在sql中重新排列列的字

我必须比较下面两个数据表

表a

cust text

1名董事（公司受托人）@委托人

2Director@Settlor@股东

表b

cust text

1Settlor@Director（公司受托人）

2Settlor@Director@股东

我正在使用，但返回了不匹配项，文本字段中的b'ze单词不按顺序排列

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任何快速解决方案，请？

要开始，您可以使用字符串拆分函数（假设您的值由文本中的@分隔），然后比较每个客户的值

用于比较值的查询 查询输出 注意，ExtraText为表a返回空值，因为它不存在

+-------+--------------------------------------+-------------+------------------------------+-------+----------------------------------------+-------------+------------------------------+
| CustA |                TextA                 | ItemNumberA |            ItemA             | CustB |                 TextB                  | ItemNumberB |            ItemB             |
+-------+--------------------------------------+-------------+------------------------------+-------+----------------------------------------+-------------+------------------------------+
| 1     | Director (Corporate Trustee)@Settlor | 1           | Director (Corporate Trustee) |     1 | Settlor@Director (Corporate Trustee)   |           2 | Director (Corporate Trustee) |
| 1     | Director (Corporate Trustee)@Settlor | 2           | Settlor                      |     1 | Settlor@Director (Corporate Trustee)   |           1 | Settlor                      |
| 2     | Director@Settlor@Shareholder         | 1           | Director                     |     2 | Settlor@Director@Shareholder@ExtraText |           2 | Director                     |
| 2     | Director@Settlor@Shareholder         | 2           | Settlor                      |     2 | Settlor@Director@Shareholder@ExtraText |           1 | Settlor                      |
| 2     | Director@Settlor@Shareholder         | 3           | Shareholder                  |     2 | Settlor@Director@Shareholder@ExtraText |           3 | Shareholder                  |
| NULL  | NULL                                 | NULL        | NULL                         |     2 | Settlor@Director@Shareholder@ExtraText |           4 | ExtraText                    |
+-------+--------------------------------------+-------------+------------------------------+-------+----------------------------------------+-------------+------------------------------+

及

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这是SQL Server 2008、2008R2还是2012。它们都是不同的，可能需要不同的解决方案。另外，您进行比较的目的是什么？是否要知道以任何顺序存在相同分隔值的所有行？您想知道每个匹配的

cust

ID缺少哪些值吗？etcAll拥有标准化数据的更多原因…@iamdave-its sql server 2012。我想比较文本列中每个匹配的custid对应值。@天哪，但不管顺序是否正确，都要这样做吗？所以表A中的cust 1 text应该=表B中的cust 1 text，对吗？

+-------+--------------------------------------+-------------+------------------------------+-------+----------------------------------------+-------------+------------------------------+
| CustA |                TextA                 | ItemNumberA |            ItemA             | CustB |                 TextB                  | ItemNumberB |            ItemB             |
+-------+--------------------------------------+-------------+------------------------------+-------+----------------------------------------+-------------+------------------------------+
| 1     | Director (Corporate Trustee)@Settlor | 1           | Director (Corporate Trustee) |     1 | Settlor@Director (Corporate Trustee)   |           2 | Director (Corporate Trustee) |
| 1     | Director (Corporate Trustee)@Settlor | 2           | Settlor                      |     1 | Settlor@Director (Corporate Trustee)   |           1 | Settlor                      |
| 2     | Director@Settlor@Shareholder         | 1           | Director                     |     2 | Settlor@Director@Shareholder@ExtraText |           2 | Director                     |
| 2     | Director@Settlor@Shareholder         | 2           | Settlor                      |     2 | Settlor@Director@Shareholder@ExtraText |           1 | Settlor                      |
| 2     | Director@Settlor@Shareholder         | 3           | Shareholder                  |     2 | Settlor@Director@Shareholder@ExtraText |           3 | Shareholder                  |
| NULL  | NULL                                 | NULL        | NULL                         |     2 | Settlor@Director@Shareholder@ExtraText |           4 | ExtraText                    |
+-------+--------------------------------------+-------------+------------------------------+-------+----------------------------------------+-------------+------------------------------+

SET ANSI_NULLS ON
GO

SET QUOTED_IDENTIFIER ON
GO

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