如何找到在SQL

Question

連續數據我有一個SQL表以下數據：

我想找到沒有與ID 3點連續的數據組。結果是

如何編寫query.please幫助。

Answer 1

這裏是查詢其只選擇這裏居然連續3行是行：

SELECT a.* 
FROM 
TABLE as a 
inner join 
TABLE as b on (a.no+1=b.no and a.id=b.id) 
inner join 
TABLE as c on (a.no+2=c.no and a.id=c.id) 
order by a.id, a.no 

爲您的數據將提供：

4 a1 4 
5 a1 3 
1 a2 2 
2 a2 4 
3 a3 2 
4 a3 3 

行（6，A1,1），（3 ，a2,5），（5，a3,4）沒有被選擇，因爲沒有（8，A1）（5，a2）和（7，A3）

Answer 2

DECLARE @temp TABLE (NO int,ID VARCHAR(2),QTY int) 
INSERT INTO @temp 
SELECT 1,'A1',5 UNION ALL 
SELECT 4,'A1',4 UNION ALL 
SELECT 5,'A1',3 UNION ALL 
SELECT 6,'A1',1 UNION ALL 
SELECT 7,'A1',0 UNION ALL 
SELECT 9,'A1',5 UNION ALL 
SELECT 12,'A1',3 UNION ALL 
SELECT 1,'A2',2 UNION ALL 
SELECT 2,'A2',4 UNION ALL 
SELECT 3,'A2',5 UNION ALL 
SELECT 4,'A2',1 UNION ALL 
SELECT 7,'A2',4 UNION ALL 
SELECT 9,'A2',5 UNION ALL 
SELECT 1,'A3',0 UNION ALL 
SELECT 3,'A3',2 UNION ALL 
SELECT 4,'A3',3 UNION ALL 
SELECT 5,'A3',4 UNION ALL 
SELECT 6,'A3',2; 

WITH tmpa AS 
(
    SELECT * 
     , NO - ROW_NUMBER() OVER(PARTITION BY ID ORDER BY ID) AS grp 
    FROM @temp 
) 
, tmpb AS 
(
    SELECT * 
     , COUNT(*) OVER(PARTITION BY ID,grp) AS grpcount 
    FROM tmpa 
) 
SELECT NO,ID,QTY FROM tmpb WHERE grpcount>1; 

結果是

4 A1 4 
5 A1 3 
6 A1 1 
7 A1 0 
1 A2 2 
2 A2 4 
3 A2 5 
4 A2 1 
3 A3 2 
4 A3 3 
5 A3 4 
6 A3 2 

我從此鏈接中找到此查詢。 Find 「n」 consecutive free numbers from table

http://sqlfiddle.com/#!1/a2633/2

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