使用Azure upload_blob上传Starlette UploadFile报错：非路径类对象

通过客户端向Python编写的GraphQL服务器发送multipart文件，服务器接收为<starlette.datastructures.UploadFile object>。尝试使用Azure Storage的upload_blob方法上传时，出现错误：expected str, bytes or os.PathLike object, not UploadFile。已按照Starlette文档使用file.read()读取文件内容，以下是实现代码及报错堆栈。

原实现代码 #

@mutation.field("fileUpload")
async def resolve_fileUpload(_, info, file):
    print(f"File - {file.filename}")
    bytes_file = await file.read()
    container_client = blob_service_client.get_container_client(
        '4160000000')
    if not container_client.exists():
        container_client.create_container()

    with open(file, "rb") as file:
        result = container_client.upload_blob(
            name='avatar', data=bytes_file)
    return {
        "status": 200,
        "error": "",
        "fileUrl": "www.test.com"
    }

报错堆栈信息 #

[2023-06-07T16:30:47.294Z] expected str, bytes or os.PathLike object, not UploadFile

GraphQL request:3:3
2 |   __typename
3 |   fileUpload(file: $file) {
  |   ^
4 |   __typename
Traceback (most recent call last):
  File "C:\Users\sumch\AppData\Local\Programs\Python\Python310\lib\site-packages\graphql\execution\execute.py", line 528, in await_result
    return_type, field_nodes, info, path, await result
  File "C:\Users\sumch\OneDrive\Projects\Flutter Projects\bol\bol-api\bol-api\user_operations\mutations.py", line 86, in resolve_fileUpload        
    with open(file, "rb") as file:
TypeError: expected str, bytes or os.PathLike object, not UploadFile

The above exception was the direct cause of the following exception:

Traceback (most recent call last):
  File "C:\Users\sumch\AppData\Local\Programs\Python\Python310\lib\site-packages\graphql\execution\execute.py", line 1036, in await_result
    return build_response(await result, errors)  # type: ignore
  File "C:\Users\sumch\AppData\Local\Programs\Python\Python310\lib\site-packages\graphql\execution\execute.py", line 403, in set_result
    results[response_name] = await result
  File "C:\Users\sumch\AppData\Local\Programs\Python\Python310\lib\site-packages\graphql\execution\execute.py", line 535, in await_result
    self.handle_field_error(error, return_type)
  File "C:\Users\sumch\AppData\Local\Programs\Python\Python310\lib\site-packages\graphql\execution\execute.py", line 569, in handle_field_error    
    raise error
  File "C:\Users\sumch\AppData\Local\Programs\Python\Python310\lib\site-packages\graphql\execution\execute.py", line 528, in await_result
    return_type, field_nodes, info, path, await result
  File "C:\Users\sumch\OneDrive\Projects\Flutter Projects\bol\bol-api\bol-api\user_operations\mutations.py", line 86, in resolve_fileUpload        
    with open(file, "rb") as file:
graphql.error.graphql_error.GraphQLError: expected str, bytes or os.PathLike object, not UploadFile

GraphQL request:3:3
2 |   __typename
3 |   fileUpload(file: $file) {
  |   ^
4 |   __typename

错误原因 #

报错核心是代码中with open(file, "rb") as file:这一行：open()函数仅接受文件路径字符串、字节或路径对象，无法直接处理UploadFile实例。而且你已经通过await file.read()将文件内容读取到bytes_file变量中，这段open操作完全多余，反而引发了类型错误。

另外，Azure的upload_blob方法本身支持直接传入字节数据，无需额外打开文件。

修正后的代码 #

@mutation.field("fileUpload")
async def resolve_fileUpload(_, info, file):
    print(f"File - {file.filename}")
    bytes_file = await file.read()
    container_client = blob_service_client.get_container_client('4160000000')
    
    if not container_client.exists():
        container_client.create_container()

    # 直接使用读取到的字节数据上传，移除多余的open操作
    result = container_client.upload_blob(name='avatar', data=bytes_file)
    
    return {
        "status": 200,
        "error": "",
        "fileUrl": "www.test.com"
    }

大文件优化方案（可选） #

如果处理大文件，为避免一次性将文件内容加载到内存，可以直接传入UploadFile的底层类文件对象实现流式上传：

@mutation.field("fileUpload")
async def resolve_fileUpload(_, info, file):
    print(f"File - {file.filename}")
    container_client = blob_service_client.get_container_client('4160000000')
    
    if not container_client.exists():
        container_client.create_container()

    # 直接传入UploadFile的底层文件对象，流式上传节省内存
    result = container_client.upload_blob(name='avatar', data=file.file)
    
    return {
        "status": 200,
        "error": "",
        "fileUrl": "www.test.com"
    }

内容的提问来源于stack exchange，提问作者Sumchans