我需要使用SQL查詢從完整路徑解析文件名和文件路徑。從完整路徑解析文件名和路徑
例如, FULLPATH - \ SERVER \ d $ \ EXPORTFILES \ EXPORT001.csv
FileName Path
EXPORT001.csv \\SERVER\D$\EXPORTFILES\
我需要使用SQL查詢從完整路徑解析文件名和文件路徑。從完整路徑解析文件名和路徑
例如, FULLPATH - \ SERVER \ d $ \ EXPORTFILES \ EXPORT001.csv
FileName Path
EXPORT001.csv \\SERVER\D$\EXPORTFILES\
使用此 -
DECLARE @full_path VARCHAR(1000)
SET @full_path = '\\SERVER\D$\EXPORTFILES\EXPORT001.csv'
SELECT LEFT(@full_path,LEN(@full_path) - charindex('\',reverse(@full_path),1) + 1) [path],
RIGHT(@full_path, CHARINDEX('\', REVERSE(@full_path)) -1) [file_name]
Declare @filepath Nvarchar(1000)
Set @filepath = 'D:\ABCD\HIJK\MYFILE.TXT'
--Using Left and Right
Select LEFT(@filepath,LEN(@filePath)-CHARINDEX('\',REVERSE(@filepath))+1) Path,
RIGHT(@filepath,CHARINDEX('\',REVERSE(@filepath))-1) FileName
-- Using Substring
Select SUBSTRING(@filepath,1,LEN(@filepath)-CHARINDEX('\',REVERSE(@filepath))+1) Path,
REVERSE(SUBSTRING(REVERSE(@filepath),1,CHARINDEX('\',REVERSE(@filepath))-1)) FileName
這是最簡單的方法
DECLARE @full_path VARCHAR(1000)
SET @full_path = '\\SERVER\D$\EXPORTFILES\EXPORT001.csv'
SELECT LEFT(@full_path, LEN(@full_path) - CHARINDEX('\', REVERSE(@full_path)) - 1),
RIGHT(@full_path, CHARINDEX('\', REVERSE(@full_path)) - 1)
這裏有一個鏈接,有人提出了幾項與此相關的功能:
回答斯特凡·斯泰格爾基於評論:
Create FUNCTION GetFileName
(
@fullpath nvarchar(260)
)
RETURNS nvarchar(260)
AS
BEGIN
DECLARE @charIndexResult int
SET @charIndexResult = CHARINDEX('\', REVERSE(@fullpath))
IF @charIndexResult = 0
RETURN NULL
RETURN RIGHT(@fullpath, @charIndexResult -1)
END
GO
測試代碼:
DECLARE @fn nvarchar(260)
EXEC @fn = dbo.GetFileName 'AppData\goto\image.jpg'
PRINT @fn -- prints image.jpg
EXEC @fn = dbo.GetFileName 'c:\AppData\goto\image.jpg'
PRINT @fn -- prints image.jpg
EXEC @fn = dbo.GetFileName 'image.jpg'
PRINT @fn -- prints NULL
我做了很多ETL工作,我一直在尋找,我可以用和qub1n'ssolution作品除了值非常好沒有反斜槓的功能。這裏是qub1n解決方案的一小調整,而不反斜槓將處理字符串:
Create FUNCTION fnGetFileName
(
@fullpath nvarchar(260)
)
RETURNS nvarchar(260)
AS
BEGIN
IF(CHARINDEX('\', @fullpath) > 0)
SELECT @fullpath = RIGHT(@fullpath, CHARINDEX('\', REVERSE(@fullpath)) -1)
RETURN @fullpath
END
樣品:
SELECT [dbo].[fnGetFileName]('C:\Test\New Text Document.txt') --> New Text Document.txt
SELECT [dbo].[fnGetFileName]('C:\Test\Text Docs\New Text Document.txt') --> New Text Document.txt
SELECT [dbo].[fnGetFileName]('New Text Document.txt') --> New Text Document.txt
SELECT [dbo].[fnGetFileName]('\SERVER\D$\EXPORTFILES\EXPORT001.csv') --> EXPORT001.csv
這裏是一個LINK到SqlFiddle
這就是解決方案! ;) – SQueek
使用REVERSE
更容易看到
DECLARE @full_path VARCHAR(1000)
SET @full_path = '\\SERVER\D$\EXPORTFILES\EXPORT001.csv'
select REVERSE(LEFT(REVERSE(@full_path),CHARINDEX('\',REVERSE(@full_path))-1)) as [FileName],
replace(@full_path, REVERSE(LEFT(REVERSE(@full_path),CHARINDEX('\',REVERSE(@full_path))-1)),'') as [FilePath]
select
LTRIM(
RTRIM(
REVERSE(
SUBSTRING(
REVERSE(Filename),0,CHARINDEX('\',REVERSE(Filename),0))
)))
from TblFilePath
感謝您提供快速回復。有效! – Don
爲什麼不只是:RIGHT(YOUR_PATH,CHARINDEX('\',REVERSE(YOUR_PATH))-1) –
我已經測試過它,它似乎足夠。唯一的缺點是,這隻適用於反斜槓(如果可用的話在.NET中爲Path.GetFileName(path))。 – qub1n