SQL 已链接服务器记录

Question

我正在研究 SQL Server 2012 我有包含此信息的门票

Ticket No| Date | Closed
T001     | D1   | 
T002     | D2   |
T003     | D3   | Yes
T004     | D4   | Yes 
T005     | D5   |
T006     | D6   | Yes

我想按 "case" 对它们进行分组，收盘后的第一张票是打开 "case" 以及随后直到收盘的所有票都是 "case"[= 的一部分14=]

Case No | Ticket No| Date | Closed
C001    | T001     | D1   | 
C001    | T002     | D2   |
C001    | T003     | D3   | Yes
C002    | T004     | D4   | Yes 
C003    | T005     | D5   |
C003    | T006     | D6   | Yes

我已经尝试了好几个小时（使用滞后、等级等），但我找不到方法来做到这一点。我找不到一种方法来生成所需的案例号你有好主意吗？非常感谢

这里是查询：

SELECT
    CONCAT('C' , ID_TICKET) As ID_CASE_TEMP
    ,ID_TICKET
    ,DT_TICKET
    ,IIF(IsNull(LAG(IsNull(LB_CLOSING,'No')) OVER (ORDER BY DT_TICKET),'Yes') = 'Yes','Yes','No') As LB_OPENNING
    ,IsNull(LB_CLOSING,'No') As LB_CLOSING
    ,LAG(ID_TICKET) OVER (ORDER BY DT_TICKET) As PREVIOUS_TICKET
    ,LAG(DT_TICKET) OVER (ORDER BY DT_TICKET) As PREVIOUS_TICKET_DATE
    ,LAG(IsNull(LB_CLOSING,'No')) OVER (ORDER BY DT_TICKET) As PREVIOUS_TICKET_CLOSING
    ,ROW_NUMBER() OVER (ORDER BY DT_TICKET) As CD_ROW_NUMBER
FROM 
    TICKETS

Answer 1

这是使用 LAG window 函数

的一种方法

SELECT 'C' + RIGHT(Concat('00', 1+Sum(CASE WHEN prev_val = 'Yes' THEN 1 ELSE 0 END)
                                        OVER (ORDER BY [date])), 3) AS [Case No], 
       [Ticket No],
       [Date],
       [Closed]
FROM   (SELECT *, Lag(closed) OVER( ORDER BY [date]) AS prev_val 
        FROM   table1) a

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