总和不等于并从SQL查询结果中删除重复项

Question

我的基本查询：

SELECT project_id
       name
       stories_produced
       on_date
      FROM project_prod
      WHERE on_date IN ('2017-03-01', '2017-06-10')
      ORDER BY project_id

它可以为我提供这些输出：

输出示例：

id       name                   stories_produced        on_date
1042    project 1               1001 (wanted)           2017-03-01
1042    project 1               1801 (wanted)           2017-06-10
1568    project 2               355  (wanted)           2017-06-10
1405    project 3               1    (not wanted)       2017-03-10
1405    project 3               1    (not wanted)       2017-06-10

Obs：（id，on_date）受到限制，这意味着在特定日期始终只能有一个项目生产的记录。

具有相同ID且在两个日期中都存在并且具有不同生产价值（需要）]的重复记录

单个记录，仅在其中一个日期上存在（需要）

问题：*

具有相同ID且在两个日期中都存在并且具有等于

生产值（不需要）的重复记录

我当前的查询，需要更改

      select project_id 
        name 
        CASE
          WHEN max(stories_produced) - min(stories_produced) = 0    
          THEN max(stories_produced)
          ELSE max(stories_produced) - min(stories_produced)
        END AS 'stories_produced'
        from project_prod
      WHERE on_date IN ('2017-03-01', '2017-06-10')
      group by project_id;
输出示例：

id       name                   stories_produced        
1042    project 1               800 (wanted)
1568    project 2               355  (wanted)
1405    project 3               1    (not wanted)
CASE当前正在not

处理第三个约束（重复的记录，具有相同的id，并且在两个日期中都存在，并且具有相等的生产值（不需要）]）] >>

是否有可能满足此条件？

我的基本查询：从project_prod中选择project_id名称为story_produced on_date，在on_date IN（'2017-03-01'，'2017-06-10'）ORDER BY project_id可以获取...

<<

一个选项使用not exists：

select
    p.project_id,
    p.name,
    p.stories_produced,
    p.on_date,
  from project_prod p
  where 
    on_date in ('2017-03-01', '2017-06-10')
    and not exists (
        select 1
        from project_prod p1
        where 
            p1.on_date in ('2017-03-01', '2017-06-10')
            and p1.on_date <> p.date
            and p1.id = p.id 
            and p1.stories_produced <> p.stories_produced

    )
  order by project_id 

Answer 1

一个选项使用not exists：