SQL选择每日间隔之间的日期范围

Question

我有一个数据库

它每小时更新一次

我想选择仅 1天间隔的数据 离。

1. 星期一
2. 星期二
3. 星期三
4. ....

我不想要选择这些间隔之间的所有数据

我不知道如何做到这一点

Select * from highscores where ...........

我想得到的结果：

id player_id total economy research military date (2017-09-07 (first record of that day)

id player_id total economy research military date (2017-09-08 (first record of that day)

id player_id total economy research military date (2017-09-09 (first record of that day)

id player_id total economy research military date (2017-09-10 (first record of that day)

id player_id total economy research military date (2017-09-11 (first record of that day)

.....

由于

这是解决方案：

SELECT 
    h.id, h.player_id, h.total, h.economy, h.research, h.military, h.date
FROM 
    highscores h
WHERE 
    h.id = (SELECT hs.id 
            FROM highscores hs
            WHERE UNIX_TIMESTAMP(DATE_FORMAT(h.date,"%Y-%m-%d 00:00:00")) = UNIX_TIMESTAMP(DATE_FORMAT(hs.date,"%Y-%m-%d 00:00:00"))
              AND h.player_id = hs.player_id
            ORDER BY date ASC
            LIMIT 1)

Answer 1

您可能正在寻找子查询，请勿使用select *

SELECT h.id, h.player_id, h.total, h.economy, h.reasearch, h.military, h.date
FROM highscores h
WHERE h.id = (
    SELECT hs.is FROM highscores hs
    WHERE UNIX_TIMESTAMP(DATE_FORMAT(h.date,"%Y-%m-%d 00:00:00")) = UNIX_TIMESTAMP(DATE_FORMAT(hs.date,"%Y-%m-%d 00:00:00"))
    AND h.player_id = hs.player_id
    ORDER BY date ASC
    LIMIT 1
)

使用子查询，您可以为每条记录选择一天的开头，
按日期排序，最早的第一个，所以当天的开始，并且只有当玩家匹配时 将其限制为1，因为您只需要1个匹配