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我在 SQL Server 表中有以下文本字段：

1!1,3!0,23!0,288!0,340!0,521!0,24!0,38!0,26!0,27!0,281!0,19!0,470!0,568!0,601!0,2!1,251!0,7!2,140!0,285!0,11!2,33!0

  1. 只想检索感叹号 (!) 之前的部分。所以1!1我只想要1，3!0我只想要3，23!0我只想要23。

  2. 还想只检索感叹号 (!) 之后的部分。所以1!1我只想要1，3!0我只想要0，23!0我只想要0。

第 1 点和第 2 点都应该插入到 SQL Server 表的不同列中。

4

3 回答 3

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我喜欢 SQL Server 的 XML 功能。这是解析数据的好方法。试试这个：

--Load the original string
DECLARE @string nvarchar(max) = '1!2,3!4,5!6,7!8,9!10';

--Turn it into XML
SET @string = REPLACE(@string,',','</SecondNumber></Pair><Pair><FirstNumber>') + '</SecondNumber></Pair>';
SET @string = '<Pair><FirstNumber>' + REPLACE(@string,'!','</FirstNumber><SecondNumber>');

--Show the new version of the string
SELECT @string AS XmlIfiedString;

--Load it into an XML variable
DECLARE @xml XML = @string;

--Now, First and Second Number from each pair...
SELECT
  Pairs.Pair.value('FirstNumber[1]','nvarchar(1024)') AS FirstNumber,
  Pairs.Pair.value('SecondNumber[1]','nvarchar(1024)') AS SecondNumber
FROM @xml.nodes('//*:Pair') Pairs(Pair);

上面的查询将字符串转换为 XML，如下所示：

<Pair><FirstNumber>1</FirstNumber><SecondNumber>2</SecondNumber></Pair> ...

然后对其进行解析以返回如下结果：

FirstNumber | SecondNumber
----------- | ------------
          1 |            2
          3 |            4
          5 |            6
          7 |            8
          9 |           10
于 2013-01-10T15:41:01.573 回答
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我完全同意那些抱怨这类数据的人。然而，事实是我们通常无法控制来源的格式。

这是我的方法...

首先，您需要一个标记器。这个效率很高（可能是最快的非 CLR）。见http://www.sqlservercentral.com/articles/Tally+Table/72993/

CREATE FUNCTION [dbo].[DelimitedSplit8K]
--===== Define I/O parameters
        (@pString VARCHAR(8000), @pDelimiter CHAR(1))
--WARNING!!! DO NOT USE MAX DATA-TYPES HERE!  IT WILL KILL PERFORMANCE!
RETURNS TABLE WITH SCHEMABINDING AS
 RETURN
--===== "Inline" CTE Driven "Tally Table" produces values from 1 up to 10,000...
     -- enough to cover VARCHAR(8000)
  WITH E1(N) AS (
                 SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL
                 SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL
                 SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1
                ),                          --10E+1 or 10 rows
       E2(N) AS (SELECT 1 FROM E1 a, E1 b), --10E+2 or 100 rows
       E4(N) AS (SELECT 1 FROM E2 a, E2 b), --10E+4 or 10,000 rows max
 cteTally(N) AS (--==== This provides the "base" CTE and limits the number of rows right up front
                     -- for both a performance gain and prevention of accidental "overruns"
                 SELECT TOP (ISNULL(DATALENGTH(@pString),0)) ROW_NUMBER() OVER (ORDER BY (SELECT NULL)) FROM E4
                ),
cteStart(N1) AS (--==== This returns N+1 (starting position of each "element" just once for each delimiter)
                 SELECT 1 UNION ALL
                 SELECT t.N+1 FROM cteTally t WHERE SUBSTRING(@pString,t.N,1) = @pDelimiter
                ),
cteLen(N1,L1) AS(--==== Return start and length (for use in substring)
                 SELECT s.N1,
                        ISNULL(NULLIF(CHARINDEX(@pDelimiter,@pString,s.N1),0)-s.N1,8000)
                   FROM cteStart s
                )
--===== Do the actual split. The ISNULL/NULLIF combo handles the length for the final element when no delimiter is found.
 SELECT ItemNumber = ROW_NUMBER() OVER(ORDER BY l.N1),
        Item       = SUBSTRING(@pString, l.N1, l.L1)
   FROM cteLen l
;
GO

然后你像这样消费它......

DECLARE @Wtf VARCHAR(1000) = '1!1,3!0,23!0,288!0,340!0,521!0,24!0,38!0,26!0,27!0,281!0,19!0,470!0,568!0,601!0,2!1,251!0,7!2,140!0,285!0,11!2,33!0'

SELECT   LEFT(Item, CHARINDEX('!', Item)-1)
        ,RIGHT(Item, CHARINDEX('!', REVERSE(Item))-1)
FROM [dbo].[DelimitedSplit8K](@Wtf, ',')

发布的函数和解析逻辑当然可以集成到一个函数中。

于 2013-01-10T15:27:32.067 回答
0

我同意规范化数据是最好的方法。但是，这里是解析数据的 XML 解决方案

DECLARE @str VARCHAR(1000) = '1!1,3!0,23!0,288!0,340!0,521!0,24!0,38!0,26!0,27!0,281!0,19!0,470!0,568!0,601!0,2!1,251!0,7!2,140!0,285!0,11!2,33!0'
    ,@xml XML

SET @xml  = CAST('<row><col>' + REPLACE(REPLACE(@str,'!','</col><col>'),',','</col></row><row><col>') + '</col></row>' AS XML)

SELECT  
     line.col.value('col[1]', 'varchar(1000)') AS col1
    ,line.col.value('col[2]', 'varchar(1000)') AS col2
FROM    @xml.nodes('/row') AS line(col)
于 2013-01-10T15:42:34.997 回答