如何在 jenkinsfile 中为 docker.build 设置自定义上下文

Question

I'm trying to adapt a docker build for jenkins. I'm following our docker-compose file and I'm creating a Jenkinsfile that is creating each container and linking them together. The problem I'm running into is that the docker-compose files declare a context that is not where the Dockerfile is. As far as I understand, jenkins will set the context to where the Dockerfile is, which puts files to be copied in a different relative location depending on whether the jenkinsfile or docker-compose file is building.

The folder structure is:

workspace
    |-docker
        |-db
           |-Dockerfile
           |-entrypoint.sh

This is how the Dockerfile declares the COPY instruction for the file in question

COPY docker/db/entrypoint.sh /

This is how my jenkinsfile builds the file. Which to my knowledge puts the context at that directory

docker.build("db", "${WORKSPACE}/docker/db")

the docker-compose file declares it like:

db:
build:
  context: .
  dockerfile: docker/db/Dockerfile

which puts the context at the root of the project.

Is there any way to tell a jenkinsfile to use the same context as the docker-compose file so that the Dockerfile's COPY instruction can remain unchanged and valid for both Jenkins and docker-compose? If that's not possible, does anyone know of any alternative solutions?

Answer 1

It turns out that I was pulling a previous version (1.6) of the docker-pipeline-plugin. The function in question has been since updated (1.7) to allow the second parameter to designate a Dockerfile location outside the context.

The updated statement in my Jenkinsfile is:

return docker.build("db", "-f docker/db/Dockerfile .")

And this allows my container to build without modifying the expected context of the developer's docker-compose or Dockerfiles.

Answer 2

The build() method builds the Dockerfile in the current directory by default. This can be overridden by providing a directory path containing a Dockerfile as the second argument of the build() method, for example:

node {
    checkout scm
    def testImage = docker.build("test-image", "./dockerfiles/test") 

    testImage.inside {
        sh 'make test'
    }
}

Answer 3

To specify the path of the dockerfile the dir{} block worked for me

stage('docker build') {
            steps {
                dir('/path/to/docker/file/'){
                    script{
                        dockerImage = docker.build("imageName")
                    }
                }
            }
        }

Answer 4

The best way is to use the dir block because you'll need to perform actions like COPY, where you must be in the location

So Use this Code

stage('Build') {
        steps {
            dir('/path/to/docker/file/'){ // put the path of Dockerfile
                script{
                    my_images = docker.build("Your_Image")
                }
            }
        }
    }

Answer 5

只需在 docker docker-compose上 Omnit context项，它就会默认为 Dockerfile 的位置。

Answer 6

以下语法应该有效： docker.build('ContainerTag''-f DockerFilePath Context' 示例：docker.build('quay.io/AppContianer:1.0.0' '-f dockerfileDir/Dockerfile.test dockerfileDir')