makefile - 如何在 Makefile 中有动态目标？

Question

看看下面的这个 Makefile。

compose:
    docker-compose up myapp

compose-shell:
    docker-compose run myapp /bin/bash

compose-shellplus:
    docker-compose run myapp make shell

compose-test:
    docker-compose run myapp make test

compose-migrate:
    docker-compose run myapp make migrate

compose-load:
    docker-compose run myapp make load

compose-export:
    docker-compose run myapp make export

compose-flush:
    docker-compose run myapp make flush

# run tests
test:
    python manage.py test --settings=$(PROJECT_SETTINGS)

# install depedencies (and virtualenv for linux)
install:
ifndef WIN
    -virtualenv -p python3 .venv
endif
    pip install -r requirements.txt

# handle django migrations
migrate:
    python manage.py makemigrations --settings=$(PROJECT_SETTINGS)
    python manage.py migrate --settings=$(PROJECT_SETTINGS)

# handle statics
static:
    python manage.py collectstatic --settings=$(PROJECT_SETTINGS)

shell:
    python manage.py shell_plus --settings=$(PROJECT_SETTINGS)

load:
    python manage.py loaddata db.json --settings=${PROJECT_SETTINGS}

export:
    python manage.py dumpdata --indent 2 --natural-foreign --natural-primary -e sessions -e admin -e contenttypes -e auth.Permission  > db.json --settings=${PROJECT_SETTINGS}

flush:
    python manage.py sqlflush --settings=${PROJECT_SETTINGS}

有没有更有效的方法来做到这一点？

例如:

compose-${target_name_after_dash}:
    docker-compose run myapp make ${target_name_after_dash}

Answer 1

在发布到 SO 之前，最好尝试在 the documentation 中找到答案。这是您可以使用 GNU make 做的最基本的事情之一。

使用 pattern rule :

compose-%:
        docker-compose run myapp make $*