Question

我正在使用Postgres 9.1来尝试获取一些数据我有三张（虚构/消毒）表：

Table "public.students"
    Column    |            Type             | Modifiers 
--------------+-----------------------------+-----------
 id           | uuid                        | not null
 name         | character varying           | 
 birth_date   | date                        | 
 last_exam_at | timestamp without time zone | 
 created_at   | timestamp without time zone | 
 code         | character varying           | 
 gender       | character varying           | 
Indexes:
    "students_id_key" UNIQUE CONSTRAINT, btree (id)
Referenced by:
    TABLE "exams" CONSTRAINT "exams_student_id_fkey"
                  FOREIGN KEY (student_id) REFERENCES students(id)

Table "public.exams_essays"
  Column  | Type | Modifiers 
----------+------+-----------
 exam_id  | uuid | 
 essay_id | uuid | 

Table "public.exams"
      Column       |            Type             | Modifiers 
-------------------+-----------------------------+-----------
 id                | uuid                        | not null
 student_id        | uuid                        | 
 created_at        | timestamp without time zone | 
 completed_at      | timestamp without time zone | 
 course            | character varying           | 

Table "public.essays"
     Column      |            Type             | Modifiers 
-----------------+-----------------------------+-----------
 id              | uuid                        | not null
 essay_type      | character varying           | not null
 filename        | character varying           |

我正在尝试获取以created_at::date和student_id分组的以下信息：

考试总数
历史论文数量（exam.course = 'history'）
英文论文数量（exam.course = 'english'）

这些查询中的每一个都不是很难单独完成，但将它们放在一起证明是困难的。

Answer 1

SELECT COUNT(DISTINCT EX.exam_id) TotalExams,
       SUM(CASE WHEN E.course = 'history' THEN 1 ELSE 0 END) HistoryEssays,
       SUM(CASE WHEN E.course = 'english' THEN 1 ELSE 0 END) EnglishEssays
FROM public.exams AS EX
LEFT JOIN public.exams_essays AS EE
    ON EX.exam_id = EE.essay_id

Answer 2

SELECT created_at::date, student_id
      ,count(*) AS exams
      ,sum(CASE WHEN course = 'history' THEN essays ELSE 0 END) AS essays_hist
      ,sum(CASE WHEN course = 'english' THEN essays ELSE 0 END) AS essays_engl
FROM   public.exams x
LEFT   JOIN (
   SELECT exam_id AS id, count(*) AS essays
   FROM   public.exams_essays
   GROUP  BY exam_id
   ) s USING (id)
GROUP  BY created_at::date, student_id;

我在连接之前聚合了n-table ，从而避免了行开头的乘法。使查询更简单（IMO）和更快。

使用多个条件的计数查询

2 个答案: