Changelog from specific day to day

Question

Using this script to get day by day changelog

echo "CHANGELOG"
echo ----------------------
git log --no-merges --format="%cd" --date=short | sort -u -r | while read DATE ; do
    echo
    echo [$DATE]
    GIT_PAGER=cat git log --no-merges --format=" * %s" --since="2019-12-15 00:00:00" --until="$DATE 24:00:00"
done

This will echo ECHO even if i don't want to get those commits,

My question here i don't want to get those date before 2019-12-15. I only want to print the date from 2019-12-15 to this day.

Right now i'm getting what i want but whit the previous date empty, i know it's related to the while condition.

So my question is to only get the date and the commit since 2019-12-15.

at the moment i'm getting All the date before 2019-12-15 ( without their commits.

And also getting all the commits + date since 2019-12-15 ( This is perfect)

[2019-17-12]
----COMMITS--
[2019-16-12]
----COMMITS--
[2019-15-12]
--COMMITS--
[2019-14-12]
[2019-13-12]
[2019-12-12]
[2019-11-12]

I want it like this

[2019-17-12]
----COMMITS--
[2019-16-12]
----COMMITS--
[2019-15-12]
--COMMITS--

Thank you

Answer 1

Capture the output first. If it's empty, exit with 0. If it's not, print the date and the output.

echo "CHANGELOG"
echo ----------------------
git log --no-merges --format="%cd" --date=short | sort -u -r | while read DATE ; do
    output=$(GIT_PAGER=cat git log --no-merges --format=" * %s" --since="2019-12-15 00:00:00" --until="$DATE 24:00:00")
    if [[ -z $output ]];then
        exit 0
    else
        echo
        echo [$DATE]
        echo "$output"
    fi
done

And after reading @phd comment, I realize that you can just add --since="2019-12-15 00:00:00" to the first git log.

echo "CHANGELOG"
echo ----------------------
git log --no-merges --format="%cd" --date=short --since="2019-12-15 00:00:00"| sort -u -r | while read DATE ; do
    echo
    echo [$DATE]
    GIT_PAGER=cat git log --no-merges --format=" * %s" --since="2019-12-15 00:00:00" --until="$DATE 24:00:00"
done