我有以下查询:
select date(updated_at) as data, COUNT(id) as numar
from `coupons`
where `user_id` = 5 and `won_by` != 0 and `updated_at` >= '2016-04-01'
group by DAY(updated_at), month(updated_at), year(updated_at)
结果如下:
2016-04-01- 229
2016-04-03- 30
2016-04-04- 6
2016-04-07- 1
2016-04-08- 1
2016-04-10- 1
我可以做些什么来接收这样的事情:
2016-04-01- 229
2016-04-02- 0
2016-04-03- 30
2016-04-04- 6
2016-04-05- 0
2016-04-06- 0
2016-04-07- 1
2016-04-08- 1
2016-04-10- 1
答案 0 :(得分:1)
我发现这样做的最好方法是简单地创建(并维护)具有单个列的辅助表,其中包含您关心的所有日期。类似的东西:
CREATE TABLE date_join( date date not null主键 );
然后以任何方便的方式插入每个日期的记录(手动,如果它是一次性的,作为日常过程的一部分,通过存储过程等)。
此时,它只是date_join和初始查询的左连接,使用CASE语句将NULL转换为0:
SELECT dj.date, q.numar
FROM date_join dj
LEFT JOIN (select date(updated_at) as date, COUNT(id) as numar
from `coupons`
where `user_id` = 5 and `won_by` != 0 and `updated_at` >= '2016-04-01'
group by DATE(updated_at)
) q
ON dj.date = q.date
ORDER BY dj.date;