SQLite 5 表，连接，求和，GroupBy

Question

我正在拼命处理一个问题，但无法解决。我有一个包含五个 table 和相应列的 SQLite 数据库：

Tab1 = {Job_ID, Company_ID, Source_ID}
Tab1_Category = {JOb_ID, CAtegory_ID}
类别 = {ID, First_level, Second_level}
Tab2 = {Job_ID, Log_Date, 点击次数, 应用程序}
来源 = {ID, 姓名}

我创建了一个示例数据库：

CREATE TABLE TAB1 (
  `job_id` INTEGER,
  `company_id` INTEGER,
  `source_id` INTEGER
);

INSERT INTO TAB1
  (`job_id`, `company_id`, `source_id`)
VALUES
  ('1', '222', '2'),
  ('2', '222', '1'),
  ('3', '222', '1'),
  ('4', '222', '1'),
  ('5', '255', '3');
  
CREATE TABLE TAB1_CATEGORY (
  `job_id` INTEGER,
  `category_id` INTEGER
);

INSERT INTO TAB1_CATEGORY
  (`job_id`, `category_id`)
VALUES
  ('1', '31'),
  ('2', '36'),
  ('3', '33'),
  ('3', '35'),
  ('4', '32'),
  ('4', '31'),
  ('5', '34');
  
CREATE TABLE CATEGORY (
  `id` INTEGER,
  `first_level` VARCHAR(3),
  `second_level` VARCHAR(3)
);

INSERT INTO CATEGORY
  (`id`, `first_level`, `second_level`)
VALUES
  ('30', 'sss', 'aaa'),
  ('31', 'sss', 'aaa'),
  ('32', 'sss', 'bbb'),
  ('33', 'ggg', 'ccc'),
  ('34', 'ggg', 'ddd'),
  ('35', 'ggg', 'eee'),
  ('36', 'hhh', 'fff');
  
CREATE TABLE SOURCE (
  `id` INTEGER,
  `name` VARCHAR(3)
);

INSERT INTO SOURCE
  (`id`, `name`)
VALUES
  ('1', 'mmm'),
  ('2', 'nnn'),
  ('3', 'ooo');

CREATE TABLE TAB2 (
  `job_id` INTEGER,
  `log_date` VARCHAR(10),
  `clicks` INTEGER,
  `applications` INTEGER
);

INSERT INTO TAB2
  (`job_id`, `log_date`, `clicks`, `applications`)
VALUES
  ('1', '01-01-1999', '6', '2'),
  ('1', '02-01-1999', '7', '3'),
  ('1', '03-01-1999', '9', '1'),
  ('2', '02-01-1999', '4', '1'),
  ('2', '05-01-1999', '8', '2'),
  ('3', '03-01-1999', '9', '0'),
  ('4', '05-01-1999', '5', '3'),
  ('4', '06-01-1999', '4', '1'),
  ('5', '01-01-1999', '1', '0'),
  ('5', '03-01-1999', '3', '1');

我需要一次查询得到以下结果>

所有 JOB_ID (Tab1) 和 Company_ID (Tab1) 的列表，其中 First_level（来自 table 类别）是“ggg”或“sss”和名称（来自 table 来源）是 "mmm"
每个 Job_ID
不同 Second_level 的总和（来自 table 类别）
每个 company_ID 的申请总和（一个 company_ID 可以有很多 Job_ids）

这是我目前所做的，但没有按照我想要的方式工作>

SELECT t1.job_id, t1.company_id,
SUM(t2.clicks), SUM(t2.applications), COUNT(DISTINCT c.second_level)
FROM TAB1 t1
JOIN SOURCE s ON s.id = t1.source_id
JOIN TAB1_CATEGORY tc ON t1.job_id = tc.job_id 
JOIN CATEGORY c ON tc.category_id = c.id
JOIN TAB2 t2 ON t1.job_id = t2.job_id
WHERE c.first_level IN ('ggg', 'sss') AND s.NAME ='mmm'
GROUP BY t1.job_id

我得到的是所有 clicks/applications 的总和，而不是每个 job_id。 :

job_id	company_id	SUM(t2.clicks)	SUM(t2.applications)	COUNT(DISTINCT c.second_level)
3	222	18	0	2
4	222	18	8	2

这就是我想要得到的：

job_id	company_id	SUM(t2.clicks)	SUM(t2.applications)	COUNT(DISTINCT c.second_level)	Total Appl per company
3	222	9	0	2	4
4	222	9	4	2	4

Answer 1

首先，您必须在 TAB2 内聚合，然后加入（使用 INNER 加入）。
您还需要 SUM() window 列的函数 Total Appl per company:

SELECT t1.JOB_ID, t1.COMPANY_ID,
       t2.total_clicks, t2.total_apps,
       COUNT(DISTINCT c.SECOND_LEVEL) count_second_level,
       SUM(t2.total_apps) OVER (PARTITION BY t1.COMPANY_ID) [Total Appl per company]
FROM TAB1 t1
INNER JOIN SOURCE s ON s.ID = t1.SOURCE_ID
INNER JOIN TAB1_CATEGORY tc ON t1.JOB_ID = tc.JOB_ID 
INNER JOIN CATEGORY c ON tc.CATEGORY_ID = c.ID 
INNER JOIN (
  SELECT JOB_ID, SUM(CLICKS) total_clicks, SUM(APPLICATIONS) total_apps
  FROM TAB2
  GROUP BY JOB_ID
) t2 ON t1.JOB_ID = t2.JOB_ID
WHERE c.FIRST_LEVEL IN ('ggg', 'sss') AND s.NAME ='mmm'
GROUP BY t1.JOB_ID, t1.COMPANY_ID, t2.total_clicks, t2.total_apps

参见demo。
结果：

> job_id | company_id | total_clicks | total_apps | count_second_level | Total Appl per company
> -----: | ---------: | -----------: | ---------: | -----------------: | ---------------------:
>      3 |        222 |            9 |          0 |                  2 |                      4
>      4 |        222 |            9 |          4 |                  2 |                      4

SQLite 5 表，连接，求和，GroupBy

SQLite 5 tables, Join, Sum, GroupBy

sqlite

join

group-by

inner-join

left-join