Sql server 4带计数查询的表联接

我使用的SQL Server数据库有4个表，如下所示：

Product表：

Product\u id

是表的主键：

| Product_id | Product_name | Description |

项目表：

项目id

是表的主键，

产品id

是

产品

表的外键

| Project_id | Project_name | Project_start_date | Product_id |

| Participant_id | Participant_name | Participant_email | Project_ID |

Participant表：

Participant\u id

是表的主键，

Project\u id

是

Project

表的外键

| Project_id | Project_name | Project_start_date | Product_id |

| Participant_id | Participant_name | Participant_email | Project_ID |

Response表：

Response\u id

是表的主键，

participant\u id

是

participant

表的外键

| Project_id | Project_name | Project_start_date | Product_id |

| Participant_id | Participant_name | Participant_email | Project_ID |

我的问题是，我是否可以将所有4个表连接在一起，列出产品名称、项目名称、参与者电子邮件，并获得每个项目的参与者数量以及每个项目的所有回复

编辑： 我觉得最好把代码放在这里。这是我尝试过的，所以请不要告诉我只使用加入

这是我从所有参与者那里得到的代码：

SELECT
    projects.Project_ID,
    count(*) total_Participants,
    projects.Project_Name,
    projects.Product_ID,
    products.Product_Name
FROM
    Project projects
INNER JOIN Participant participant on
    projects.Project_ID = participant.Project_ID
INNER JOIN Products products ON
    products.Product_ID = projects.Product_ID
group by
    projects.Project_ID,
    projects.Project_Name,
    projects.Product_ID,
    products.Product_Name,
    projects.Project_ID
ORDER BY
    projects.Project_ID DESC

这是我得到的全部回复代码：

SELECT
    projects.Project_ID,
    count(*) total_Participants,
    projects.Project_Name,
    projects.Product_ID,
    products.Product_Name
FROM
    Project projects
INNER JOIN Participant participant on
    projects.Project_ID = participant.Project_ID
INNER JOIN Products products ON
    products.Product_ID = projects.Product_ID
INNER JOIN Response Response ON
    Response.Participant_ID = participant.Participant_ID
group by
    projects.Project_ID,
    projects.Project_Name,
    projects.Product_ID,
    products.Product_Name,
    projects.Project_ID
ORDER BY
    projects.Project_ID DESC

---

这是一种我可以使用一个查询语句来获取参与者的电子邮件、总参与者和总响应的方法吗？

如果没有您预期结果的示例，这是我能想到的最好的方法

一些样本数据：

declare @product table
(
    product_id int,
    product_name nvarchar(20)
);
insert into @product (product_id, product_name) values
(1, 'Product ABC'),
(2, 'Product DEF'),
(3, 'Product GHI');


declare @project table
(
    project_id int,
    project_name nvarchar(20),
    product_id int
);
insert into @project (project_id, project_name, product_id) values
(1, 'Project 001', 1),
(2, 'Project 002', 1),
(3, 'Project 003', 2),
(4, 'Project 004', 3);


declare @participant table
(
    participant_id int,
    participant_name nvarchar(20),
    participant_email nvarchar(20),
    project_id int
);
insert into @participant (participant_id, participant_name, participant_email, project_id) values
(1, 'Andy', 'andy@host.org', 1),
(2, 'Beatrice', 'beatrice@host.org', 1),
(3, 'Charles', 'charles@host.org', 2),
(4, 'Charles', 'charles@host.org', 4),
(5, 'David', 'david@host.org', 3),
(6, 'Eve', 'eve@host.org', 2);


declare @response table
(
    response_id int,
    response_content nvarchar(100),
    participant_id int
);
insert into @response (response_id, response_content, participant_id) values
(1, 'please', 1),
(2, 'provide', 2),
(3, 'sample', 2),
(4, 'data', 3),
(5, 'next', 5),
(6, 'time', 4),
(7, 'thank', 6),
(8, 'you', 6),
(9, '>.<', 6);

这给了我：

product_name         project_name         participant_email    participant_count response_count
-------------------- -------------------- -------------------- ----------------- --------------
Product ABC          Project 001          andy@host.org        2                 3
Product ABC          Project 001          beatrice@host.org    2                 3
Product ABC          Project 002          charles@host.org     2                 4
Product ABC          Project 002          eve@host.org         2                 4
Product DEF          Project 003          david@host.org       1                 1
Product GHI          Project 004          charles@host.org     1                 1

---

是的，这是可能的，尝试一下，当你有一个尝试时，你被卡住了。并包括样本数据（作为DDL/DML）和预期结果以及您的尝试。是的。它被称为一个

JOIN

，你可以用它来

JOIN

表格。@Moon，你为什么想要参与者的电子邮件？当您试图获取项目级信息时。我建议您提供一份样本数据和预期结果。那会有帮助的