Docker build-arg和copy

Question

试图复制文件夹内容,它适用于我硬编码路径,如:

COPY ./my-folder /path/to/location

但是需要能够改变这条路径,所以我尝试使用这样的构建参数:

COPY ${folderVariable} /path/to/location

然后用

--build-arg folderVariable=./my-folder

但是当我只想要"my-folder"的内容时,它会将所有内容复制到与"my-folder"相同的文件夹中

Answer 1

在使用之前,您需要使用ARGin 来定义它Dockerfile:

FROM alpine:3.3

ARG folderVariable=./my-folder # Optional default value to be `./my-folder`

COPY ${folderVariable} /opt/my-folder

建立它像:

docker build --build-arg folderVariable=./folder-copy -t test .

更多细节请参考:https://github.com/docker/docker/blob/master/docs/reference/builder.md#arg

Answer 2

扩展仍然不适用于这种COPY --from=$var ...情况。但是您可以创建中间图像作为别名，如下所示：

ARG docsBranch=4.5
ARG docsFullPath=registry.myCompany.pro/group/project-docs/docs:$docsBranch

# Lifehack
FROM $docsFullPath as docs

FROM node:10.21.0-buster-slim
WORKDIR /app

# Now we can use docs instead of $docsFullPath
COPY --from=docs /app/html/en ./documentation/en