python - 处理后将文件从 azure 存储容器上传到另一个容器时出现问题

Question

场景是:我在 Azure 存储中有一个 CSV 文件。我想处理该文件的一列(例如，在记录接收的每分钟分离并创建一个新文件)，然后将新文件存储在另一个 azure 存储容器上。

在下面的代码中，我读取一个文件并处理它并创建单独的文件，但是当我想上传时收到此错误:[Errno 2] 没有这样的文件或目录:我的代码是:

import os, uuid
from azure.storage.blob import BlobServiceClient, BlobClient, ContainerClient, __version__
import pandas as pd

try:
    print("Azure Blob Storage v" + __version__ + " - Python quickstart sample")

    accountName = "***"
    accountKey = "*****"
    containerName = "sourcedata"
    blobName = "testdataset.csv"
    urlblob = "https://***.blob.core.windows.net/sorcedata/testdataset.csv"
    connect_str = "******"

    blobService = BlobServiceClient(account_name=accountName, account_key=accountKey, account_url=urlblob)


    # Create the BlobServiceClient object which will be used to create a container client
    blob_service_client = BlobServiceClient.from_connection_string(connect_str)

    # Create a unique name for the container
    container_name = str(uuid.uuid4())

    # Create the container
    container_client = blob_service_client.create_container(container_name)

    df = pd.read_csv(urlblob)

    # create datetime column
    df['datetime'] = pd.to_datetime(df.received_time, format='%M:%S.%f')

    # groupby with Grouper, and save to csv
    for g, d in df.groupby(pd.Grouper(key='datetime', freq='1min')):
        # Create file name
        filename = str(g.time()).replace(':', '')
        # remove datetime column and save CSV file 
        d.iloc[:, :-1].to_csv(f'{filename}.csv', index=False)
        # Create a blob client using the local file name as the name for the blob
        blob_client = blob_service_client.get_blob_client(container=container_name, blob=filename)
        print("\nUploading to Azure Storage as blob:\n\t" + filename)
        # Upload the created file
        with open(filename, "rb") as data:
            blob_client.upload_blob(data)

  

except Exception as ex:
    print('Exception:')
    print(ex)

Answer 1

你应该这样编写代码:

with open(filename + ".csv", "rb") as data:

filename 只是您的文件名，不带后缀。它是不完整的，所以当python打开这个文件时，它找不到该文件。

结果图像: