我想导入在json文件（my_info.json）中定义的变量作为bazel规则的属性。我试过这个方法（https://docs.bazel.build/versions/5.3.1/skylark/tutorial-sharing-variables.html），但不想使用.bzl文件，直接导入变量到BUILD. bazel的属性。
我想使用那些从my_info.json导入的变量作为其他BULD.bazel文件的属性。
projects/python_web/BUILD.bazel

load("//projects/tools/parser:config.bzl", "MY_REPO","MY_IMAGE")
container_push(
  name = "publish",
  format = "Docker",
  registry = "registry.hub.docker.com",
  repository = MY_REPO,
  image = MY_IMAGE,
  tag = "1",
)

在Bazel slack中询问类似的问题时，我被告知，无法直接将变量导入Bazel，需要解析json变量并将其写入.bzl文件。
我也试过这段代码，但没有写在config.bzl文件中。
my_info.json

{
    "MYREPO" : "registry.hub.docker.com",
    "MYIMAGE" : "michael/monorepo-python-web"
}

WORKSPACE.bazel

load("//projects/tools/parser:jsonparser.bzl", "load_my_json")
load_my_json(
    name = "myjson"
)

projects/tools/parser/jsonparser.bzl

def _load_my_json_impl(repository_ctx):
    json_data = json.decode(repository_ctx.read(repository_ctx.path(Label(":my_info.json"))))
    config_lines = ["%s = %s" % (key, repr(val)) for key, val in json_data.items()]
    repository_ctx.file("config.bzl", "\n".join(config_lines))

load_my_json = repository_rule(
    implementation = _load_my_json_impl,
    attrs = {},
)

projects/tools/parser/BUILD.bazel

load("@aspect_bazel_lib//lib:yq.bzl", "yq")
load(":config.bzl", "MYREPO", "MY_IMAGE")

yq(
    name = "convert",
    srcs = ["my_info2.json"],
    args = ["-P"],
    outs = ["bar.yaml"],
)

正在执行：% bazel build projects/tools/parser：convert

ERROR: Traceback (most recent call last):
        File "/Users/michael.taquia/Documents/Personal/Projects/bazel/bazel-projects/multi-language-bazel-monorepo/projects/tools/parser/BUILD.bazel", line 2, column 22, in <toplevel>
                load(":config.bzl", "MYREPO", "MY_IMAGE")
Error: file ':config.bzl' does not contain symbol 'MYREPO'

在进行故障排除时，我看到执行调用了jsonparser.bzl，但从未进入**_load_my_json_impl函数（基于print语句），也没有向config.bzl**写入任何内容。
注意：已在macOS 12.6（21 G115）达尔文Kernel版本21.6.0上测试

There is a better way to do that? A code snippet will be very useful.