繁体   English   中英

SQL查询将具有相同索引的多个项目分组到同一行

[英]SQL Query to group multiple items with the same index into the same row

因此,我有三个表,其中包含有关用户及其部门的数据,我需要创建一个查询,以组织用户之间的数据。 第一个表包含用户的名称和ID,如下所示:

user_name   user_id
-----------------------
bob         1
chuck       2
bill        3

下表对用户及其部门ID进行了索引,如下所示:

user_id     department_id
-----------------------------
1           x
1           y
2           x
3           z
2           z
2           y

最后一张表格根据其ID列出了部门名称:

department_id   department_name
-------------------------------
x               research
y               development
z               advertising

如何编写一个查询,列出所有用户以及他们所在的每个部门? 它看起来应该像这样:

user_name   user_id     department_1    department_2    department_3
----------------------------------------------------------------------------
bob         1           research        development
chuck       2           research        development     advertising
bill        3           advertising

我尝试了许多不同的方法,但找不到任何可以提供此结果的方法。 我的输出当前列出重复的用户(如果他们有多个部门),例如:

user_name   user_id     department_1    department_2    department_3
----------------------------------------------------------------------------
bob         1           research        research        research
bob         1           development     development     development
chuck       2           research        research        research
chuck       2           development     development     development
chuck       3           advertising     advertising     advertising
bill        3           advertising     advertising     advertising

当前代码:

WITH DATA AS
(
SELECT 
Users.user_name AS Name,
Users.user_id AS Id,
Departments.department_name AS Department1,
Departments.department_name AS Department2,
Departments.department_name AS Department3
FROM Users
    JOIN Department_Index ON Users.user_id = Department_Index.user_id
    JOIN Departments ON Departments.department_id = Department_Index.department_id
WHERE Departments.department_id = Department_Index.department_id
)

SELECT
Name,Id,Department1,Department2,Department3
FROM DATA

我也尝试过这样的事情

WITH DATA AS
(
SELECT 
Users.user_name AS Name,
Users.user_id AS Id,
( SELECT Departments.department_name
  FROM Departments
     JOIN Department_Index ON Department_Index.user_id = Users.user_id
  WHERE Departments.department_name = Department_Index.department_name
) AS Departments
FROM Users
)

SELECT
Name,Id,Departments[0],Departments[1],Departments[2]
FROM DATA

但这也不起作用。

任何帮助表示赞赏!

在您告诉我们表名之前,我已经写了这个。

关键是使用row_number()窗口函数对部门名称进行排序,然后再加入此CTE 3次。

with ud_numbered as
(
   select u.user_id, ud.department_id, d.department_name
       row_number() over (partition by u.user_id order by ud.department_id asc) as dept_no
   from user_table u
   left join user_departement ud on u.user_id = ud1.user_id
   left join department d on ud.department_id = d.department_id
)
select u.user_id, u.user_name, n1.department_name as department_1,
                               n2.department_name as department_2, 
                               n3.department_name as department_3
from user_table u
left join ud_numbered n1 on u.user_id = n1.user_id and n1.dept_no = 1
left join ud_numbered n2 on u.user_id = n2.user_id and n2.dept_no = 2
left join ud_numbered n3 on u.user_id = n3.user_id and n3.dept_no = 3

我个人将使用ROW_NUMBER()PIVOT来解决这个问题。 ROW_NUMBER()将为每个部门提供一个人的唯一标识符,从1开始,然后您可以使用该标识符进行修改。 所使用的聚合函数,在这种情况下, MAX是因为任意的,如规定的标识符是唯一的,所以你正在服用的MAX单值:

WITH UserData AS
(   SELECT  u.[user_id], 
            u.[user_name],
            d.department_name,
            RowNumber = ROW_NUMBER() OVER(PARTITION BY u.[user_id] ORDER BY d.department_id)
    FROM    Users AS u
            INNER JOIN Department_Index AS di
                ON di.user_id = u.user_id
            INNER JOIN Departments AS d
                ON d.department_id = di.department_id
)
SELECT  pvt.[user_id],
        pvt.[user_name],
        Department1 = pvt.[1],
        Department2 = pvt.[2],
        Department3 = pvt.[3]
FROM    Data AS d
        PIVOT 
        (   MAX(department_name)
            FOR RowNumber IN ([1], [2], [3])
        ) AS pvt;

暂无
暂无

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM