如何获得特定字符后的所有字符字符串SQL

问题描述:

我有一个包含有以下形式的字符串行一列:如何获得特定字符后的所有字符字符串SQL

#S6TF3.01,#S6TF3.09,#S6TF3.10,#S6TF3.13 

我希望能够得到所有行的结果如下:

01,09,10,13 

我使用T-SQL,并尝试以下操作:

SUBSTRING(E.REJECT_WF_NOS, CHARINDEX('.', E.REJECT_WF_NOS) + 1, LEN(E.REJECT_WF_NOS)) 
+0

如果是这样构成的数据库也不是那么好... – decadenza

+0

如果背后总有小数2号,我会用string_split在2016年,以用逗号分割,然后颠倒你的行和子串2,然后将其逆转回去:) – plaidDK

我可能会分裂因为结构不是固定的你似乎只想要小数点后的值。这会将逗号分隔,然后在小数点后取所有内容,然后将其重新连接到一个字符串中。

ONLINE DEMO

declare @var varchar(64) = 'S6TF3.01,#S6TF3.09,#S6TF3.10,#S6TF3.13' 

SELECT 
    STUFF((
      SELECT ',' + substring(item,charindex('.',Item) + 1,32) 
      FROM dbo.DelimitedSplit8K(@var,',') 
      FOR XML PATH(''), TYPE).value('.', 'NVARCHAR(MAX)'), 1, 1, '') 

Here Is the Function from 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 
+1

干得好。但是,我会将nvarchar(max)更改为varchar(64)。该值>数据类型定义了输出字符串的数据类型,这意味着您将在varchar(64)执行时返回nvarchar(max)。并且转换将隐含。 –

下面是一个简单的版本2016,如果你想它rowbased

select reverse(substring(reverse(value),1,2)),* from 
string_split('#S6TF3.01,#S6TF3.09,#S6TF3.10,#S6TF3.13',',') 

结果多行 - 2016年

enter image description here **

如果你想在一排而已,这可能是2016年的方式

DECLARE @MyTable TABLE 
(
ID int , 
Strings varchar(10) 
) 

INSERT INTO @MyTable (ID,Strings) 
select 1,reverse(substring(reverse(value),1,2)) as SplittedValues 
from string_split('#S6TF3.01,#S6TF3.09,#S6TF3.10,#S6TF3.13',',') 

---Select * from @MyTable 

SELECT ID, NewVales = STUFF((SELECT N', ' + Strings 
FROM @MyTable AS p2 
WHERE p2.ID = p.ID 
ORDER BY Strings 
FOR XML PATH(N'')), 1, 2, N'') 
FROM @MyTable AS p 
GROUP BY ID 
ORDER BY ID; 

结果1排 - 2016

enter image description here

对于t他甚至不需要分配器;你可以使用parsename来“分割”这些值。

declare @string varchar(100) = '#S6TF3.01,#S6TF3.09,#S6TF3.10,#S6TF3.13'; 

select newString = stuff 
((select ','+substring(item, 1, charindex(',', item)-1) 
    from (values (4),(3),(2),(1)) t(n) 
    cross apply (values (substring(@string, charindex('.',@string)+1, 8000)+',')) s(string) 
    cross apply (values (parsename(string,n))) split(item) 
    for xml path('')), 1,1,''); 

这将是比delimitedsplit8k更快,不需要的SQL Server 2016在SQL服务器2017年,你可以进一步简化如下:

select string_agg(substring(item, 1, charindex(',', item)-1),',') 
from (values (4),(3),(2),(1)) t(n) 
cross apply (values (substring(@string, charindex('.',@string)+1, 8000)+',')) s(string) 
cross apply (values (parsename(string,n))) split(item) 
+0

只是一个注释,该函数不需要SQL Server 2016年我喜欢parsename虽然如果有不超过4个值。 – scsimon

如果你有这样的行数据由行类似如下表结构:

ID DATA 
1 #S6TF3.01 
2 #S6TF3.09 
3 #S6TF3.10 
4 #S6TF3.13 

如果你始终像数据如上然后只写功能如下:

SELECT SUBSTRING(DATA, 8, 2) DATA from #TEMP 

结果:

ID DATA  Result 
1 #S6TF3.01 01 
2 #S6TF3.09 09 
3 #S6TF3.10 10 
4 #S6TF3.13 13