[英]How to split a single column values to multiple column values?
我在将单列值拆分为多列值时遇到问题。
例如:
Name
------------
abcd efgh
ijk lmn opq
asd j. asdjja
asb (asdfas) asd
asd
我需要这样的输出:
first_name last_name
----------------------------------
abcd efgh
ijk opq
asd asdjja
asb asd
asd null
中间名可以省略(不需要中间名) 列已经创建,需要从单个Name
列插入数据。
您的方法无法正确处理大量名称,但是...
SELECT CASE
WHEN name LIKE '% %' THEN LEFT(name, Charindex(' ', name) - 1)
ELSE name
END,
CASE
WHEN name LIKE '% %' THEN RIGHT(name, Charindex(' ', Reverse(name)) - 1)
END
FROM YourTable
马丁的替代品
select LEFT(name, CHARINDEX(' ', name + ' ') -1),
STUFF(name, 1, Len(Name) +1- CHARINDEX(' ',Reverse(name)), '')
from somenames
样品表
create table somenames (Name varchar(100))
insert somenames select 'abcd efgh'
insert somenames select 'ijk lmn opq'
insert somenames select 'asd j. asdjja'
insert somenames select 'asb (asdfas) asd'
insert somenames select 'asd'
insert somenames select ''
insert somenames select null
;WITH Split_Names (Name, xmlname)
AS
(
SELECT
Name,
CONVERT(XML,'<Names><name>'
+ REPLACE(Name,' ', '</name><name>') + '</name></Names>') AS xmlname
FROM somenames
)
SELECT
xmlname.value('/Names[1]/name[1]','varchar(100)') AS first_name,
xmlname.value('/Names[1]/name[2]','varchar(100)') AS last_name
FROM Split_Names
并检查下面的链接以供参考
http://jahaines.blogspot.in/2009/06/converting-delimited-string-of-values.html
您需要的是一个拆分的用户定义函数。 有了这个,解决方案看起来像
With SplitValues As
(
Select T.Name, Z.Position, Z.Value
, Row_Number() Over ( Partition By T.Name Order By Z.Position ) As Num
From Table As T
Cross Apply dbo.udf_Split( T.Name, ' ' ) As Z
)
Select Name
, FirstName.Value
, Case When ThirdName Is Null Then SecondName Else ThirdName End As LastName
From SplitValues As FirstName
Left Join SplitValues As SecondName
On S2.Name = S1.Name
And S2.Num = 2
Left Join SplitValues As ThirdName
On S2.Name = S1.Name
And S2.Num = 3
Where FirstName.Num = 1
这是一个示例拆分函数:
Create Function [dbo].[udf_Split]
(
@DelimitedList nvarchar(max)
, @Delimiter nvarchar(2) = ','
)
RETURNS TABLE
AS
RETURN
(
With CorrectedList As
(
Select Case When Left(@DelimitedList, Len(@Delimiter)) <> @Delimiter Then @Delimiter Else '' End
+ @DelimitedList
+ Case When Right(@DelimitedList, Len(@Delimiter)) <> @Delimiter Then @Delimiter Else '' End
As List
, Len(@Delimiter) As DelimiterLen
)
, Numbers As
(
Select TOP( Coalesce(DataLength(@DelimitedList)/2,0) ) Row_Number() Over ( Order By c1.object_id ) As Value
From sys.columns As c1
Cross Join sys.columns As c2
)
Select CharIndex(@Delimiter, CL.list, N.Value) + CL.DelimiterLen As Position
, Substring (
CL.List
, CharIndex(@Delimiter, CL.list, N.Value) + CL.DelimiterLen
, CharIndex(@Delimiter, CL.list, N.Value + 1)
- ( CharIndex(@Delimiter, CL.list, N.Value) + CL.DelimiterLen )
) As Value
From CorrectedList As CL
Cross Join Numbers As N
Where N.Value <= DataLength(CL.List) / 2
And Substring(CL.List, N.Value, CL.DelimiterLen) = @Delimiter
)
我最近用过:
select
substring(name,1,charindex(' ',name)-1) as Col1,
substring(name,charindex(' ',name)+1,len(name)) as Col2
from TableName
这是我在 SQLite 数据库上执行此操作的方法:
SELECT SUBSTR(name, 1,INSTR(name, " ")-1) as Firstname,
SUBSTR(name, INSTR(name," ")+1, LENGTH(name)) as Lastname
FROM YourTable;
希望能帮助到你。
SELECT
SUBSTRING_INDEX(SUBSTRING_INDEX(rent, ' ', 1), ' ', -1) AS currency,
SUBSTRING_INDEX(SUBSTRING_INDEX(rent, ' ', 3), ' ', -1) AS rent
FROM tolets
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