --pretty=oneline -10
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 --reverse
cat -n
adds line numbers (commit index?)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.
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.
Simply log everything reverse -1 means list one log
git log --reverse -1