如果我不知道zip的内容，如何在bazel中正确解压缩文件? [英] How do I unzip a file in bazel properly if I don't know the contents of the zip?

问题描述

我要定义一个规则来解压缩给定的zip文件.但是，我不知道zip的内容，因此，例如，我不能在genrule中指定outs.这似乎是一个常见的问题，并且四处搜寻会导致我遇到类似情况的人，但是我还没有看到解决该问题的具体示例.

I was to define a rule that unzips a given zip file. However, I don't know the contents of the zip, so I cannot specify outs in a genrule, for example. This seems like a common problem, and googling around leads me to people who have encountered similar scenarios, but I haven't yet seen a specific example of how to solve this.

我想要类似的东西:

genrule(
  name="unzip",
  src="file.zip",
  outs=glob(["**"]), # except you're not allowed to use glob here
  cmd = "unzip $(location file)",
)

推荐答案

您可以使用工作区规则，用于为所有内容压缩的zip文件创建一个BUILD文件.

You could use a Workspace Rule to create a BUILD file for the zip that globs everything.

WORKSPACE文件中的类似内容

new_http_archive(
    name = "my_zip",
    url = "http://example.com/my_zip.zip",
    build_file_content = """
        filegroup(
            name = "srcs",
            srcs = glob(["*"]),
            visibility = ["//visibility:public"]
        )
    """
)

然后从BUILD文件中，您可以使用@my_zip//:srcs

Then from a BUILD file you can reference this as an input using @my_zip//:srcs

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