Git merge conflict to always take the newest file

Question

How can I make a git conflict ALWAYS get resolved by taking the newest file (latest changed timestamp), completley without a prompt?

I am building a syncscript with

Answer 1

Based on @Lilleman answer which is good but has a bug: if a file is modified on local master which is older than origin/master the expected behaviour is to pick origin/master but if you add a commit on top of local master it will pick the local (older version) of the file. git log needs to be executed with an extra argument which is the file path.

here a better version including variable branch names it's a bit hacky but it works

merge

#!/bin/sh
cd `findup .git`
PATH=$PATH:$( dirname "${BASH_SOURCE[0]}" )
HEAD=`git rev-parse --abbrev-ref HEAD`

echo "* merge=newest" > .gitattributes
sed -ie '/merge "newest"/,+2d' .git/config
echo "[merge \"newest\"]" >> .git/config
echo -e "\tname = Merge by newest commit" >> .git/config
echo -e "\tdriver = git-merge-newest %O %A %B %L %P $HEAD $1" >> .git/config
git merge $1
sed -ie '/merge "newest"/,+2d' .git/config
rm .gitattributes
cd - 

git-merge-newest

#!/bin/sh
if git merge-file -p -q "$2" "$1" "$3" > /dev/null;
        then git merge-file "$2" "$1" "$3";
        else
                MINE=$(git log --format="%ct" --no-merges "$6" -1 $5);
                THEIRS=$(git log --format="%ct" --no-merges "$7" -1 $5);
                if [ $MINE -gt $THEIRS ]; then
                  git merge-file -q --ours "$2" "$1" "$3" >/dev/null
                else
                  git merge-file -q --theirs "$2" "$1" "$3">/dev/null
                fi
fi

findup used

#!/bin/sh

pwd="`pwd`"
start="$pwd"
while [ ! "$pwd" -ef .. ]; do
[ -e "$1" ] && echo -n "$pwd" && exit
cd .. || exit 1
pwd="`pwd`"
done
exit 1