“码头工人构建"恰好需要 1 个参数 [英] "docker build" requires exactly 1 argument(s)

问题描述

我正在尝试从特定的 Dockerfile 构建映像，并同时对其进行标记；我正在关注 在线说明docker build，但我收到以下错误:

I am trying to build an image from a specific Dockerfile, and tag it at the same time; I am following the online instructions fordocker build, but I get the following error:

码头工人构建"只需要 1 个参数

"docker build" requires exactly 1 argument(s)

我的目录结构:

project/
    foo/
    MyDockerfile

这是我运行的命令:

docker build -f/full/path/to/MyDockerfile -t proj:myapp

我尝试了上述命令的各种组合，但结果始终是上面给出的错误消息.为什么会发生这种情况 - 正如我按照文档所说的那样?

I have tried various combinations of the above command, but the results are always the error message given above. Why is this happening - as I am following what the documentation says?

推荐答案

参数 -f 更改 Dockerfile 的名称(当它不同于常规的 Dockerfile 时).它不适用于将完整路径传递给 docker build.路径作为第一个参数.

Parameter -f changes the name of the Dockerfile (when it's different than regular Dockerfile). It is not for passing the full path to docker build. The path goes as the first argument.

语法是:

docker build [PARAMS] PATH

所以在你的情况下，这应该有效:

So in your case, this should work:

docker build -f MyDockerfile -t proj:myapp/full/path/to/

或者如果你在项目目录中，你只需要使用一个点:

or in case you are in the project directory, you just need to use a dot:

docker build -f MyDockerfile -t proj:myapp .

这篇关于“码头工人构建"恰好需要 1 个参数的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！