docker: specifying a Dockerfile - "docker build" requires exactly 1 argument

Question

This question already has answers here: question from:https://stackoverflow.com/questions/65932455/docker-specifying-a-dockerfile-docker-build-requires-exactly-1-argument

Answer 1

As mentioned in the comments, -f doesn't set the path to the Dockerfile, but the name of the file used as a Dockerfile. you would need to do something like

docker build -t <tag> <path to the folder of the Dockerfile named X> -f X

More details, as mentioned, here.