MySQL查询，用于计算某些数字的连续出现次数

还有一个问题，但我想以前没人问过

我有一张像这样的桌子

+----+------+---------------------+------+-----------+--------------+
| id | unit |       datetime      | idle | idlecount | boutduration |
+----+------+---------------------+------+-----------+--------------+
| 1  |  A   | 2009-12-04 08:10:05 |  139 |           |              |
| 2  |  A   | 2009-12-04 08:20:05 |  107 |           |              |
| 3  |  A   | 2009-12-04 08:30:05 |  0   |           |              |
| 4  |  A   | 2009-12-04 08:40:05 |  144 |           |              |
| 5  |  A   | 2009-12-04 08:50:05 |  0   |           |              |
| 6  |  A   | 2009-12-04 09:00:05 |  0   |           |              |
| 7  |  A   | 2009-12-04 09:10:05 |  58  |           |              |
| 8  |  A   | 2009-12-04 09:20:05 |  0   |           |              |
| 9  |  A   | 2009-12-04 09:30:05 |  0   |           |              |
| 10 |  A   | 2009-12-04 09:40:05 |  0   |           |              |
| 11 |  A   | 2009-12-04 09:50:05 |  0   |           |              |
| 12 |  A   | 2009-12-04 10:00:05 |  107 |           |              |
| 13 |  A   | 2009-12-04 10:10:05 |  0   |           |              |
| 14 |  A   | 2009-12-04 10:20:05 |  144 |           |              |
| etc...

+----+------+---------------------+------+-----------+--------------+
| id | unit |       datetime      | idle | idlecount | boutduration |
+----+------+---------------------+------+-----------+--------------+
| 1  |  A   | 2009-12-04 08:10:05 |  139 |           |              |
| 2  |  A   | 2009-12-04 08:20:05 |  107 |           |              |
| 3  |  A   | 2009-12-04 08:30:05 |  0   |           |              |
| 4  |  A   | 2009-12-04 08:40:05 |  144 |           |              |
| 5  |  A   | 2009-12-04 08:50:05 |  0   |     2     |   00:20:00   |
| 6  |  A   | 2009-12-04 09:00:05 |  0   |           |              |
| 7  |  A   | 2009-12-04 09:10:05 |  58  |           |              |
| 8  |  A   | 2009-12-04 09:20:05 |  0   |     4     |   00:40:00   |
| 9  |  A   | 2009-12-04 09:30:05 |  0   |           |              |
| 10 |  A   | 2009-12-04 09:40:05 |  0   |           |              |
| 11 |  A   | 2009-12-04 09:50:05 |  0   |           |              |
| 12 |  A   | 2009-12-04 10:00:05 |  107 |           |              |
| 13 |  A   | 2009-12-04 10:10:05 |  0   |           |              |
| 14 |  A   | 2009-12-04 10:20:05 |  144 |           |              |
| etc...

我需要计算列idle中连续出现的零的数量，并确定每个序列的持续时间。单一事件是不相关的。所以结果应该是这样的

+----+------+---------------------+------+-----------+--------------+
| id | unit |       datetime      | idle | idlecount | boutduration |
+----+------+---------------------+------+-----------+--------------+
| 1  |  A   | 2009-12-04 08:10:05 |  139 |           |              |
| 2  |  A   | 2009-12-04 08:20:05 |  107 |           |              |
| 3  |  A   | 2009-12-04 08:30:05 |  0   |           |              |
| 4  |  A   | 2009-12-04 08:40:05 |  144 |           |              |
| 5  |  A   | 2009-12-04 08:50:05 |  0   |           |              |
| 6  |  A   | 2009-12-04 09:00:05 |  0   |           |              |
| 7  |  A   | 2009-12-04 09:10:05 |  58  |           |              |
| 8  |  A   | 2009-12-04 09:20:05 |  0   |           |              |
| 9  |  A   | 2009-12-04 09:30:05 |  0   |           |              |
| 10 |  A   | 2009-12-04 09:40:05 |  0   |           |              |
| 11 |  A   | 2009-12-04 09:50:05 |  0   |           |              |
| 12 |  A   | 2009-12-04 10:00:05 |  107 |           |              |
| 13 |  A   | 2009-12-04 10:10:05 |  0   |           |              |
| 14 |  A   | 2009-12-04 10:20:05 |  144 |           |              |
| etc...

+----+------+---------------------+------+-----------+--------------+
| id | unit |       datetime      | idle | idlecount | boutduration |
+----+------+---------------------+------+-----------+--------------+
| 1  |  A   | 2009-12-04 08:10:05 |  139 |           |              |
| 2  |  A   | 2009-12-04 08:20:05 |  107 |           |              |
| 3  |  A   | 2009-12-04 08:30:05 |  0   |           |              |
| 4  |  A   | 2009-12-04 08:40:05 |  144 |           |              |
| 5  |  A   | 2009-12-04 08:50:05 |  0   |     2     |   00:20:00   |
| 6  |  A   | 2009-12-04 09:00:05 |  0   |           |              |
| 7  |  A   | 2009-12-04 09:10:05 |  58  |           |              |
| 8  |  A   | 2009-12-04 09:20:05 |  0   |     4     |   00:40:00   |
| 9  |  A   | 2009-12-04 09:30:05 |  0   |           |              |
| 10 |  A   | 2009-12-04 09:40:05 |  0   |           |              |
| 11 |  A   | 2009-12-04 09:50:05 |  0   |           |              |
| 12 |  A   | 2009-12-04 10:00:05 |  107 |           |              |
| 13 |  A   | 2009-12-04 10:10:05 |  0   |           |              |
| 14 |  A   | 2009-12-04 10:20:05 |  144 |           |              |
| etc...

我认为这需要使用，但我以前从未使用过，并且对MySQL文档感到有点害怕。尽管如此，我还是梦想着一个单一的查询解决方案

---

干杯，Tom

你可以通过相关子查询来实现这一点。第一个获取每个之后的下一个非零值。第二个计算中间的值。以下是获取空闲计数的示例：

select t.*,
       (select (case when count(*) > 1 then count(*) end)
        from t t3
        where t3.id >= t.id and
              (t3.id < t.nextNonZeroIdle or nextNonZeroIdle is NULL) and
              t3.idle = 0
       ) as IdleCOunt
from (select t.id, t.unit, t.datetime, t.idle,
             (select id
              from t t2
              where t2.unit = t.unit and
                    t2.id > t.id and
                    t2.idle > 0
              order by id
              limit 1
             ) as nextNonZeroIdle
      from t
     ) t
from t;

---

哇，我还在想你的答案。但我有一种预感，这个查询可能需要几个小时才能遍历我的表。它大约有170万行；