“ docker build”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;我正在按照在线说明进行操作 docker build
,但出现以下错误:
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:
" docker build"恰好需要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
我尝试过以上命令的各种组合,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-正如我按照文档中的说明一样?
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(与常规 Dockerfile
不同)。它不是用于传递完整路径到 docker build
。路径是第一个参数。
Parameter -f
changes name of the Dockerfile (when it's different than regular Dockerfile
). It is not for passing full path to docker build
. Path goes as first argument.
语法是:
docker build [PARAMS]路径
因此,在您的情况下,这应该可行:
So in your case this should work:
docker build -f MyDockerfile -t proj:myapp / full / path / to /
或者如果您在项目目录中,您只需要使用一个点:
or in case you are in project directory, you just need to use a dot:
docker build -f MyDockerfile -t proj:myapp。
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