SQL：根据先前的ID查找

问题描述

| 我有一个如下表：

MysqL> select * from tries;
+----+--------+-----------+
| id | person | succeeded |
+----+--------+-----------+
|  1 |      1 |         1 |
|  2 |      1 |         1 |
|  3 |      2 |         0 |
|  4 |      4 |         1 |
|  5 |      2 |         1 |
|  6 |      2 |         0 |
|  7 |      3 |         0 |
|  8 |      3 |         0 |
|  9 |      3 |         0 |
| 10 |      1 |         0 |
| 11 |      4 |         1 |
| 12 |      4 |         1 |
+----+--------+-----------+

我希望那些（至少）有一个

try

成功的人，紧接着是一个Failed1ѭ失败的人（分别由

1

和

0

给予）。当我说“跟随”时，是指同一个人的上一次尝试，并以较低的ѭ5给出。 因此，在这种情况下： 人2的成功尝试是

id = 5

，失败的是

id = 3

，因此符合条件。 人1没有立即跟随失败的成功，因此未能通过标准 人3没有成功，因此不符合条件 人4没有失败，因此未通过条件 我将如何编写这样的查询？     

解决方法

SELECT t1.person,MIN(t1.id) as SuccessID
    FROM tries t1
    WHERE t1.succeeded = 1
        AND t1.person IN (SELECT t2.person
                              FROM tries t2
                              WHERE t2.succeeded = 0
                                  AND t2.id < t1.id)
    GROUP BY t1.person

    ,

Select ...
From tries As T
    Join tries As T2
        On T2.id = T.id + 1
            And T2.succeeded = 0
Where T.succeeded = 1

如果我们不能认为Ids是完全连续的：

Select ...
From tries As T
    Join    (
            Select T1.id,Min(T2.Id) As NextId
            From tries As T1
                Join tries As T2
                    On T2.id > T.id
            Group By T1.id
            ) As TriesAndNext
        On TriesAndNext.Id = T.Id
    Join tries As TNext
        On TNext.Id = TriesAndNext.NextId
            And TNext.succeeded = 0
Where T.succeeded = 1