Use a executable if it exists, otherwise, use another-CodePudding

As you know may noticed, Docker changed the name of compose from docker-compose to docker compose

I have a Makefile that calls docker-compose

run:
    docker-compose up --build

However I want to make my Makefile portable, I was wondering if it is possible to the Makefile first tries if docker-compose exists, if not, uses docker compose

Is it possible?

CodePudding user response：

If you want it to be the most portable then you'd implement it in the shell, something like:

run:
        test -n "$$(command -v docker-compose)" \
            && docker-compose up --build \
            || docker compose up --build

If you're willing to use make-specific features you can do something a bit fancier such as:

ifeq ($(shell command -v docker-compose;),)
    COMPOSE := docker compose
else
    COMPOSE := docker-compose
endif

run:
        $(COMPOSE) up --build