Question

我有下表-案例

ID  Date Created    Date Resolved
1   1/1/2015    
2   1/1/2015        1/1/2015
3   2/1/2015
4   1/1/2015        2/1/2015
5   4/1/2015

我必须编写一个SQL查询，该查询返回创建的案例数与每天解决的案例数预期结果：

Day Created Resolved
1/1/2015    3   1
2/1/2015    1   1
3/1/2015    0   0
4/1/2015    1   0

我正在尝试类似的事情：

SELECT 
  DateCreated as Day, 
  COUNT(DateCreated) as Created, 
  DateResolved as Resolved 
FROM cases 
GROUP BY Day;

但是有些错误，有人可以帮助我理解我必须实现的逻辑吗？

Answer 1

您应该对每个结果使用单独的查询，然后使用UNION将它们组合在一起。

尝试一下：

    SELECT day, SUM(created) as created, SUM(solved) as solved FROM
    (

    SELECT c1.date_created as day, COUNT(c1.id) as created, 0 as solved FROM cases c1 GROUP BY c1.date_created

    UNION 

    SELECT c2.date_solved as day, 0 as created, COUNT(c2.id) as solved FROM cases c2 WHERE c2.date_solved IS NOT NULL GROUP BY c2.date_solved
    ) as unionTable

    GROUP BY day;

致谢！

Answer 2

您可以使用union all和group by解决大部分问题：

select dte, sum(created), sum(resolved)
from ((select date_created as dte, 1 as created, 0 as resolved
       from t
      ) union all
      (select date_resolved as dte, 0 as created, 1 as resolved
       from t
      )
     ) t
group by dte;

这不会产生该行：

3/1/2015    0   0

因为日期不在原始数据中。有多种生成方法，具体取决于数据库。或者，您可能有一个日历表，周围摆满了您可能需要的所有日期。

编辑：

在MySQL中，我建议：

select d.dte,
       (select count(*)
        from t
        where t.date_created = d.dte
       ) as created,
       (select count(*)
        from t
        where t.date_resolved = d.dte
       ) as resolved
from (select date('2015-01-01') as dte union all
      select date('2015-02-01') as dte union all
      select date('2015-03-01') as dte union all
      select date('2015-04-01') as dte 
     ) d;

注意：对于t(date_created)和t(date_resolved)上的索引，这可能是最快的查询。

Answer 3

您没有提及正在使用的数据库，所以我给您一个通用的答案。您可以使用FULL OUTER JOIN来连接等式的两边。例如：

select
  coalesce(c.d, r.d) as date,
  c.cnt as created,
  r.cnt as resolved
from (
  select date(date_created) as d, count(*) as cnt 
  from my_table group by date(date_created)
) c
full outer join ( 
  select date(date_resolved) as d, count(*) as cnt
  from my_table group by date(date_resolved)
) r on c.d = r.d

Answer 4

对于SQL Server，这是一个选项。

SELECT A.[DateCreated] [Day]
    , COUNT(id) [Created]
    , (SELECT COUNT( [B].[ID] ) FROM cases B WHERE B.[DateResolved] = A.[DateCreated] ) [Resolved]
FROM cases A
GROUP BY [A].[DateCreated]

这不会创建03/01/2015数据行，因为它不在初始数据集中。

返回已启动案例和已解决案例的表sql

4 个答案: