我的客户将姓名数据作为名称字符串发送给我，其中包括单个条目中的姓氏、名字和中间名。我需要将它们分成姓氏、名字和中间名。我在网上找到了一些脚本，但它们不能满足我的目的，因为它们要么 (1) 使用不同的格式，要么 (2) 不能很好地处理边缘情况。请参阅以下示例：
- 南丁格尔，佛罗伦萨 -> 佛罗伦萨南丁格尔
- 邦德，詹姆斯邦德 -> 詹姆斯邦德邦德
- 雅培，埃德温 A. -> 埃德温 A. 雅培
有人可以帮我编写一个 SQL Server 脚本，将一个字符串拆分成我正在寻找的各个部分吗？
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14198 次
5 回答
4
请注意以下事项：
- 始终请求标准化数据以确保最高的数据质量。我试图列举姓氏、名字和中间名组合的所有可能情况，但我确信我没有得到所有这些情况。
- 我的脚本需要以下格式：LastName@DELIMITER1@DELIMITER2FirstName@DELIMITER2MiddleName，但可以轻松更改为其他格式。
- 此脚本不会像 Dr. 那样分隔图块，也不会处理后缀。
- 感谢 MemKills 对我扩展的测试数据集的想法。
>
DECLARE @DELIMITER1 varchar(1), @DELIMITER2 varchar(1), @MAX_LENGTH int
SET @DELIMITER1 = ','
SET @DELIMITER2 = ' '
SET @MAX_LENGTH = 50
SELECT [Name],
SUBSTRING(Name,1,CHARINDEX(@DELIMITER1,Name) -1) AS LastName, -- Less one char for @DELIMITER1
SUBSTRING(Name,CHARINDEX(@DELIMITER1,Name)+ 2,@MAX_LENGTH) AS FirstAndMiddle, -- Plus two for @DELIMITER1 and @DELIMITER2
CASE
-- Middle name follows two-name first names like Mary Ann
WHEN LEN(SUBSTRING(NAME, CHARINDEX(@DELIMITER1,Name)+ 2,@MAX_LENGTH)) - LEN(REPLACE(SUBSTRING(NAME, CHARINDEX(@DELIMITER1,Name)+ 2,@MAX_LENGTH), @DELIMITER2, '')) > 0
THEN SUBSTRING(Name, LEN(Name) - CHARINDEX(@DELIMITER2, REVERSE(Name))+2, @MAX_LENGTH)
ELSE NULL
END AS MiddleName,
CASE
-- Count the number of @DELIMITER2. Choose the string between the @DELIMITER1 and the final @DELIMITER2.
WHEN LEN(SUBSTRING(NAME, CHARINDEX(@DELIMITER1,Name)+ 2,@MAX_LENGTH)) - LEN(REPLACE(SUBSTRING(NAME, CHARINDEX(@DELIMITER1,Name)+ 2,@MAX_LENGTH), @DELIMITER2, '')) > 0
Then SUBSTRING(Name, CHARINDEX(@DELIMITER1,Name)+ 2,
(LEN(SUBSTRING(NAME, CHARINDEX(@DELIMITER1,Name)+ 2,@MAX_LENGTH))
- LEN(SUBSTRING(Name, LEN(Name) - CHARINDEX(@DELIMITER2, REVERSE(Name))+2, @MAX_LENGTH))))
ELSE SUBSTRING(Name,CHARINDEX(@DELIMITER1,Name)+ 2,@MAX_LENGTH)
END AS FirstName
FROM
(
SELECT [Name] = 'Zzz, A' UNION ALL
SELECT 'de Zzz, Aaa' UNION ALL
SELECT 'Zzz, Aaaa' UNION ALL
SELECT 'Zzz, A B' UNION ALL
SELECT 'Zzz, Aaaa Bbbb' UNION ALL
SELECT 'de Zzz, Aaaa' UNION ALL
SELECT 'de Zzz, Aaaa B' UNION ALL
SELECT 'van Zzz, Aaaa B' UNION ALL
SELECT 'Yyy-Zzz, Aaaa B' UNION ALL
SELECT 'd''Zzz, Aaaa B' UNION ALL
SELECT 'Zzz, Aaaa Bbbb C' UNION ALL
SELECT 'Zzz, Aaaa Bbbb Cccc'
) AS X
于 2014-08-05T15:30:23.297 回答
1
试试这个代码。我觉得效率更高一些。请随时修改或改进它。谢谢。
DECLARE @FullName VARCHAR(60),
@FirstName VARCHAR(30),
@LastName VARCHAR(30),
@MiddleInitialPrep VARCHAR(60) = null,
@MiddleInitial VARCHAR(1) = null
SET @FullName = 'Dr. John Edward Doe III'
-- NAME CLEAN UP TO REMOVE PREFIXES AND SUFFIXES
SET @FullName = REPLACE(@FullName, 'Mr. ', '')
SET @FullName = REPLACE(@FullName, 'Mr ', '')
SET @FullName = REPLACE(@FullName, 'Mrs. ', '')
SET @FullName = REPLACE(@FullName, 'Mrs ', '')
SET @FullName = REPLACE(@FullName, 'Ms. ', '')
SET @FullName = REPLACE(@FullName, 'Ms ', '')
SET @FullName = REPLACE(@FullName, 'Miss ', '')
SET @FullName = REPLACE(@FullName, 'Dr. ', '')
SET @FullName = REPLACE(@FullName, 'Dr ', '')
