最大遞歸已用盡

Question

我需要一種有效的方法來傳入參數[StartingNumber]並從[StartingNumber]中進行計數，直到找到一個缺失的數字。

我用下面的SQL來獲取下一個數字：

 DECLARE @StartOffset int 
     SET @StartOffset = 23 
     ; With Missing as (
      select @StartOffset as N where not exists(
    select * from [QUEUE] where QueueNum = @StartOffset AND ismelutash = 1) 
     ), Sequence as 
    (  select @StartOffset as N from [QUEUE] where QueueNum = @StartOffset 
      union all 
      select b.QueueNum from [QUEUE] b inner join Sequence s 
on b.QueueNum = s.N + 1 and b.ismelutash = 1 
     ) 
     select COALESCE((select N from Missing),(select MAX(N)+1 from Sequence)) 

它已經工作了一段時間，但現在當我運行它，我得到「終止聲明。在語句完成之前，最大遞歸100已經耗盡。

任何人有任何想法？由於

編輯：

我加MAXRECURSION但它只是負載和不返回數據：

DECLARE @StartOffset int 
SET @StartOffset = 50 

DECLARE @isMelutash int 
SET @isMelutash = 0 


; With QueueFilters as (
    select queuenum from queue where ismelutash = 1 
), Missing as (
    select @StartOffset as N where not exists(select * from QueueFilters where queuenum = @StartOffset) 
), Sequence as (
    select @StartOffset as N from QueueFilters where queuenum = @StartOffset 
    union all 
    select b.queuenum from QueueFilters b inner join Sequence s on b.queuenum = s.N + 1 
) 
select COALESCE((select N from Missing),(select MAX(N)+1 from Sequence)) 
**OPTION(MAXRECURSION 150)** 

Answer 1

你可以在當前的代碼中使用MAXRECURSION選項

然而，不需要逐行遞增（使用Itzik Ben-Gan的方法）。這將檢測哪裏沒有行上排隊ismelutash箱子= 1太，因爲它使用的帳簿表作爲參考序列

;WITH 
    Pass0 as (select 1 as C union all select 1), --2 rows 
    Pass1 as (select 1 as C from Pass0 as A, Pass0 as B),--4 rows 
    Pass2 as (select 1 as C from Pass1 as A, Pass1 as B),--16 rows 
    Pass3 as (select 1 as C from Pass2 as A, Pass2 as B),--256 rows 
    Pass4 as (select 1 as C from Pass3 as A, Pass3 as B),--65536 rows 
    Pass5 as (select 1 as C from Pass4 as A, Pass4 as B),--4,294,967,296 rows 
    Tally as (select row_number() over(order by C) as Number from Pass5) 
select TOP 1 
    Number 
from 
    Tally T 
    LEFT JOIN --corrected, oops. 
    [QUEUE] Q ON T.Number = Q.QueueNum AND Q.ismelutash = 1 
where 
    T.Number >= @StartOffset AND T.Number <= 1000000 
    AND 
    Q.QueueNum IS NULL 
ORDER BY 
    T.Number 

編輯：在端

原始代碼與MAXRECURSION提示。 This link也較好

 DECLARE @StartOffset int 
     SET @StartOffset = 23 
     ; With Missing as (
      select @StartOffset as N where not exists(
    select * from [QUEUE] where QueueNum = @StartOffset AND ismelutash = 1) 
     ), Sequence as 
    (  select @StartOffset as N from [QUEUE] where QueueNum = @StartOffset 
      union all 
      select b.QueueNum from [QUEUE] b inner join Sequence s 
on b.QueueNum = s.N + 1 and b.ismelutash = 1 
     ) 
     select COALESCE((select N from Missing),(select MAX(N)+1 from Sequence)) 
     OPTION (MAXRECURSION 0) 

Answer 2

我相信這是等價的，它不使用遞歸：

SELECT COALESCE(MIN(QueueNum) 
     , (SELECT Max(QueueNum) + 1 
      FROM [Queue] 
      WHERE QueueNum > @StartOffset) 
     , @StartOffset) 
    FROM [QUEUE] 
    WHERE QueueNum >= @StartOffset 
    AND ismelutash != 1