我有這樣的場景：SQL Server的LTRIM或子獲得的電子郵件地址唯一
John Doe [email protected]
John [email protected]
我想離開那隻能分析出此如下：
John Doe
John
換句話說，除去僅限電子郵件地址。
我已經試過這樣：
SELECT
(CASE WHEN CHARINDEX(‘ ‘, REVERSE(column1)) > 0 THEN Left(column1, CHARINDEX(‘ ‘, REVERSE(column1)) – 1)
ELSE column1
END) AS column1
FROM Book1
你在正確的軌道上......只要加入len函數拖放 「減1」
SELECT (CASE WHEN CHARINDEX(' ', REVERSE(column1)) > 0 THEN Left(column1, len (column1) - CHARINDEX(' ', REVERSE(column1)))
ELSE column1
END) AS column1
FROM Book1
DECLARE @Str VARCHAR(MAX) = 'John Doe [email protected]'
SELECT LEFT(@Str, LEN(@Str) - LEN(REVERSE(SUBSTRING(REVERSE(@Str), 1, CHARINDEX(' ', REVERSE(@Str)) - 1))))
DECLARE @MyTable TABLE
(
ID INT IDENTITY(1,1) PRIMARY KEY,
Col1 NVARCHAR(200) NOT NULL
);
INSERT @MyTable(Col1) VALUES(N'John Doe [email protected]');
INSERT @MyTable(Col1) VALUES(N'John [email protected]');
INSERT @MyTable(Col1) VALUES(N'[email protected]');
INSERT @MyTable(Col1) VALUES(N' [email protected]');
SELECT *,
CASE
WHEN oa.SpaceIndex > 0 THEN RIGHT(t.Col1, oa.SpaceIndex - 1)
ELSE t.Col1
END AS EmailAddress,
CASE
WHEN oa.SpaceIndex > 0 THEN LEFT(t.Col1, LEN(t.Col1) - oa.SpaceIndex)
-- or WHEN oa.SpaceIndex > 0 THEN NULLIF(LEFT(t.Col1, LEN(t.Col1) - oa.SpaceIndex), '')
END AS FullName
FROM @MyTable t
OUTER APPLY(SELECT CHARINDEX(N' ', REVERSE(t.Col1)) AS SpaceIndex) oa
結果：
ID Col1 SpaceIndex EmailAddress FullName
-- -------------------------- ---------- ----------------- --------
1 John Doe [email protected] 18 [email protected] John Doe
2 John [email protected] 18 [email protected] John
3 [email protected] 0 [email protected] NULL
4 [email protected] 18 [email protected]
我用我試過的東西編輯過帖子。謝謝邁克爾。 –
http://stackoverflow.com/questions/19146268/sql-server-rtrim-ltrim-and-charindex/19146361#19146361 –