Bazel | How to copy resources to build directory?

Question

I am making some openGL project, and want to just copy one of my directory into build directory (I store my textures there).

So basically this is what my project structure looks like:

|-WORKSPACE
|-/src/
|  -BUILD
|  -main.cpp
|  -*some folders here*
|-/resources/
|  -BUILD
|  -*some folders here*

All i want is to remain the same relation between directories

This is what i tried:

# src/BUILD file - I use it to build the whole program

cc_binary(
    name = "OpenGL_Project",
    srcs = ["main.cpp"],
    deps = ["//src/renderer:renderer", "//src/scene", "//src/input", "//src/gui"],
    data = ["//resources:resources"]
)

genrule(
    name = "copy_resources",
    srcs = ["//resources"],
    outs = ["resources"],
    cmd = "cp -r $(SRCS) $(OUTS)"
)

And

# resources/BUILD file

filegroup(
    name = "resources",
    srcs = glob(["shaders/**","textures/**"]),
    visibility = ["//visibility:public"],
)

I don't get any errors during build, i tried cleaning it using

bazel clean --expunge

and building again - but it didn't seem to work. Important to add, there is NO resources folder at build directory at all, not that it's in the wrong place.

Do you guys have any ideas what's wrong ?

Answer 1

bzlws_copy does the job. See here

Add to your WORKSPACE file:

load("@bazel_tools//tools/build_defs/repo:http.bzl", "http_archive")
load("@bazel_tools//tools/build_defs/repo:git.bzl", "git_repository")

http_archive(
    name = "bzlws",
    strip_prefix = "bzlws-f929e5380f441f50a77776d34a7df8cacdbdf986",
    url = "https://github.com/zaucy/bzlws/archive/f929e5380f441f50a77776d34a7df8cacdbdf986.zip",
    sha256 = "5bebb821b158b11d81dd25cf031b5b26bae97dbb02025df7d0e41a262b3a030b",
)

Add to your BUILD file:

bzlws_copy(
    name = "copy_resources",
    out = "resources/{FILENAME}",
    force = True,
    tags = ["manual"],
    srcs = [
        "//resources",
    ],
)