找到Bazel输出文件？
当我运行Bazel Build时，Bazel报告了目标的输出文件。我该如何编程查询？
在此示例中，我想获得bazel-out /...路径。
$ bazel build foo/bar:registration
INFO: Analyzed target //foo/bar:registration (546 packages loaded, 13485 targets configured).
INFO: Found 1 target...
Target //foo/bar:registration up-to-date:
bazel-out/k8-fastbuild-ST-a82eca4c6b7c/bin/foo/bar/registration/registration_/registration
When I run bazel build, bazel reports the target's output file. How can I programmatically query this?
In this example, I want to get the bazel-out/... path.
$ bazel build foo/bar:registration
INFO: Analyzed target //foo/bar:registration (546 packages loaded, 13485 targets configured).
INFO: Found 1 target...
Target //foo/bar:registration up-to-date:
bazel-out/k8-fastbuild-ST-a82eca4c6b7c/bin/foo/bar/registration/registration_/registration
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通过使用
cquery命令的 starlark”> starlark输出模式。例如，要获得类似于构建发出的输出，请使用：将来 ，
cquery也可能有一个专用选项，可以处理简单的情况，而无需在命令行上涉及Starlark。This can be accomplished with great flexibility by using the
cquerycommand's Starlark output mode. For instance, to obtain output similar to that emitted bybuild, use:In the future, there may also be a dedicated option in
cquerythat handles the simple cases without the need to involve Starlark on the command line.更新：自Bazel 5.3以来，确实有一种较短的方法。
Update: there's indeed a shorter way since Bazel 5.3.