我本来以为可以使用 docker-compose run -f <使用这个 Docker Compose 文件>，但是它却给了我一个使用错误。
我该怎么做呢？
以下是我的输出：
ls docker-compose-file.yml
docker-compose-file.yml
me@host:~/work/my-dir$ docker-compose -f docker-compose-file.yml run
Run a one-off command on a service.
For example:
$ docker-compose run web python manage.py shell
By default, linked services will be started, unless they are already
running. If you do not want to start linked services, use
`docker-compose run --no-deps SERVICE COMMAND [ARGS...]`.
Usage:
run [options] [-v VOLUME...] [-p PORT...] [-e KEY=VAL...] [-l KEY=VALUE...] [--]
SERVICE [COMMAND] [ARGS...]
Options:
-d, --detach Detached mode: Run container in the background, print
new container name.
--name NAME Assign a name to the container
--entrypoint CMD Override the entrypoint of the image.
-e KEY=VAL Set an environment variable (can be used multiple times)
-l, --label KEY=VAL Add or override a label (can be used multiple times)
-u, --user="" Run as specified username or uid
--no-deps Don't start linked services.
--rm Remove container after run. Ignored in detached mode.
-p, --publish=[] Publish a container's port(s) to the host
--service-ports Run command with the service's ports enabled and mapped
to the host.
--use-aliases Use the service's network aliases in the network(s) the
container connects to.
-v, --volume=[] Bind mount a volume (default [])
-T Disable pseudo-tty allocation. By default `docker-compose run`
allocates a TTY.
-w, --workdir="" Working directory inside the container
docker-compose run启动一个基于docker-compose.yml文件中的某个容器的一次性容器，但是有不同的命令和其他几个差异。这意味着您需要将服务名称和命令都作为参数传递。您想要完成什么任务；您是否应该使用docker-compose up呢？ - David Maze