Expose workspace names through bazel info

[dinowernli]

It would be really nice to have bazel info expose the name of the workspace in a similar way that bazel info workspace exposes the path to the root of the workspace. Perhaps as additional output to bazel info workspace, or maybe a bazel info workspace-name?

[roysaurav]

@dinowernli @aehlig @mhlopko @buchgr Can you provide an example of the above situtation?

[kchodorow]

I'm actually working on a change right now that will make the execution root basename be the workspace name (so basename $(bazel info execution_root) would give you the workspace name). Would that be sufficient for your use case?

[dinowernli]

Yes, that would be great. Thanks.

[kchodorow]

Marking as a dup of #1681, then. Hopefully should be fixed in a couple of days (won't make it into 0.4.5, though).

[EricCousineau-TRI]

I have made a hacky workaround for people with Bazel <=0.4.5 (myself included):
https://github.com/EricCousineau-TRI/repro/blob/860323d/shell/bazel_workspace_name.py

Slow, but it appears to work.

[dpetrou-continua]

Hi. I'd like to get the workspace name from a Python script that I bazel run. The script calls bazel info using subprocess. This fails because the script's CWD when it runs is not the source tree, so bazel info execution_root fails:

WARNING: Invoking Bazel in batch mode since it is not invoked from within a workspace (below a directory having a WORKSPACE file).
ERROR: The 'info' command is only supported from within a workspace (below a directory having a WORKSPACE file).

You may wonder why I need to bazel run the script instead of executing it directly: It's because it uses imports that are pulled in via pip_parse().

Would love any guidance.
Thanks,
David

[dpetrou-continua]

Nevermind; if I run the script from the root, ala ./bazel-bin/package/target it works.

[guw]

This seems to be broken with bzlmod.

I tried (without success):

• MODULE.bazel using module(name="foo")
• WORKSPACE.bzlmod and defining workspace(name="foo")
• MODULE.bazel using module(..., repo_name="foo")
• keeping WORKSPACE using workspace(name="foo")

However, the information in execution root is always _main.

The documentation in the migration guide says:

the default repository name for your workspace is __main__

The documentation for module function says:

By default, the name of the repo is the name of the module.

My goal is detecting the workspace name. I have been following guidance given in the issue here but seems to be no longer applicable.

There are three possibilities:

• __main__
• _main
• name of the module

[fmeum]

@guw The term "workspace name" is extremely overloaded at this point. Which name do you really need and what do you need it for?

[guw]

The immediate need is to allow a human to identify a workspace in an IDE (this is my "foo").

I am currently also using it to customize classpath resolution within IDEs in case of override_repository. But that needs to be investigated separately as I found additional bits of breaking changes.

So let's define workspace name as per the initial ask, which I assume is workspace( name = "foo" ).

[fmeum]

As a human-readable identifier, I would recommend either the name of the containing directory (which Bazel uses to name the bazel-foo convenience symlink, also used by IntelliJ) or the name of the main module (which you can get with native.module_name in the main repository).

[guw]

Thanks a lot @fmeum. Would native.module_name return the repo_name or just name of the module metadata?

[fmeum]

Thanks a lot @fmeum. Would native.module_name return the repo_name or just name of the module metadata?

It always returns the name, i.e. the one you would also find the module under in a registry.