的MySQL得到表

Question

失蹤的ID我在MySQL有這個表，例如：

ID | Name 
1 | Bob 
4 | Adam 
6 | Someguy 

如果您發現，沒有ID號（2,3和5）。

我該如何編寫一個查詢，以便MySQL只回答丟失的ID，在這種情況下：「2,3,5」？

Answer 1

SELECT a.id+1 AS start, MIN(b.id) - 1 AS end 
    FROM testtable AS a, testtable AS b 
    WHERE a.id < b.id 
    GROUP BY a.id 
    HAVING start < MIN(b.id) 

希望這個環節也有助於 http://www.codediesel.com/mysql/sequence-gaps-in-mysql/

Answer 2

一個比較有效的查詢：

SELECT (t1.id + 1) as gap_starts_at, 
     (SELECT MIN(t3.id) -1 FROM my_table t3 WHERE t3.id > t1.id) as gap_ends_at 
FROM my_table t1 
WHERE NOT EXISTS (SELECT t2.id FROM my_table t2 WHERE t2.id = t1.id + 1) 
HAVING gap_ends_at IS NOT NULL 

Answer 3

上面的查詢將會給兩列，所以你可以嘗試這種在單個列拿到丟失號碼

select start from 
(SELECT a.id+1 AS start, MIN(b.id) - 1 AS end 
    FROM sequence AS a, sequence AS b 
    WHERE a.id < b.id 
    GROUP BY a.id 
    HAVING start < MIN(b.id)) b 
UNION 
select c.end from (SELECT a.id+1 AS start, MIN(b.id) - 1 AS end 
    FROM sequence AS a, sequence AS b 
    WHERE a.id < b.id 
    GROUP BY a.id 
    HAVING start < MIN(b.id)) c order by start; 

Answer 4

要添加一點到伊萬的答案，第是版本顯示數字開始時如果1不存在：

SELECT 1 as gap_starts_at, 
     (SELECT MIN(t4.id) -1 FROM testtable t4 WHERE t4.id > 1) as gap_ends_at 
FROM testtable t5 
WHERE NOT EXISTS (SELECT t6.id FROM testtable t6 WHERE t6.id = 1) 
HAVING gap_ends_at IS NOT NULL limit 1 
UNION 
SELECT (t1.id + 1) as gap_starts_at, 
     (SELECT MIN(t3.id) -1 FROM testtable t3 WHERE t3.id > t1.id) as gap_ends_at 
FROM testtable t1 
WHERE NOT EXISTS (SELECT t2.id FROM testtable t2 WHERE t2.id = t1.id + 1) 
HAVING gap_ends_at IS NOT NULL;