git - How do I programmatically determine if there are uncommited changes? -

in makefile, i'd perform actions if there uncommited changes (either in working tree or index). what's cleanest , efficient way that? command exits return value of 0 in 1 case , non-zero in other suit purposes.

i can run git status , pipe output through grep, feel there must better way.

update: op daniel stutzbach points out in comments simple command git diff-index worked him:

git diff-index --quiet head -- 

you can see "how check if command succeeded?" if using in bash script:

git diff-index --quiet head -- || echo "untracked"; // 

note: commented anthony sottile

git diff-index head ... fail on branch has no commits (such newly initialized repository).
1 workaround i've found git diff-index $(git write-tree) ...

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"programmatically" means never ever rely on porcelain commands.
always rely on plumbing commands.

see "checking dirty index or untracked files git" alternatives (like git status --porcelain)

you can take inspiration from new "require_clean_work_tree function" written speak ;) (early october 2010)

require_clean_work_tree () {     # update index     git update-index -q --ignore-submodules --refresh     err=0      # disallow unstaged changes in working tree     if ! git diff-files --quiet --ignore-submodules --             echo >&2 "cannot $1: have unstaged changes."         git diff-files --name-status -r --ignore-submodules -- >&2         err=1     fi      # disallow uncommitted changes in index     if ! git diff-index --cached --quiet head --ignore-submodules --             echo >&2 "cannot $1: index contains uncommitted changes."         git diff-index --cached --name-status -r --ignore-submodules head -- >&2         err=1     fi      if [ $err = 1 ]             echo >&2 "please commit or stash them."         exit 1     fi }