查询返回太多结果

Question

SQL 查询 return 预期 a.id = 366

的 29 个结果

    select a.name, c.name, MAX(B.date), MAX(b.renew_date) as MAXDATE
    from boson_course c
    inner join boson_coursedetail b on (c.id = b.course_id)
    inner join boson_coursedetail_attendance d on (d.coursedetail_id = b.id)
    inner join boson_employee a on (a.id = d.employee_id)
    where a.id = 366
    GROUP BY a.name, c.name
    order by MAX(b.renew_date), MAX(b.date) desc;

SQL 代码下方 returns 34 个结果，两个不同的 Provides 提供相同课程的多个结果。我知道这些额外的结果是因为我将 e.name 添加到要 return 的列表中。但所需要的只是具有最新日期和提供商名称的 29 个条目。

    select a.name, c.name, e.name, MAX(B.date), MAX(b.renew_date) as MAXDATE
    from boson_course c
    inner join boson_coursedetail b on (c.id = b.course_id)
    inner join boson_coursedetail_attendance d on (d.coursedetail_id = b.id)
    inner join boson_employee a on (a.id = d.employee_id)
    inner join boson_provider e on b.provider_id = e.id
    where a.id = 366
    GROUP BY a.name, c.name, e.name
    order by MAX(b.renew_date), MAX(b.date) desc;

任何人都可以将此代码修改为 return 每个课程具有 MAX(renew_date) 的单个 DISTINCT 提供商名称。

Answer 1

尝试使用 distinct on:

select distinct on (a.name, c.name, e.name), a.name, c.name, e.name,
       B.date, b.renew_date as MAXDATE
from boson_course c
inner join boson_coursedetail b on (c.id = b.course_id)
inner join boson_coursedetail_attendance d on (d.coursedetail_id = b.id)
inner join boson_employee a on (a.id = d.employee_id)
inner join boson_provider e on b.provider_id = e.id
where a.id = 366
ORDER BY a.name, c.name, e.name, B.date desc
order by MAX(b.renew_date), MAX(b.date) desc;

Answer 2

这个returns正好一个行每个不同的组合(a.name, c.name):
最新的 renew_date.
其中，最新的date（可能与全局max(date)不同！）。
其中，字母顺序第一的e.name：

SELECT DISTINCT ON (a.name, c.name)
       a.name AS a_name, c.name AS c_name, e.name AS e_name
     , b.renew_date, b.date
FROM   boson_course       c
JOIN   boson_coursedetail b on c.id = b.course_id
JOIN   boson_coursedetail_attendance d on d.coursedetail_id = b.id
JOIN   boson_employee     a on a.id = d.employee_id
JOIN   boson_provider     e on b.provider_id = e.id
WHERE  a.id = 366
ORDER  BY a.name, c.name
     , b.renew_date DESC NULLS LAST
     , b.date DESC NULLS LAST
     , e.name;

结果按a_name、c_name排序在前。如果您需要 原始排序顺序 ，请将其包装在子查询中：

SELECT *
FROM  (<query from above>) sub
ORDER  BY renew_date DESC NULLS LAST
        , date DESC NULLS LAST
        , a_name, c_name, e_name;

DISTINCT ON 的解释：

Select first row in each GROUP BY group?

为什么 DESC NULL LAST？

PostgreSQL sort by datetime asc, null first?

旁白：不要使用基本类型名称，例如 date 广告列名称。此外，name 几乎从来都不是一个好名字。如您所见，我们必须使用别名来使此查询有用。关于命名约定的一些一般建议：

How to implement a many-to-many relationship in PostgreSQL?

查询返回太多结果

Query returning too many results

sql

postgresql

greatest-n-per-group