Question

我得到了一个很好的答案＆＃34;从完整路径中解析文件名和路径＆＃34;这是由@AgentSQL提供的。

答案简单而有效，但我无法理解它。代码如下：

DECLARE @full_path VARCHAR(1000)
SET @full_path = '\\SERVER\D$\EXPORTFILES\EXPORT001.csv'

SELECT LEFT(@full_path,LEN(@full_path) - charindex('\',reverse(@full_path),1) + 1) [path], 
       REVERSE(LEFT(REVERSE(@full_path),CHARINDEX('\', REVERSE(@full_path), 1) - 1)) [file_name]

我在单独使用时理解左，反向和charindex的功能，但如果有人能帮我理解其中的代码，我真的很感激。

Answer 1

DECLARE @full_path VARCHAR(1000)
SET @full_path = '\\abc\files\india\users\dknthasdx\Dropped CUSIPs_12_10_2014.xlsx'
select REVERSE(LEFT(REVERSE(@full_path),CHARINDEX('\', REVERSE(@full_path), 1) - 1)) [FileName]



--1) REVERSE(@full_path) --Reverses the whole path : xslx.4102_01_21_sPISUC depporD\sznahtnam\sresu\aidni\selif\lns\\

--2) CHARINDEX('\', REVERSE(@full_path), 1) -- Returns 31 in this case : The index  first occurence in REVERSE PATH OF /

--3) LEFT(REVERSE(@full_path),CHARINDEX('\', REVERSE(@full_path), 1) - 1) -- return the expression till first occurence of /

--4) REVERSE(LEFT(REVERSE(@full_path),CHARINDEX('\', REVERSE(@full_path), 1) - 1)) [FileName] -- reverse the String again (reversed earlier) to get the file name

Answer 2

  --two additional variants  
DECLARE @full_path VARCHAR(1000)
    SET @full_path = '\\SERVER\D$\EXPORTFILES\EXPORT001.csv'
    --1 variant
    SELECT  REPLACE(@full_path,RIGHT(@full_path, CHARINDEX('\', REVERSE(@full_path)) - 1), '') [path] ,
            RIGHT(@full_path, CHARINDEX('\', REVERSE(@full_path)) - 1) [file_name]  

    --2 variant
    SELECT  LEFT(@full_path,(LEN(@full_path) - CHARINDEX('\', REVERSE(@full_path)) + 1)) [path] ,
            RIGHT(@full_path, CHARINDEX('\', REVERSE(@full_path)) - 1) [file_name]

我刚刚得到了“从完整路径解析文件名和路径”的解决方案。有人可以帮我理解代码吗？

2 个答案: