Question

我正在尝试计算两个数据库之间的时间戳，但由于不是我的设计缺陷，因此有一个时间戳重叠。

SELECT date(time + INTERVAL 8 HOUR) as day, COUNT(DISTINCT comment)
FROM  news.data
GROUP BY day
    UNION ALL
SELECT date(time + INTERVAL 8 HOUR) as day, COUNT(DISTINCT comment)
FROM`news-backup`.`data`
GROUP BY day
ORDER BY year(day) desc, day(day) DESC
LIMIT 20

似乎发生了什么，两个数据库的范围都有一些时间戳，因此它们会为某些日期生成单独的计数。所以它会从新闻和新闻备份中算出今天的数量

EX:
date      count
2013-1-15  10
2013-1-15  13
2013-1-14  8
2013-1-13  15

我想要的是

EX:
date      count
2013-1-15  23
2013-1-14  8
2013-1-13  15

这是一个踢球者，我在视图中需要它，因此有一些限制（不允许子查询）。思考？不，我不能改变DB之间发生的数据转储序列

Answer 1

您不能在视图中放置子查询，但可以在视图中放置视图。

所以：

create view1 as 
    SELECT date(time + INTERVAL 8 HOUR) as day, 'current' as which, COUNT(DISTINCT comment) as cnt
    FROM  news.data
    GROUP BY day
        UNION ALL
    SELECT date(time + INTERVAL 8 HOUR) as day, 'backup' as which, COUNT(DISTINCT comment) as cnt
    FROM`news-backup`.`data`
    GROUP BY day, which

我不确定你合并它们的逻辑是：

create view2 as
    select day, max(cnt)  -- sum(cnt)?  prefer current or backup?
    from view1
    group by day
    ORDER BY day desc

禁止子查询的文档是here。请务必搜索“SELECT语句不能包含”。

如果您有一个包含所有日期的表，则以下“荒谬”的SQL可能有效：

select c.date,
       coalesce( (select count(distinct comment) from news.data where date(time + INTERVAL 8 HOUR) = c.date),
                 (select count(distinct comment) from news_backup.data where date(time + INTERVAL 8 HOUR) = c.date)
               ) as NumComments
from calendar c

此版本假设您首先需要“新”，然后是备份。如果你想要总和，那么你可以添加它们。

计算视图中两个数据库之间的偏移量的时间戳

1 个答案: