SQL拆分字符串

时间:2017-12-06 10:45:20

标签: sql sql-server sql-server-2014

我在表格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,但无法从正确的方向通过右边的'OU ='将图案分割得正确。看起来很乱,看不到像在C#中那样分割字符串的优雅方式。 - SQL 2014。

2 个答案:

答案 0 :(得分:1)

您可以在SQL中使用Split String函数将astring拆分为数组,然后使用'OU ='substring访问行。其语法如下

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

或者可以使用此处描述的方法

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

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

答案 1 :(得分:0)

看起来您想要OU=的倒数第二个实例。这是你如何做到的。

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@table替换为您的列名和表名。

如果你想要一个灵巧的方式来分割东西,这里有一个功能可以做到非常快,这要归功于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

以下是在您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,我们只会将row_number()仅应用于以OU开头并按顺序排序的项目。因此,row_number()= 2(最后一个)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