writing a one-liner to step back one commit in a git repo

Question

I want to step backwards, commit-by-commit, to find some css styling that I overwrote. I'm aware of git-bisect; I used it and it didn't help me find my old work.

I can get a table of commit hashes in historical order with this command:

$ git log --pretty=oneline | awk '{print $1;}'

f3791f1da33b9e92a3a5292a0542834aca0908ef
2720a9ec4bb5b4f7b4dd02b056c375e71a44be3c
a9bb455608db70c76545e5862b89a02f9e9f590d

So I just need to grab the first one, and do a git checkout with it. I've tried to use head, but I don't know enough about pass syntax to pass the output correctly to it:

$ $(git log --pretty=oneline | awk '{print $1;}') | head -1
bash: 0dcca7e530fca682552341718462af150fe27dc5: command not found

$ $(git log --pretty=oneline | awk '{print $1;}') > head -1
bash: 0dcca7e530fca682552341718462af150fe27dc5: command not found

How can I get the top of the output from my previous command

I'd like this as a one-liner, so that I can add it to my git aliases, but if it's not, that's fine; I'll make a bash function.

Answer 1

To check out the parent commit, run

git checkout HEAD~1

This can be repeated ad nauseam, until you reach the first commit.

If you need to explore different parents leading up to a merge, use ^ instead:

git checkout HEAD^2

will check out the second parent of the current merge commit.

See “Specifying revisions” in the git-rev-parse documentation for details.