如何在SQL Server中拆分虚线分隔的值？

如何仅获取表中数据的中间部分？我尝试了以下代码，但这只删除了正确的部分。。。我的输出应该只是中间部分

例如，当我选择数据

1-021514-1

时，输出应该是

021514

，不带左右破折号

select LEFT(ticketid, CHARINDEX('-', ticketid + '-') + 4) 
from Table

我的数据是：

|TicketID   |
------------
|1-021514-1 |
|10-021514-1|
|2-021514-1 |
|4-021414-1 |

试试这个：

select right(left(ticketid, charindex('-', ticketid, charindex('-', ticketid, 0) + 1) - 1), len(left(ticketid, charindex('-', ticketid, charindex('-', ticketid, 0) + 1) - 1)) - charindex('-', left(ticketid, charindex('-', ticketid, charindex('-', ticketid, 0) + 1) - 1), 0)) from Table

请尝试：

SELECT
  LEFT(st, CHARINDEX('-', st)-1) TicketID
from
(
    SELECT
        SUBSTRING(TicketID,  CHARINDEX('-',TicketID)+1, 10000) st
    FROM Table
)x

试试这个

with t as (select TicketID  as val)
select t.*,
       LEFT(val, charindex('-', val) - 1),
   SUBSTRING(val, charindex('-', val)+1, len(val) - CHARINDEX('-', reverse(val)) - charindex('-', val)),
       REVERSE(LEFT(reverse(val), charindex('-', reverse(val)) - 1))
from t;

SELECT STUFF(
                STUFF(TicketID,1,CHARINDEX('-',TicketID,1),'')
                ,CHARINDEX('-',STUFF(TicketID,1,CHARINDEX('-',TicketID,1),''),1)
                ,LEN(TicketID)
                ,'') 
        from Table1

（或）

使用下面的函数

CREATE FUNCTION dbo.SplitStrings_CTE(@List nvarchar(max), @Delimiter nvarchar(1))
RETURNS @returns TABLE(val nvarchar(max), [level] int, PRIMARY KEY CLUSTERED([level]))
AS
BEGIN
;WITH cte AS
 (
  SELECT SUBSTRING(@List, 0, CHARINDEX(@Delimiter, @List)) AS val,
         CAST(STUFF (@List + @Delimiter, 1, CHARINDEX(@Delimiter, @List), '') AS nvarchar(max)) AS stval,
         1 AS [level]
  UNION ALL
  SELECT SUBSTRING(stval, 0, CHARINDEX(@Delimiter, stval)),
         CAST(STUFF (stval, 1, CHARINDEX(@Delimiter, stval), '') AS nvarchar(max)),
         [level] + 1
  FROM cte
  WHERE stval != ''
  )
  INSERT @returns
  SELECT REPLACE(val, ' ', '') AS val, [level]
  FROM cte
  RETURN
END

试试这个

with t as (select TicketID  as val)
select t.*,
       LEFT(val, charindex('-', val) - 1),
   SUBSTRING(val, charindex('-', val)+1, len(val) - CHARINDEX('-', reverse(val)) - charindex('-', val)),
       REVERSE(LEFT(reverse(val), charindex('-', reverse(val)) - 1))
from t;

SELECT STUFF(
                STUFF(TicketID,1,CHARINDEX('-',TicketID,1),'')
                ,CHARINDEX('-',STUFF(TicketID,1,CHARINDEX('-',TicketID,1),''),1)
                ,LEN(TicketID)
                ,'') 
        from Table1

嗨Ron试试这个

   declare @string varchar(25)='1-021514-1'
        declare @val varchar(25)
        SELECT @val= SUBSTRING(@string,  CHARINDEX('-', @string)+1,  ((CHARINDEX('-',@string,(charindex('-',@string)+1))-CHARINDEX('-', @string))-1))
        select @val

---

可能您正在寻找一个拆分函数。是传递给LEFT或SUBSTRING函数的示例1-021514无效长度参数。由于一个破折号，我犯了这个错误。请给我这个建议