Azure -文件共享中具有给定扩展名的文件

Question

我正在使用this连接到Azure文件共享并上载文件。我想选择扩展名文件的扩展名，但我不能。我得到了一个错误显示如下。如果我删除.txt，一切都会正常工作。有没有办法在上传时指定文件扩展名？

错误：

Exception: ResourceNotFoundError: The specified parent path does not exist.

代码：

def main(blobin: func.InputStream):
    file_client = ShareFileClient.from_connection_string(conn_str="<con_string>", 
                                                            share_name="data-storage", 
                                                            file_path="outgoing/file.txt")


    f = open('/home/temp.txt', 'w+')
    f.write(blobin.read().decode('utf-8'))
    f.close()

    # Operation on file here

    f = open('/home/temp.txt', 'rb')
    string_to_upload = f.read()
    f.close()

    file_client.upload_file(string_to_upload)

Answer 1

我认为您收到此错误的原因是因为您的文件服务共享中不存在outgoing文件夹。我拿了你的代码，并运行它，无论有没有扩展，在这两种情况下，我得到了相同的错误。

然后我创建了一个文件夹，并尝试上传文件，我成功地做到了。

下面是我使用的最终代码：

from azure.storage.fileshare import ShareFileClient, ShareDirectoryClient

conn_string = "DefaultEndpointsProtocol=https;AccountName=myaccountname;AccountKey=myaccountkey;EndpointSuffix=core.windows.net"

share_directory_client = ShareDirectoryClient.from_connection_string(conn_str=conn_string, 
                                                        share_name="data-storage",
                                                        directory_path="outgoing")

file_client = ShareFileClient.from_connection_string(conn_str=conn_string, 
                                                        share_name="data-storage", 
                                                        file_path="outgoing/file.txt")

# Create folder first.
# This operation will fail if the directory already exists.
print "creating directory..."
share_directory_client.create_directory()
print "directory created successfully..."

# Operation on file here
f = open('D:\\temp\\test.txt', 'rb')
string_to_upload = f.read()
f.close()

#Upload file
print "uploading file..."
file_client.upload_file(string_to_upload)
print "file uploaded successfully..."