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zip - 如何在 bazel 中压缩文件

转载 作者:行者123 更新时间:2023-12-01 09:10:26 25 4
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我有一组文件作为我的存储库的一部分。如何从 bazel 中的这些文件中生成 zip 文件。我找到了 tar.gz 等的规则,但找不到如何获得 zip 存档的方法。

找到提到 zipper 的引用资料,但无法弄清楚如何加载和使用它。对bazel更有经验的人可以提供帮助吗?

最佳答案

zipper 实用程序位于 @bazel_tools//tools/zip:zipper ,这是它的用法:

Usage: zipper [vxc[fC]] x.zip [-d exdir] [[zip_path1=]file1 ... [zip_pathn=]filen]
v verbose - list all file in x.zip
x extract - extract files in x.zip to current directory, or
an optional directory relative to the current directory
specified through -d option
c create - add files to x.zip
f flatten - flatten files to use with create or extract operation
C compress - compress files when using the create operation
x and c cannot be used in the same command-line.

For every file, a path in the zip can be specified. Examples:
zipper c x.zip a/b/__init__.py= # Add an empty file at a/b/__init__.py
zipper c x.zip a/b/main.py=foo/bar/bin.py # Add file foo/bar/bin.py at a/b/main.py

If the zip path is not specified, it is assumed to be the file path.

所以它可以像这样用在 genrule 中:
$ tree
.
├── BUILD
├── dir
│   ├── a
│   ├── b
│   └── c
└── WORKSPACE

1 directory, 5 files


$ cat BUILD
genrule(
name = "gen_zip",
srcs = glob(["dir/*"]),
tools = ["@bazel_tools//tools/zip:zipper"],
outs = ["files.zip"],
cmd = "$(location @bazel_tools//tools/zip:zipper) c $@ $(SRCS)",
)


$ bazel build :files.zip
INFO: Analyzed target //:files.zip (7 packages loaded, 41 targets configured).
INFO: Found 1 target...
Target //:files.zip up-to-date:
bazel-bin/files.zip
INFO: Elapsed time: 0.653s, Critical Path: 0.08s
INFO: 1 process: 1 linux-sandbox.
INFO: Build completed successfully, 2 total actions


$ unzip -l bazel-bin/files.zip
Archive: bazel-bin/files.zip
Length Date Time Name
--------- ---------- ----- ----
0 2010-01-01 00:00 dir/a
0 2010-01-01 00:00 dir/b
0 2010-01-01 00:00 dir/c
--------- -------
0 3 files

它可以类似地用于 Starlark:
def _some_rule_impl(ctx):

zipper_inputs = []
zipper_args = ctx.actions.args()
zipper_args.add("c", ctx.outputs.zip.path)
....
ctx.actions.run(
inputs = zipper_inputs,
outputs = [ctx.outputs.zip],
executable = ctx.executable._zipper,
arguments = zipper_args,
progress_message = "Creating zip...",
mnemonic = "zipper",
)


some_rule = rule(
implementation = _some_rule_impl,
attrs = {
"deps": attr.label_list(),
"$zipper": attr.label(default = Label("@bazel_tools//tools/zip:zipper"), cfg = "host", executable=True),
},
outputs = {"zip": "%{name}.zip"},
)

关于zip - 如何在 bazel 中压缩文件,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/57914937/

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