我有一个这样的SQL表:

| SomeID         | OtherID     | Data
+----------------+-------------+-------------------
| abcdef-.....   | cdef123-... | 18,20,22
| abcdef-.....   | 4554a24-... | 17,19
| 987654-.....   | 12324a2-... | 13,19,20

是否有一个查询,我可以执行查询,如SELECT OtherID, SplitData where SomeID = 'abcdef-.......,返回单独的行,就像这样:

| OtherID     | SplitData
+-------------+-------------------
| cdef123-... | 18
| cdef123-... | 20
| cdef123-... | 22
| 4554a24-... | 17
| 4554a24-... | 19

基本上把我的数据在逗号处分割成单独的行?

我知道将逗号分隔的字符串存储到关系数据库中听起来很愚蠢,但消费者应用程序中的正常用例使这种方法非常有用。

我不想在应用程序中进行拆分,因为我需要分页,所以我想在重构整个应用程序之前探索选项。

它是SQL Server 2008(非r2)。


当前回答

函数

CREATE FUNCTION dbo.SplitToRows (@column varchar(100), @separator varchar(10))
RETURNS @rtnTable TABLE
  (
  ID int identity(1,1),
  ColumnA varchar(max)
  )
 AS
BEGIN
    DECLARE @position int = 0;
    DECLARE @endAt int = 0;
    DECLARE @tempString varchar(100);
    
    set @column = ltrim(rtrim(@column));

    WHILE @position<=len(@column)
    BEGIN       
        set @endAt = CHARINDEX(@separator,@column,@position);
            if(@endAt=0)
            begin
            Insert into @rtnTable(ColumnA) Select substring(@column,@position,len(@column)-@position);
            break;
            end;
        set @tempString = substring(ltrim(rtrim(@column)),@position,@endAt-@position);

        Insert into @rtnTable(ColumnA) select @tempString;
        set @position=@endAt+1;
    END;
    return;
END;

用例

select * from dbo.SplitToRows('T14; p226.0001; eee; 3554;', ';');

或者只是一个有多个结果集的选择

DECLARE @column varchar(max)= '1234; 4748;abcde; 324432';
DECLARE @separator varchar(10) = ';';
DECLARE @position int = 0;
DECLARE @endAt int = 0;
DECLARE @tempString varchar(100);

set @column = ltrim(rtrim(@column));

WHILE @position<=len(@column)
BEGIN       
    set @endAt = CHARINDEX(@separator,@column,@position);
        if(@endAt=0)
        begin
        Select substring(@column,@position,len(@column)-@position);
        break;
        end;
    set @tempString = substring(ltrim(rtrim(@column)),@position,@endAt-@position);

    select @tempString;
    set @position=@endAt+1;
END;

其他回答

你可以使用SQL Server中的递归函数:


示例表:

CREATE TABLE Testdata
(
    SomeID INT,
    OtherID INT,
    String VARCHAR(MAX)
);

INSERT Testdata SELECT 1,  9, '18,20,22';
INSERT Testdata SELECT 2,  8, '17,19';
INSERT Testdata SELECT 3,  7, '13,19,20';
INSERT Testdata SELECT 4,  6, '';
INSERT Testdata SELECT 9, 11, '1,2,3,4';

查询

WITH tmp(SomeID, OtherID, DataItem, String) AS
(
    SELECT
        SomeID,
        OtherID,
        LEFT(String, CHARINDEX(',', String + ',') - 1),
        STUFF(String, 1, CHARINDEX(',', String + ','), '')
    FROM Testdata
    UNION all

    SELECT
        SomeID,
        OtherID,
        LEFT(String, CHARINDEX(',', String + ',') - 1),
        STUFF(String, 1, CHARINDEX(',', String + ','), '')
    FROM tmp
    WHERE
        String > ''
)
SELECT
    SomeID,
    OtherID,
    DataItem
FROM tmp
ORDER BY SomeID;
-- OPTION (maxrecursion 0)
-- normally recursion is limited to 100. If you know you have very long
-- strings, uncomment the option

