SQL拆分字符串

Question

我在一个表中有以下两个字符串：

CN=ABCDEFG,OU=7SOE,OU=TEMP,OU=Sydney,OU=Australia,DC=LDAP,DC=COM
CN=ABCDEFGHIJ,OU=7SOEFHU,OU=TEST,OU=TEMP,OU=London,OU=Europe,DC=LDAP,DC=COM

我想从这两个中提取以下信息：

Sydney
London

我开始使用 RIGHT、SUBSTRING 和 CHARINDEX，但无法通过右侧的“OU=”将其拆分为正确的模式。看起来很乱，看不到像在 C# 中那样优雅的分割字符串的方法。 - SQL 2014。

Answer 1

您可以使用 SQL 中的拆分字符串函数将您的字符串拆分为数组，然后使用 'OU=' 子字符串访问行。语法如下

SELECT value FROM STRING_SPLIT('Lorem,ipsum,dolor,sit,amet.', ',');

或者可以使用这里描述的方法

Split data in sql with multiple delimiters ( , and - ) with owns rule

和

https://raresql.com/2013/01/10/sql-server-split-string-into-rows-based-on-multiple-delimiters/

Answer 2

看起来您想要OU= 的倒数第二个实例。以下是您可以这样做的方法。

declare @table table (c1 varchar(256))
insert into @table
values
('CN=ABCDEFG,OU=7SOE,OU=TEMP,OU=Sydney,OU=Australia,DC=LDAP,DC=COM'),
('CN=ABCDEFGHIJ,OU=7SOEFHU,OU=TEST,OU=TEMP,OU=London,OU=Europe,DC=LDAP,DC=COM')


select 
    right(left(c1, len(c1) - charindex('UO',reverse(c1)) - 2),charindex('UO',reverse(left(c1, len(c1) - charindex('UO',reverse(c1)) - 2))) - 2)
from @table

只需将c1 和@table 分别替换为您的列名和表名即可。

如果您想要一种巧妙的拆分方式，这里有一个功能可以非常快速地完成，感谢Jeff Moden。

CREATE FUNCTION [dbo].[DelimitedSplit8K] (@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

在你的情况下如何使用它：

select 
    t.c1
    ,x.ItemNumber                   --this represents the item number in the string.
    ,x.Item                         --this is the value of that item
    ,RN = row_number() over (partition by c1 order by x.ItemNumber desc)
from 
    @table t
    cross apply
    dbo.DelimitedSplit8K(c1,',') x
--where
--  left(x.Item,2) = 'OU'           --uncomment this out to see how the RN will be used in the CTE below
order by
    t.c1, 
    x.ItemNumber

因此，为了只获得您想要的结果，我们将仅将row_number() 应用于以OU 开头的那些项目，并按降序排列。因此， = 2（倒数第二个）的 row_number() 将是您想要的值。

;with cte as(
select 
    *
    ,RN = row_number() over (partition by c1 order by x.ItemNumber desc)
from 
    @table
    cross apply
    dbo.DelimitedSplit8K(c1,',') x
where
    left(x.Item,2) = 'OU')

select
    c1,
    YourExtract = right(Item,len(Item) - 3)
from
    cte
where
    RN = 2