Bazel：输出目录的genrule

我刚刚开始与 Bazel 合作。所以，我提前道歉，我没能弄清楚这一点。

我正在尝试运行一个命令，将一堆文件输出到一个目录，并使该目录可用于后续目标。我有两种不同的尝试：

1. 使用正则
2. 写我自己的规则

我天真地希望能用一个来做到这一点genrule。但是，您似乎不能说“我不知道这个命令将输出什么”并将目录放入outs。现在我正在尝试编写一个可以使用的规则ctx.actions.declare_directory但我还没有完全理解。我似乎无法理解tools从我的工作空间进入我的规则。

我的 genrule 尝试看起来像这样：

genrule(
    name = "doit",
    srcs = [
        "doitConfigA",
        "doitConfigB",
    ],
    cmd = 'HOME=. ./$(location path/to/doit) install',

    # Neither of the below outs work - seems like bazel wants to know
    # exactly this list of files. I don't know the files that
    # will be output ahead of time.

    # This one looks at the `out_dir` that I already have and
    # expects the files to be the same which they might not be
    outs = glob(["out_dir/**/*.*"]),

    # this fails with:
    # "declared output 'out_dir' was not 
    # created by genrule. This is probably because the genrule actually 
    # didn't create this output, or because the output was a directory 
    # and the genrule was run remotely (note that only the contents of 
    # declared file outputs are copied from genrules run remotely)"
    outs = ['out_dir'],
    tools = ['path/to/doit'],
)

我的自定义规则尝试如下所示：

def _impl(ctx):
  dir = ctx.actions.declare_directory("out_dir")

  ctx.actions.run_shell(
      outputs=[dir],
      progress_message="Running doit install ...",
      command="HOME=. ./path/to/doit install",
      tools=[ctx.attr.tools],
  )

doit = rule(
    implementation=_impl,
    attrs={
      "tools": attr.label_list(allow_files=True),
    },
    outputs={"out": "out_dir"},
)

然后，运行我的doit规则，我的 BUILD 文件如下所示：

doit(
  name = 'doit',
  tools = ['path/to/doit'],
)

在我的规则中，该命令运行，但它不喜欢我尝试使用目录outs， 它似乎。在我的自定义规则中，我似乎无法告诉 Bazel 我想使用./path/to/doit作为我工作空间中的工具，例如expected type 'File' for 'tools' element but got type 'list' instead ...

看来我必须缺少一些基本的东西，因为这肯定是运行命令并将一堆未知内容输出到目录的常见情况？

的输出genrule 必须是固定的文件列表 https://docs.bazel.build/versions/master/be/general.html#genrule.outs。作为解决方法，您可以从输出目录创建 zip。

我用这种方法来操纵输出yarn install通常的方法不可行的地方：

genrule(
  name = "node_modules",
  srcs = [
    "package.json",
    "yarn.lock",
  ],
  cmd = " && ".join([
    "yarn install --pure-lockfile",
    "zip -r $@ node_modules",
  ]),
  outs = [
    "node_modules.zip",
  ],
)

然后是消耗 zip 的规则：

# Rule that generates a list of the folders in node_modules
genrule(
  name = "node_modules_ls",
  srcs = [
    ":node_modules",
  ],
  cmd = " && ".join([
    "unzip $(location :node_modules) -d . ",
    "ls > $@",
  ]),
  outs = [
    "out.txt",
  ],
)