0

我想按日期计算项目数，但有时该表不包含带有某些日期的行，并且结果中有一个漏洞，例如：

+------------+--------+
| date       | number |
+------------+--------+
| 2021-03-12 |      1 |
| 2021-03-15 |      1 |
| 2021-03-17 |      1 |
| 2021-03-18 |      1 |
+------------+--------+

我想要这个结果：

+------------+--------+
| date       | number |
+------------+--------+
| 2021-03-12 |      1 |
| 2021-03-13 |      0 |
| 2021-03-14 |      0 |
| 2021-03-15 |      1 |
| 2021-03-16 |      0 |
| 2021-03-17 |      1 |
| 2021-03-18 |      1 |
+------------+--------+

所以，我创建了一个临时表WITH，它可以工作：

WITH recursive Date_Ranges AS (
       select DATE(NOW() - INTERVAL 3 MONTH) AS Date
       union
       select Date + interval 1 day
       from Date_Ranges
       where Date < DATE(NOW())
   )
    SELECT *, 0 AS number FROM Date_Ranges

接下来，我想检索统计信息，我尝试使用以下查询：

WITH recursive Date_Ranges AS (
   select DATE(NOW()- INTERVAL 3 MONTH) AS Date
   union
   select Date + interval 1 day
   from Date_Ranges
   where Date < DATE(NOW())
)
select *, '0' AS value from Date_Ranges
UNION DISTINCT 
SELECT DATE(time) AS date, count(*) AS number
                    FROM my_table
                    WHERE time >= NOW()- INTERVAL 3 MONTH
                    AND time <= DATE(NOW())
                    GROUP BY DATE(ca_time)

但是UNION因为值不同所以不起作用，所以mysql保留2行：

example:
+------------+--------+
| date       | number |
+------------+--------+
| 2021-03-12 |      0 |
| 2021-03-13 |      0 |
| 2021-03-14 |      0 |
| 2021-03-15 |      0 |
| 2021-03-16 |      0 |
| 2021-03-17 |      0 |
| 2021-03-18 |      0 |
| 2021-03-12 |      2 |
| 2021-03-13 |      6 |
| 2021-03-14 |      7 |
| 2021-03-15 |      8 |
| 2021-03-16 |      9 |
| 2021-03-17 |      5 |
| 2021-03-18 |      0 |
+------------+--------+

0 的行如何与第二个联合合并？

谢谢。

编辑：

我试过这个查询，但没有结果的日期不在这里：

WITH recursive dates AS (
   SELECT DATE(NOW() - INTERVAL 3 MONTH) AS Date
   UNION ALL
   SELECT Date + interval 1 day
   FROM dates
   WHERE Date < DATE(NOW())
)
SELECT date, COALESCE(COUNT(1), 0) AS number
FROM dates
LEFT JOIN geo
ON dates.date = DATE(geo.ca_time)
WHERE departmentCode = '89'
GROUP BY geo.ca_time
ORDER BY geo.ca_time

结果：

+------------+--------+
| date       | number |
+------------+--------+
| 2021-02-21 |      1 |
| 2021-02-22 |     11 |
| 2021-02-27 |      8 |
...
4

2 回答 2

1

UNION ALL 版本

WITH recursive Date_Ranges AS (
   select DATE(NOW()- INTERVAL 3 MONTH) AS Date
   union
   select Date + interval 1 day
   from Date_Ranges
   where Date < DATE(NOW())
)   
select date, count(*) - 1 AS number
from (
   select date
   from Date_Ranges
   union all
   select DATE(time) 
   from my_table
   where time >= NOW() - INTERVAL 3 MONTH
             AND time <= DATE(NOW())
) t
group by date

db<>小提琴

于 2021-05-17T13:54:24.207 回答
0

加入你的 cte 和你的桌子：

WITH recursive Date_Ranges AS (
   select DATE(NOW()- INTERVAL 3 MONTH) AS Date
   union
   select Date + interval 1 day
   from Date_Ranges
   where Date < DATE(NOW())
)

select dr.date ,coalesce(sum(t.num),0)
from Date_Ranges dr
left join my_table t
 on dr.date = t.datecol
group by dr.date

db<>在这里摆弄

于 2021-05-17T13:40:23.747 回答