Can docker compose skip build if the image is pres

问题:

Is it possible to run docker-compose up in such a way that it will only build if the image isn't present in the repository?

I'm working on a test scenario that will run in two different environments: locally on a laptop, and on a server as part of test automation. My docker-compose file looks something like this:

services:
  my-service1:
    build: "./local-repo1"
    image: "image1"
  my-service2:
    build: "./local-repo2"
    image: "image2"
  ...

If I run it locally where the local-repo directories exist it runs fine. If I try to run it on the server where it instead pulls from a docker repository it complains that it cannot find the build path. If I take out the build property it runs fine on the server, but then it won't run locally unless I build the images beforehand.

Is there a way to make it only try to build if the image doesn't already exist? If not I can try some workarounds, but I'd prefer to use just one docker-compose file that handles each case.

回答1:

You can use docker-compose pull to fetch images. Then if they are present already, Compose will not try to build them again.

To be really sure to avoid rebuilds, you can use --no-build.

docker-compose pull
docker-compose up -d --no-build

Your real problem is that you are specifying a build context, but then trying to use docker-compose without that build context being present.

When docker-compose runs, even if it has no plan to do a build, it will verify the build context at least exists. If it doesn't, then it will fail.

All you need to do to satisfy this requirement is create an empty directory for any missing build context. That should make docker-compose happy enough to run.

mkdir -p local-repo1 local-repo2