SET @FullName = REPLACE(@FullName, ' Jr.', '')
SET @FullName = REPLACE(@FullName, ' Jr', '')
SET @FullName = REPLACE(@FullName, ' Sr.', '')
SET @FullName = REPLACE(@FullName, ' Sr', '')
SET @FullName = REPLACE(@FullName, ' III', '')
SET @FullName = REPLACE(@FullName, ' II', '')
-- RETRIEVE FIRST AND LAST NAMES
SET @FirstName = LEFT(@FullName, NULLIF(CHARINDEX(' ', @FullName) - 1, -1))
SET @LastName = RIGHT(@FullName, ISNULL(NULLIF(CHARINDEX(' ', REVERSE(@FullName)) - 1, -1), LEN(@FullName)))
-- ISOLATE MIDDLE INITIAL
SET @MiddleInitialPrep = REPLACE(@FullName, @FirstName, '')
SET @MiddleInitialPrep = REPLACE(@MiddleInitialPrep, @LastName, '')
SET @MiddleInitial = REPLACE(@MiddleInitialPrep, ' ', '')
SELECT @FirstName First_Name, @MiddleInitial Middle_Initial, @LastName Last_Name
于 2014-11-04T16:14:23.987 回答
1
下面的代码适用于姓氏、名字的 M 名字符串。用您的名称字符串列名称替换“名称”。由于当有中间首字母时，您有一个句号作为最后一个字符，所以您将在每行（2、6 和 8）中用 3 替换 2，并将“RIGHT（Name，1）”更改为“RIGHT （姓名，2）”在第 8 行。
SELECT SUBSTRING(Name, 1, CHARINDEX(',', Name) - 1) LastName ,
CASE WHEN LEFT(RIGHT(Name, 2), 1) <> ' '
THEN LTRIM(SUBSTRING(Name, CHARINDEX(',', Name) + 1, 99))
ELSE LEFT(LTRIM(SUBSTRING(Name, CHARINDEX(',', Name) + 1, 99)),
LEN(LTRIM(SUBSTRING(Name, CHARINDEX(',', Name) + 1, 99)))
- 2)
END FirstName ,
CASE WHEN LEFT(RIGHT(Name, 2), 1) = ' ' THEN RIGHT(Name, 1)
ELSE NULL
END MiddleName
于 2015-06-10T02:02:32.617 回答
0
很好的解决方案。我做了一些修改以适应我的情况，其中分隔符是空格，中间名只是中间名的首字母（有时不存在）。以下解决方案甚至可以解析多间距名称，例如：“Jo Ann Taylor Haynes”，没有中间首字母。
SET @DELIMITER1 = ' '
SET @DELIMITER2 = ' '
SET @MAX_LENGTH = 50
SELECT [Name],
SUBSTRING(Name,1,CHARINDEX(@DELIMITER1,Name) -1) AS LastName,
SUBSTRING(Name,CHARINDEX(@DELIMITER1,Name)+ 1,@MAX_LENGTH) AS FirstAndMiddle,
CASE
WHEN LEN(SUBSTRING(NAME, CHARINDEX(@DELIMITER1,Name)+ 1,@MAX_LENGTH)) - LEN(REPLACE(SUBSTRING(NAME, CHARINDEX(@DELIMITER1,Name)+ 1,@MAX_LENGTH), @DELIMITER2, '')) = 1
THEN SUBSTRING(Name, LEN(Name) - CHARINDEX(@DELIMITER2, REVERSE(Name))+1, @MAX_LENGTH)
ELSE NULL
END AS MiddleName,
CASE
WHEN LEN(SUBSTRING(NAME, CHARINDEX(@DELIMITER1,Name)+ 1,@MAX_LENGTH)) - LEN(REPLACE(SUBSTRING(NAME, CHARINDEX(@DELIMITER1,Name)+ 1,@MAX_LENGTH), @DELIMITER2, '')) = 1
Then SUBSTRING(Name, CHARINDEX(@DELIMITER1,Name)+ 1,
(LEN(SUBSTRING(NAME, CHARINDEX(@DELIMITER1,Name)+ 1,@MAX_LENGTH))
- LEN(SUBSTRING(Name, LEN(Name) - CHARINDEX(@DELIMITER2, REVERSE(Name))+1, @MAX_LENGTH))))
ELSE SUBSTRING(Name,CHARINDEX(@DELIMITER1,Name)+ 1,@MAX_LENGTH)
END AS FirstName
于 2017-06-13T15:37:19.040 回答
-4
select substr(
'santhosh kumar kota'
,1,
instr(
'santhosh kumar kota'
,' '
,1
,1)
) as fname
,substr('santhosh kumar kota'
,instr(
'santhosh kumar kota'
,' '
,1
,1)
,(instr(
'santhosh kumar kota'
,' '
,1
,2)-instr(
'santhosh kumar kota'
,' '
,1
,1)
)
)as mname
,substr('santhosh kumar kota'
,instr(
'santhosh kumar kota'
,' '
,1
,2)
,(length('santhosh kumar kota')+1)-instr(
'santhosh kumar kota'
,' '
,1
,2)
)as lname
from dual
/
于 2017-06-20T18:10:30.700 回答