Question

我在MariaDB中有一个名为“事件”的表，下面是数据

  <connectionStrings><add name="MyEntities" connectionString="metadata=res://*/Model.MyReplicationModel.csdl|res://*/Model.MyReplicationModel.ssdl|res://*/Model.MyReplicationModel.msl;provider=System.Data.SqlClient;provider connection string=&quot;data source=********;initial catalog=**********;persist security info=True;user id=sa;password=*******;MultipleActiveResultSets=True;App=EntityFramework&quot;" providerName="System.Data.EntityClient"/></connectionStrings>

我想计算一个月中每天每天的所有“开放”状态，并使用一个参数来选择要获取的月份和年份。

例如，我的月份值为“ 07”，年份为“ 2018”。它应该返回：

+-----+----------+---------------------+
| ID  | status   |     created_at      |
+-----+----------+---------------------+
|  1  | open     | 2018-07-03 16:15:24 |
|  2  | open     | 2018-07-03 16:15:24 |
|  3  | open     | 2018-07-05 16:15:24 |
|  4  | open     | 2018-07-08 16:15:24 |
|  5  | closed   | 2018-07-15 16:15:24 |
+-----+----------+---------------------+

Answer 1

请尝试以下查询：

   select thedate,case when status is null then 0 else 1 end as count
     from (Select '2018-07-01' As [TheDate]
         Union All
         Select DateAdd(month, 1, TheDate) From dt Where [TheDate] < '2018-07-31') as dt left join tablename 
    on dt.thedate=tablename.date

Answer 2

使用下面的代码来实现结果集。

声明两个变量：

var_from_date =月的第一天的日期。 var_daydiff =每月的天数

         WITH RECURSIVE cte_days (months, days, dates, n) AS
        (
          SELECT DATE(DATE_ADD(var_from_date, INTERVAL 0 DAY)) as dates,
                    0 as n
          UNION ALL
          SELECT DATE(DATE_ADD(var_from_date, INTERVAL n+1 DAY)) as dates, 
                    n + 1 as n 
          FROM cte_days WHERE n <= var_daydiff
        )
        SELECT ch.date, IFNULL(COUNT(t.id), 0) as count
        FROM cte_days ch
        LEFT JOIN table t on ch.dates = t.date
        GROUP BY ch.date;

Answer 3

以下您可以根据需要生成日期的方式我已经给where gen_date between '2018-07-01' and '2018-12-31'了，您必须使用放置并使用case when和aggregate函数的日期条件

    select gen_date as date ,sum(case when status='open' then 1 else 0 end) as Cnt from

    (    select * from 
        (select adddate('1970-01-01',t4*10000 + t3*1000 + t2*100 + t1*10 + t0) gen_date from
         (select 0 t0 union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t0,
         (select 0 t1 union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t1,
         (select 0 t2 union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t2,
         (select 0 t3 union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t3,
         (select 0 t4 union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t4) v
        where gen_date between '2018-07-01' and '2018-12-31'
    ) as T1 left join your_table t2 on T1.gen_date= date(t2.created_at)
group by gen_date

计算特定月份每天的所有记录

3 个答案: