是否需要旧版云存储权限？

Question

我的困惑是为什么我需要包括“传统”云存储角色。我更喜欢避免说“遗留”的东西，因为听起来它们会在这些天中被弃用。我做错了吗？

这是我的情况：

我正在使用 appengine 项目中的服务帐户访问另一个项目中云存储中的文件。我正在使用 Google Python 客户端访问数据。

我分配了角色：

Storage Object Creator
Storage Object Viewer

但是当我尝试访问文件时出现错误：

<service account> does not have storage.buckets.get access

只有在我添加了“遗留角色”之后，它才最终可以访问：

Storage legacy bucket writer
Storage legacy bucket reader

这是代码：

def download_blob(bucket_name, source_blob_name, destination_file_name):
    """Downloads a blob from the bucket."""
    bucket = storage_client.get_bucket(bucket_name)
    blob = bucket.blob(source_blob_name)
    blob.download_to_filename(destination_file_name)

    print('Blob {} downloaded to {}.'.format(
          source_blob_name,
          destination_file_name))

def upload_blob(bucket_name, source_file_name, destination_blob_name):
    """Uploads a file to the bucket."""
    bucket = storage_client.get_bucket(bucket_name)
    blob = bucket.blob(destination_blob_name)
    blob.upload_from_filename(source_file_name)

    print('File {} uploaded to {}.'.format(
          source_file_name,
          destination_blob_name))

谢谢罗布

Answer 1

对于您的代码，我在下面添加了其他注释：

def download_blob(bucket_name, source_blob_name, 
destination_file_name):
    """Downloads a blob from the bucket."""
    """The following .get_bucket() requires storage.buckets.get permission."""
    bucket = storage_client.get_bucket(bucket_name)
    """The following doesn't"""
    bucket = storage_client.bucket(bucket_name)
    blob = bucket.blob(source_blob_name)
    blob.download_to_filename(destination_file_name)

    print('Blob {} downloaded to {}.'.format(
          source_blob_name,
          destination_file_name))

重新迭代：

storage_client.get_bucket(bucket_name) 需要 storage.bucket.get 权限，因为它正在执行存储桶元数据 GET 请求。

storage_cilent.bucket(bucket_name)不需要此权限，因为它不执行 GET 请求，只创建一个名称由bucket_name.

对于上传绕过 storage.buckets.get 问题：

from google.cloud import storage
storage_client = storage.Client()
bucket = storage_client.bucket(bucket_name)
blob = bucket.blob(source_blob_name)
blob.upload_from_filename(source_file_name)