Bazel: how to glob headers into one include path

In Buck, one might write:

exported_headers = subdir_glob([
    ("lib/source", "video/**/*.h"),
    ("lib/source", "audio/**/*.h"),
],
excludes = [
    "lib/source/video/codecs/*.h",
],
prefix = "MediaLib/")

This line would make those headers available under MediaLib/. What would be the equivalent in Bazel?

I ended up writing a rule to do this. It provides something similar to the output of a filegroup, and could be combined with cc_library in a macro.

def _impl_flat_hdr_dir(ctx):
    path = ctx.attr.include_path
    d = ctx.actions.declare_directory(path)
    dests = [ctx.actions.declare_file(path + "/" + h.basename)
             for h in ctx.files.hdrs]

    cmd = """
        mkdir -p {path};
        cp {hdrs} {path}/.
        """.format(path=d.path, hdrs=" ".join([h.path for h in ctx.files.hdrs]))

    ctx.actions.run_shell(
       command = cmd,
       inputs = ctx.files.hdrs,
       outputs = dests + [d],
       progress_message = "doing stuff!!!"
    )

    return struct(
       files = depset(dests)
    )

flat_hdr_dir = rule(
    _impl_flat_hdr_dir,
    attrs = {
        "hdrs": attr.label_list(allow_files = True),
        "include_path": attr.string(mandatory = True),
    },
    output_to_genfiles = True,
)