Return 在分组依据中计数 0

Question

我已经 table 调用了 Errorlogs 并提供了以下数据

+----+-------------+------------+
| ID | sps_Bereich | Date       |
+----+-------------+------------+
| 10 | helpeu1     | 2020-07-31 |
+----+-------------+------------+
| 11 | helpeu2     | 2020-07-31 |
+----+-------------+------------+
| 12 | helpeu3     | 2020-07-31 |
+----+-------------+------------+
| 13 | helpeu3     | 2020-07-31 |
+----+-------------+------------+
| 14 | helpeu4     | 2020-07-31 |
+----+-------------+------------+
| 15 | helpeu3     | 2020-07-30 |
+----+-------------+------------+
| 16 | helpeu4     | 2020-07-30 |
+----+-------------+------------+
| 17 | helpeu4     | 2020-07-30 |
+----+-------------+------------+
| 18 | helpeu4     | 2020-07-30 |
+----+-------------+------------+

我还有一个 table 叫 Bereich

+----+---------+
| ID | Name    |
+----+---------+
| 1  | helpeu1 |
+----+---------+
| 2  | helpeu2 |
+----+---------+
| 3  | helpeu3 |
+----+---------+
| 4  | helpeu4 |
+----+---------+

我想按 sps_bereich 和日期分组，我正在使用以下查询

select sps_bereich, [Date], count(*) as [Nr of errors] from [ErrorLogs] 
   where SPS_Bereich IN (
   select Name from [Bereich]
   )  
   group by SPS_Bereich, Date 
   order by date desc, SPS_Bereich asc

它给了我以下输出。

+-------------+------------+--------------+
| sps_bereich | Date       | Nr of errors |
+-------------+------------+--------------+
| helpeu1     | 2020-07-31 | 1            |
+-------------+------------+--------------+
| helpeu2     | 2020-07-31 | 1            |
+-------------+------------+--------------+
| helpeu3     | 2020-07-31 | 2            |
+-------------+------------+--------------+
| helpeu4     | 2020-07-31 | 1            |
+-------------+------------+--------------+
| helpeu3     | 2020-07-30 | 1            |
+-------------+------------+--------------+
| helpeu4     | 2020-07-30 | 3            |
+-------------+------------+--------------+

我想要的是在没有数据的地方包含 sps_bereich。在这种情况下，我想将 helpeu1,helpeu2 添加为 0 日期 2020-07-30

我想要以下输出。

+-------------+------------+--------------+
| sps_bereich | Date       | Nr of errors |
+-------------+------------+--------------+
| helpeu1     | 2020-07-31 | 1            |
+-------------+------------+--------------+
| helpeu2     | 2020-07-31 | 1            |
+-------------+------------+--------------+
| helpeu3     | 2020-07-31 | 2            |
+-------------+------------+--------------+
| helpeu4     | 2020-07-31 | 1            |
+-------------+------------+--------------+
| helpeu1     | 2020-07-30 | 0            |
+-------------+------------+--------------+
| helpeu2     | 2020-07-30 | 0            |
+-------------+------------+--------------+
| helpeu3     | 2020-07-30 | 1            |
+-------------+------------+--------------+
| helpeu4     | 2020-07-30 | 3            |
+-------------+------------+--------------+

我在想添加一个 IN 子句必须 return 我的所有值，但事实并非如此。

这里是 SQLFiddle.

Answer 1

使用左连接 -

DEMO

select A.Name, A.date,coalesce(B.cnt,0) as no_of_errors 
from
(select distinct date,name from Errorlogs cross apply Bereich)A
left join 
(select sps_Bereich, [Date],count(*) cnt
from Errorlogs group by sps_Bereich, [Date]
)B
on A.date=B.date and A.Name=B.sps_Bereich order by A.date desc, A.name asc

Answer 2

试试这个。它构建一个完整的姓名和日期网格，然后用计数填充它。

 select grid.name,grid.date,coalesce(cnt.cnt,0) from
  
    (select distinct e.date, d.name from errorlogs e, Bereich d) grid
    left outer join
    (
   select b.sps_Bereich, b.date, count(*) as cnt from [ErrorLogs] b
  
     group by b.sps_Bereich,b.date
    ) cnt  
    on grid.date=cnt.date and grid.name=cnt.sps_bereich

Answer 3

我建议将查询写成：

select b.name, d.date, count(el.id) as num_errors 
from Bereich b cross join
     (select distinct date from errorlogs) d left join
     errorlogs el
     on el.date = d.date and el.sps_Bereich = b.name 
group by b.name, d.date
order by d.date desc, b.name;

这个想法很简单。使用 cross join 在结果集中生成您想要的行。然后用left join把已有的数据带进来

Here 是一个 db<>fiddle.

Return 在分组依据中计数 0

Return Count 0 in group by

sql

sql-server

group-by

sql-server-2008