输出

 SomeID | OtherID | DataItem 
--------+---------+----------
 1      | 9       | 18       
 1      | 9       | 20       
 1      | 9       | 22       
 2      | 8       | 17       
 2      | 8       | 19       
 3      | 7       | 13       
 3      | 7       | 19       
 3      | 7       | 20       
 4      | 6       |          
 9      | 11      | 1        
 9      | 11      | 2        
 9      | 11      | 3        
 9      | 11      | 4        

当使用这种方法时,您必须确保您的值中没有包含非法XML - user1151923

我总是使用XML方法。确保使用VALID XML。我有两个函数在有效的XML和文本之间转换。(我倾向于去掉回车,因为我通常不需要它们。

CREATE FUNCTION dbo.udf_ConvertTextToXML (@Text varchar(MAX)) 
    RETURNS varchar(MAX)
AS
    BEGIN
        SET @Text = REPLACE(@Text,CHAR(10),'');
        SET @Text = REPLACE(@Text,CHAR(13),'');
        SET @Text = REPLACE(@Text,'<','&lt;');
        SET @Text = REPLACE(@Text,'&','&amp;');
        SET @Text = REPLACE(@Text,'>','&gt;');
        SET @Text = REPLACE(@Text,'''','&apos;');
        SET @Text = REPLACE(@Text,'"','&quot;');
    RETURN @Text;
END;


CREATE FUNCTION dbo.udf_ConvertTextFromXML (@Text VARCHAR(MAX)) 
    RETURNS VARCHAR(max)
AS
    BEGIN
        SET @Text = REPLACE(@Text,'&lt;','<');
        SET @Text = REPLACE(@Text,'&amp;','&');
        SET @Text = REPLACE(@Text,'&gt;','>');
        SET @Text = REPLACE(@Text,'&apos;','''');
        SET @Text = REPLACE(@Text,'&quot;','"');
    RETURN @Text;
END;

很晚了,但是试试这个:

SELECT ColumnID, Column1, value  --Do not change 'value' name. Leave it as it is.
FROM tbl_Sample  
CROSS APPLY STRING_SPLIT(Tags, ','); --'Tags' is the name of column containing comma separated values

所以我们有了这个: tbl_Sample:

ColumnID|   Column1 |   Tags
--------|-----------|-------------
1       |   ABC     |   10,11,12    
2       |   PQR     |   20,21,22

运行此查询后:

ColumnID|   Column1 |   value
--------|-----------|-----------
1       |   ABC     |   10
1       |   ABC     |   11
1       |   ABC     |   12
2       |   PQR     |   20
2       |   PQR     |   21
2       |   PQR     |   22

谢谢!

DECLARE @id_list VARCHAR(MAX) = '1234,23,56,576,1231,567,122,87876,57553,1216';
DECLARE @table TABLE ( id VARCHAR(50) );
DECLARE @x INT = 0;
DECLARE @firstcomma INT = 0;
DECLARE @nextcomma INT = 0;

SET @x = LEN(@id_list) - LEN(REPLACE(@id_list, ',', '')) + 1; -- number of ids in id_list

WHILE @x > 0
    BEGIN
        SET @nextcomma = CASE WHEN CHARINDEX(',', @id_list, @firstcomma + 1) = 0
                              THEN LEN(@id_list) + 1
                              ELSE CHARINDEX(',', @id_list, @firstcomma + 1)
                         END;
        INSERT  INTO @table
        VALUES  ( SUBSTRING(@id_list, @firstcomma + 1, (@nextcomma - @firstcomma) - 1) );
        SET @firstcomma = CHARINDEX(',', @id_list, @firstcomma + 1);
        SET @x = @x - 1;
    END;

SELECT  *
FROM    @table;

检查这个

 SELECT A.OtherID,  
     Split.a.value('.', 'VARCHAR(100)') AS Data  
 FROM  
 (
     SELECT OtherID,  
         CAST ('<M>' + REPLACE(Data, ',', '</M><M>') + '</M>' AS XML) AS Data  
     FROM  Table1
 ) AS A CROSS APPLY Data.nodes ('/M') AS Split(a);