如何在U-SQL查询中获取文件的完整路径

Question

是否可以像下面的u-sql代码一样获取文件“完整路径”，就像文件名列一样？

DECLARE @in = "D:/Sample/Data/{FileName}"; 

@drivers = 
    EXTRACT id string, 
      first_name string, 
      last_name string, 
      address string, 
      FileName string 
    FROM @in 
    USING USING Extractors.Csv(); 

OUTPUT @drivers 
TO "/ReferenceGuide/DML/QSE/Extract/Drivers.txt" 
USING Outputters.Csv(); 

从上面的u-sql查询我能够得到文件名，但我想文件的完整路径也。像“D：/ Sample/Data”

Answer 1

那么你已经在脚本的顶部有一个变量的输入路径，所以只需将其添加到输出数据集。就像这样：

DECLARE @in = "D:/Sample/Data/{FileName}"; 

@drivers = 
    EXTRACT id string, 
      first_name string, 
      last_name string, 
      address string, 
      FileName string 
    FROM @in 
    USING USING Extractors.Csv(); 


@driverswithpath = 
    SELECT 
     *, 
     @in AS 'InputPath' 
    FROM 
     @drivers; 

OUTPUT @driverswithpath 
TO "/ReferenceGuide/DML/QSE/Extract/Drivers.txt" 
USING Outputters.Csv(); 

Answer 2

您也可以使用在U型SQL字符串.NET方法，你可以使用Replace方法来重建整个文件路径，这样的事情：

DECLARE @in string = "input/{FileName}"; 

@drivers = 
    EXTRACT id string, 
      first_name string, 
      last_name string, 
      address string, 
      FileName string 
    FROM @in 
    USING Extractors.Csv(); 


@driverswithpath = 
    SELECT 
     *, 
     @in.Replace("{FileName}", FileName) AS InputPath 
    FROM 
     @drivers; 

OUTPUT @driverswithpath TO "/output/output.csv" 
USING Outputters.Csv();