"path" attribute of filegroup

[Shimin Guo]

I'm having a hard time understanding the "path" attribute of the filegroup rule.

Say I have

filegroup(

    name = "mygroup",

    srcs = [

        "//data:a_file.txt",

        "//data:some/subdirectory/another_file.txt",

    ],

    path = "mydata",

)

cc_binary(

    name = "prog",

    srcs = [ ... ],

    data = [":mygroup"],

)

What does the path attribute do here? It seems the runfiles directory generated for prog is the same where I specify the path or not.

[Greg Estren]

The "path" attribute is exposed through a FilegroupPathProvider, which isn't getting any practical use right now in Bazel (cc_binary doesn't exploit it).

Is there anything specific you need from this attribute or are you just curious?

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[Shimin Guo]

I was trying to pull files from other packages have them appear in a particular directory layout. I guess that's not what the path attribute is for?

[Brian Silverman]

No, it isn't. The way to do pull files in is using a genrule or a custom Skylark rule to copy them.

Something along the lines of this should work for the genrule approach:

files = ["//data:a_file.txt", "//data:some/subdirectory/another_file.txt"]

genrule(

  name = 'copy_the_files',

  srcs = files,

  outs = [f.replace('the_directory', 'another_place/folder') for f in files],

  cmd = '\n'.join(['mkdir -p $$(dirname $(location %s)) && cp $(location %s) $(location :%s)' % (f, f, f.replace('the_directory', 'another_place/folder')) for f in files]),

)

Depending on exactly what you're doing, writing that (or at least the part that does the actual renaming) as a skylark macro and/or putting the output names in a separate list and using zip when generating the cmd to avoid duplicating that logic might make sense.

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[Shimin Guo]

Got it. Thanks for the help.