C# .NET azure-data-lake-gen2 azure-data-lake

试图将文件从Azure Data Lake文件夹转移到另一个

发布于 2025-01-30 03:39:57 字数 810 浏览 3 评论 0原文

我有函数应用程序从数据湖读取文件，并对文件内容进行一些处理。如果失败，则将文件移至“错误”文件夹。我尝试过，但没有成功。解决方案我尝试了 so-solution

public static async Task<DataLakeFileClient> MoveDirectory(string file)
{        
    DataLakeServiceClient serviceClient = await GetDataLakeClient(); 

    DataLakeFileSystemClient filesystemClient = 
    serviceClient.GetFileSystemClient(<CONTAINER>);
    
    DataLakeFileClient fileClient = 
    filesystemClient.GetFileClient("Provision/" + file);
    
    return await fileClient.RenameAsync("Provision/Error/" + file);
}

引起404 sourcePathnotfound。

有什么提示或建议如何将文件从一个目录转移到另一个目录？

原文

I have function app read the file from data lake and do some processing of the file content. If it failed it move the file to the "Error" folder. I tried this but was unsuccessful. Solution I tried SO-solution

public static async Task<DataLakeFileClient> MoveDirectory(string file)
{        
    DataLakeServiceClient serviceClient = await GetDataLakeClient(); 

    DataLakeFileSystemClient filesystemClient = 
    serviceClient.GetFileSystemClient(<CONTAINER>);
    
    DataLakeFileClient fileClient = 
    filesystemClient.GetFileClient("Provision/" + file);
    
    return await fileClient.RenameAsync("Provision/Error/" + file);
}

Cause a 404 SourcePathNotFound.

Any tips or advice how I can move files from one directory to another?

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叹沉浮 2025-02-06 03:39:57

进行了几次更改之后，我们能够完成这项工作。在使用GetFileClient作为所需文件时，您需要使用DataLakeFileSystemClient而不是DataLakeserviceClient。以下是对我有用的代码。

DataLakeServiceClient dataLakeServiceClient = new DataLakeServiceClient("<CONNECTION STRING>");
DataLakeFileSystemClient dataLakeFileSystemClient = dataLakeServiceClient.GetFileSystemClient("<CONTAINER>");

DataLakeFileClient sourceDataLakeFileClient = dataLakeFileSystemClient.GetFileClient("provision/<FILENAME>");
return await sourceDataLakeFileClient.RenameAsync("provision/Error/<FILENAME>");

结果：

执行前

执行后

After making a few changes we could able to get this work. While using GetFileClient for the required file you need to use DataLakeFileSystemClient instead of DataLakeServiceClient. Below is the code that worked for me.

DataLakeServiceClient dataLakeServiceClient = new DataLakeServiceClient("<CONNECTION STRING>");
DataLakeFileSystemClient dataLakeFileSystemClient = dataLakeServiceClient.GetFileSystemClient("<CONTAINER>");

DataLakeFileClient sourceDataLakeFileClient = dataLakeFileSystemClient.GetFileClient("provision/<FILENAME>");
return await sourceDataLakeFileClient.RenameAsync("provision/Error/<FILENAME>");

RESULTS:

Before execution