Question

如何从NID开始每隔7天选择一个最新的UID，从今天开始算起。

如果今天是7月11日，则下表

+-----+------------+-----+
| NID |  timestamp | UID |
+-----+------------+-----+
|   1 | 1341719851 |   8 | //July 7
|   2 | 1341115051 |   8 | //July 1
|   3 | 1341547051 |   8 | //July 6
|   4 | 1341719851 |   8 | //July 8
|   5 | 1341979051 |   8 | //July 11
|   6 | 1341806251 |   9 | //July 9
|   7 | 1341460651 |   9 | //July 5
|   8 | 1341892651 |   9 | //July 10
+-----+------------+-----+

将输出：

+-----+------------+-----+
| NID |  timestamp | UID |
+-----+------------+-----+
|   2 | 1341115051 |   8 | //July 1
|   5 | 1341979051 |   8 | //July 11
|   8 | 1341892651 |   9 | //July 10
+-----+------------+-----+

在过去7天内，每位用户的最新NID为'5'和'8'，在之前的7天内，最近的NID为'2'等等...

我假设，Group By会做的伎俩;但我不知道从哪里开始。

更新

根据最佳答案，这是有效的查询：

SELECT nid, timestamp, uid, weeks_ago
FROM (
    SELECT nid, timestamp, uid, FLOOR(
        (UNIX_TIMESTAMP()- timestamp)/604800
    ) weeks_ago
    FROM `table`
    ORDER BY timestamp DESC
) x
GROUP BY uid, weeks_ago

Answer 1

select nid, max(timestamp), uid, weeks_ago
from (select nid, timestamp, uid, floor(datediff(now(), from_unixtime(timestamp))/7) weeks_ago
      from mytable) x
group by nid, uid, weeks_ago

Answer 2

  select * from (
    (
      select nid, uid,timestamp,floor(datediff(now(), from_unixtime(timestamp))/7) as   weeks,from_unixtime(timestamp) as Dt from test t1 order by dt desc limit 0,2)
 Union
  (
  select nid, uid,timestamp,
  floor(datediff(now(), from_unixtime(timestamp))/7) as weeks,from_unixtime(timestamp) as Dt
 from test t2 group by weeks having weeks>0 order by dt asc limit 0,1
    )
     ) t4

可能会解决您的问题。试试这个

按多列和时间跨度对MYSQL结果进行分组

2 个答案: