Question

我将以下字符串分为两列：

给出：

DECLARE @String VARCHAR(MAX) = 'Mak^1,Jak^2,Smith^3,Lee^4,Joseph^5'

我想将其分为两列：

column1  column2
-----------------
Mak      1
Jak      2
Smith    3
Lee      4
Joseph   5

我的尝试：

表值函数：

CREATE FUNCTION [dbo].[udf_Split]
(
@InputString VARCHAR(8000), 
@Delimiter VARCHAR(50)
)
RETURNS @Items TABLE (ID INTEGER IDENTITY(1,1), Item VARCHAR(8000))

AS
BEGIN
      IF @Delimiter = ' '
      BEGIN
            SET @Delimiter = ','
            SET @InputString = REPLACE(@InputString, ' ', @Delimiter)
      END
      IF (@Delimiter IS NULL OR @Delimiter = '')
            SET @Delimiter = ','

      DECLARE @Item VARCHAR(8000)
      DECLARE @ItemList VARCHAR(8000)
      DECLARE @DelimIndex INT

      SET @ItemList = @InputString
      SET @DelimIndex = CHARINDEX(@Delimiter, @ItemList, 0)
      WHILE (@DelimIndex != 0)
      BEGIN
            SET @Item = SUBSTRING(@ItemList, 0, @DelimIndex)
            INSERT INTO @Items VALUES (@Item)

            SET @ItemList = SUBSTRING(@ItemList, @DelimIndex+1, LEN(@ItemList)-@DelimIndex)
            SET @DelimIndex = CHARINDEX(@Delimiter, @ItemList, 0)
      END -- End WHILE

      IF @Item IS NOT NULL 
      BEGIN
            SET @Item = @ItemList
            INSERT INTO @Items VALUES (@Item)
      END

      ELSE INSERT INTO @Items VALUES (@InputString)

      RETURN

END

函数调用：

SELECT Item FROM [dbo].[udf_Split](@String ,',');

输出：

Item
--------------
Mak^1
Jak^2
Smith^3
Lee^4
Joseph^5

Answer 1

首先，请注意，SQL Server 2008 r2不在扩展支持范围内。现在是升级到较新版本的时候了。

对于单个字符串，我可能会使用一些动态SQL魔术：

DECLARE @String VARCHAR(MAX) = 'Mak^1,Jak^2,Smith^3,Lee^4,Joseph^5'

DECLARE @Sql VARCHAR(MAX) = 'SELECT Name,Id FROM (VALUES (''' + REPLACE(REPLACE(REPLACE(@String,'''',''''''), ',', '),('''), '^', ''',') + ')) V(Name, Id)';

-- @Sql now contains this:
-- SELECT Name,Id FROM (VALUES ('Mak',1),('Jak',2),('Smith',3),('Lee',4),('Joseph',5)) V(Name, Id)

EXEC(@Sql)

结果：

Name    Id
Mak     1
Jak     2
Smith   3
Lee     4
Joseph  5

Answer 2

在下面尝试此脚本

DECLARE @String VARCHAR(MAX) = 'Mak^1,Jak^2,Smith^3,Lee^4,Joseph^5';

DECLARE @TempTable AS TABLE(data VARCHAR(MAX))
INSERT INTO @TempTable
SELECT @String

;WITH CTE
AS
(
SELECT Split.A.value('.','nvarchar(1000)') AS data
FROM 
(
SELECT CAST('<S>'+REPLACE(data,',','</S><S>')+'</S>' AS XML ) AS data
FROM @TempTable
)AS A
CROSS APPLY data.nodes('S') AS Split(A)
)
SELECT LTRIM(RTRIM(SUBSTRING(data,0,CHARINDEX('^',data)))) AS column1,
       LTRIM(RTRIM(SUBSTRING(data,CHARINDEX('^',data)+1,LEN (data)))) AS column2
FROM CTE

结果

column1 column2
-------------------
Mak         1
Jak         2
Smith       3
Lee         4
Joseph      5

使用上述脚本创建表值参数函数

CREATE FUNCTION [dbo].[udf_SplitFun](@InputData VARCHAR(MAX))
RETURNS @Return TABLE ( column1 VARCHAR(200),column2 INT)
AS
BEGIN
DECLARE @TempTable AS TABLE
(
data VARCHAR(MAX)
)
INSERT INTO @TempTable
SELECT @InputData

;WITH CTE
AS
(
SELECT Split.A.value('.','nvarchar(1000)') AS data
FROM 
(
SELECT CAST('<S>'+REPLACE(data,',','</S><S>')+'</S>' AS XML ) AS data
FROM @TempTable
)AS A
CROSS APPLY data.nodes('S') AS Split(A)
)
INSERT INTO @Return(column1,column2)
SELECT LTRIM(RTRIM(SUBSTRING(data,0,CHARINDEX('^',data)))) AS column1,
       LTRIM(RTRIM(SUBSTRING(data,CHARINDEX('^',data)+1,LEN (data)))) AS column2
FROM CTE

RETURN;

END

执行如下所示的功能

DECLARE @InputData VARCHAR(MAX) = 'Mak^1,Jak^2,Smith^3,Lee^4,Joseph^5';

SELECT * FROM [dbo].[udf_SplitFun] (@InputData)
GO

Answer 3

我觉得更好的方法是摆脱可怕的WHILE并使用基于集合的方法；我们将在这里使用delimitedsplit8K（如果您使用的是2012+，请使用delimitedsplit8k_lead或使用2016+的用户，可以使用STRING_SPLIT）。

考虑到这一点，以上内容变得很简单：

DECLARE @String varchar(MAX) = 'Mak^1,Jak^2,Smith^3,Lee^4,Joseph^5';

SELECT LEFT(DS.Item,CHARINDEX('^',DS.Item)-1) AS Col1,
       STUFF(DS.Item,1, CHARINDEX('^',DS.Item),'') AS Col2
FROM dbo.DelimitedSplit8K(@String, ',') DS;

Answer 4

在最新版本的SQL Server中，您可以使用string_split()：

select left(s.value, charindex('^', value) - 1) as column1,
       stuff(s.value, 1, charindex('^', value), '') as column2
from string_split(@string, ',') s ;

您可能会发现下载拆分函数来处理此问题最为方便。

否则，我认为递归CTE是一种简单的方法：

with cte as (
      select convert(varchar(max), null) as row,
             @string as str
      union all
      select convert(varchar(max), left(str, charindex(',', str + ',') - 1)),
             convert(varchar(max), stuff(str, 1, charindex(',', str + ','), ''))
      from cte
      where str <> '' 
     )
select left(cte.row, charindex('^', cte.row) - 1) as column1,
       stuff(cte.row, 1, charindex('^', cte.row), '')
from cte
where row is not null;

Here是db <>小提琴。

Answer 5

您可以再次使用该拆分功能来按插入号拆分每行。喜欢：

SELECT SplitByCaret1.Item, SplitByCaret2.Item 
FROM [dbo].[udf_Split](@String ,',') SplitByComma
CROSS APPLY (SELECT * FROM [dbo].[udf_Split](SplitByComma.Item ,'^') Splitted WHERE Splitted.ID=1) SplitByCaret1
CROSS APPLY (SELECT * FROM [dbo].[udf_Split](SplitByComma.Item ,'^') Splitted WHERE Splitted.ID=2) SplitByCaret2

将字符串分成两列

5 个答案: