Question

我在MySQL中有一个名为Session的表，如下所示：

+-----+---------------------+---------------------+--------------------+
| id  | start               | finish              | name               |
+-----+---------------------+---------------------+--------------------+
| -12 | 2013-04-27 09:00:00 | 2013-04-27 13:00:00 | Saturday Setup 1   |
| -11 | 2013-04-27 13:00:00 | 2013-04-27 18:00:00 | Saturday Setup 2   |
| -10 | 2013-04-27 23:00:00 | 2013-04-28 08:00:00 | Saturday Night     |
|  -3 | 2013-04-28 08:00:00 | 2013-04-28 13:00:00 | Sunday Setup 1     |
|  -2 | 2013-04-28 13:00:00 | 2013-04-28 18:00:00 | Sunday Setup 2     |
|  -1 | 2013-04-28 23:00:00 | 2013-04-29 08:00:00 | Sunday Night       |
|   1 | 2013-04-29 09:00:00 | 2013-04-29 13:00:00 | Monday Setup 1     |
|   2 | 2013-04-29 13:00:00 | 2013-04-29 17:00:00 | Monday Setup 2     |
|   3 | 2013-04-29 17:00:00 | 2013-04-29 21:00:00 | Monday Setup 3     |
|   4 | 2013-04-29 23:00:00 | 2013-04-30 08:00:00 | Monday Night       |
|  10 | 2013-04-30 09:00:00 | 2013-04-30 13:00:00 | Tuesday Setup 1    |
|  11 | 2013-04-30 13:00:00 | 2013-04-30 17:00:00 | Tuesday Setup 2    |
|  12 | 2013-04-30 23:00:00 | 2013-05-01 08:00:00 | Tuesday Night      |
more rows etc...

我想创建一个查询，该查询将选择表格中的所有列，其他列显示在按开始排序的同一天（如果有）的下一个ID。

结果如下：

+-----+--------+---------------------+---------------------+--------------------+
| id  | nextid | start               | finish              | name               |
+-----+--------+---------------------+---------------------+--------------------+
| -12 |    -11 | 2013-04-27 09:00:00 | 2013-04-27 13:00:00 | Saturday Setup 1   |
| -11 |    -10 | 2013-04-27 13:00:00 | 2013-04-27 18:00:00 | Saturday Setup 2   |
| -10 |        | 2013-04-27 23:00:00 | 2013-04-28 08:00:00 | Saturday Night     |
|  -3 |     -2 | 2013-04-28 08:00:00 | 2013-04-28 13:00:00 | Sunday Setup 1     |
|  -2 |     -1 | 2013-04-28 13:00:00 | 2013-04-28 18:00:00 | Sunday Setup 2     |
|  -1 |        | 2013-04-28 23:00:00 | 2013-04-29 08:00:00 | Sunday Night       |
|   1 |      2 | 2013-04-29 09:00:00 | 2013-04-29 13:00:00 | Monday Setup 1     |
|   2 |      3 | 2013-04-29 13:00:00 | 2013-04-29 17:00:00 | Monday Setup 2     |
|   3 |      4 | 2013-04-29 17:00:00 | 2013-04-29 21:00:00 | Monday Setup 3     |
|   4 |        | 2013-04-29 23:00:00 | 2013-04-30 08:00:00 | Monday Night       |
|  10 |     11 | 2013-04-30 09:00:00 | 2013-04-30 13:00:00 | Tuesday Setup 1    |
|  11 |     12 | 2013-04-30 13:00:00 | 2013-04-30 17:00:00 | Tuesday Setup 2    |
|  12 |        | 2013-04-30 23:00:00 | 2013-05-01 08:00:00 | Tuesday Night      |
more rows etc...

任何建议都会受到赞赏......

Answer 1

它不会非常有效，因为索引不能用于评估自我加入标准DATE(..) = DATE(..)，但您实际上是在尝试获取groupwise minimum：

SELECT cur.id, nxt.id nextid, cur.start, cur.finish, cur.name
FROM (
  SELECT      s1.*, MIN(s2.start) next
  FROM        Session s1
    LEFT JOIN Session s2
           ON DATE(s1.start) = DATE(s2.start)
          AND s1.start < s2.start
  GROUP BY    s1.start
) cur LEFT JOIN Session nxt ON cur.next = nxt.start

在sqlfiddle上查看。

Answer 2

我不确定，但你可以试试这个

为会话表名称创建一个视图，它可以在同一天选择显示下一个ID的附加列
然后您可以从表格和seesion_view中选择数据。
我不确定使用加入是否正常，但您也可以尝试使用

最后你的MySql应该是5.4 +

Answer 3

这是我认为相关子查询很有用的地方：

select id,
       (select id from session s2 where date(s.start) = date(s2.start) order by start desc limit 1
       ) as nextId
       start, finish, name
from session s

Answer 4

使用开始日期来确定天数是否匹配，这样的事情应该这样做。一个subselect来获取min id，它比同一天的另一个id大。我不敢测试。

SELECT a.id, Sub1.MinId AS nextid, a.start, a.finish, a.name  
FROM SomeTable a
LEFT OUTER JOIN (SELECT a.id, MIN(b.id) AS MinId
FROM SomeTable a
INNER JOIN SomeTable b
ON DATE(a.start) = DATE(b.start)
AND a.id < b.id
GROUP BY a.id) Sub1
ON a.id = Sub1.id

或者id不在开始顺序

SELECT a.id, c.id AS nextid, a.start, a.finish, a.name  
FROM SomeTable a
LEFT OUTER JOIN (SELECT a.id, MIN(b.start) AS MinStart
FROM SomeTable a
INNER JOIN SomeTable b
ON DATE(a.start) = DATE(b.start)
AND a.id < b.id
GROUP BY a.id) Sub1
ON a.id = Sub1.id
LEFT OUTER JOIN SomeTable c
ON Sub1.MinStart = c.start

下一行的MySQL条件查询

4 个答案: