从内部连接中获取不同的行

Question

我有以下五个表，用于管理数据库中用户、组和角色之间的关系：

• 用户
• 用户组
• 群
• group_role
• 角色

一个用户属于一个组。此数据存储在 user_group 表中，该表只有两列：user_fk 和 group_fk。 一个组可以包含许多角色。此信息存储在表 group_role 中，该表只有两列：group_fk 和 role_fk。

所以我对这 5 个表编写了以下查询，连接结果为我自己提供了例如 5 行，因为我所属的组包含 5 个角色。

SELECT
user.last_name,
user.first_name,
user.user_name,
group.name,
role.name 
FROM
user 
INNER JOIN user_group ON user_group.user_fk = user.id
INNER JOIN group ON group.id = user_group.group_fk
INNER JOIN group_role ON group.id = group_role.group_fk
INNER JOIN role ON group_role.role_fk = role.id

有没有办法将查询更改为例如只为一个用户返回一行并列出他的所有组和角色？

Answer 1

简单地聚合每个用户并使用LISTAGG 来获取他们的组和角色字符串。为此，您需要 Oracle 19c，因为 Oracle 在早期版本中忘记为 LISTAGG 实现 DISTINCT。

SELECT
  user.last_name,
  user.first_name,
  user.user_name,
  LISTAGG(group.name, ', ') WITHIN GROUP (ORDER BY group.name) AS groups,
  LISTAGG(DISTINCT role.name, ', ') WITHIN GROUP (ORDER BY role.name) AS roles
FROM user 
INNER JOIN user_group ON user_group.user_fk = user.id
INNER JOIN group ON group.id = user_group.group_fk
INNER JOIN group_role ON group.id = group_role.group_fk
INNER JOIN role ON group_role.role_fk = role.id
GROUP BY user.last_name, user.first_name, user.user_name
ORDER BY user.last_name, user.first_name, user.user_name;

在早期版本中，您可以使用REGEXP_REPLACE 删除由LISTAGG 生成的重复项，前提是列表的长度不超过LISTAGG 可以处理的长度。

REGEXP_REPLACE(
  LISTAGG(role.name, ', ') WITHIN GROUP (ORDER BY role.name),
  '([^, ]+)(, \1)+($|, )',
  '\1\3')

Answer 2

您可以将listagg 聚合函数与group by 一起使用，如下所示：

SELECT user.last_name,
       user.first_name, 
       user.user_name, 
       listagg(group.name,',') within group (order by group.name) as groupss,
       listagg(role.name,',') within group (order by role.name) as roless 
FROM user  INNER JOIN user_group ON user_group.user_fk = user.id 
INNER JOIN group ON group.id = user_group.group_fk 
INNER JOIN group_role ON group.id = group_role.group_fk 
INNER JOIN role ON group_role.role_fk = role.id 
Group by  user.last_name, 
          user.first_name, 
          user.user_name;

Answer 3

我会推荐两个子查询。为什么？角色可能会重复，listagg() 不支持distinct。

所以：

select u.last_name, u.first_name, u.user_name, 
       ug.groups, ur.roles           
from user u left join
     (select ug.user_fk,
             listagg(g.name, ',') within group (order by g.name) as groups
      from user_group ug join
           group g
           on g.id = ug.group_fk 
      group by ug.user_fk
     ) ug
     on ug.user_fk = u.id left join
     (select ug.user_fk,
             listagg(r.name, ',') within group (order by r.name) as roles 
      from user_group ug join
           group g join
           on g.id = ug.group_fk join
           group_role gr
           on g.id = gr.group_fk join
           role r
           on gr.role_fk = r.id 
      group by ug.user_fk
     ) ur
     on ur.user_fk = u.id;

您也可以使用一个子查询和条件聚合来编写：

select u.last_name, u.first_name, u.user_name, 
       listagg(case when seqnum_g = 1 then ugr.name end, ',') within group (order by ug.name) as groups,
       listagg(case when seqnum_r = 1 then ugr.name end, ',') as roles within group (order by ur.name) as roles 
from user u left join
     (select ug.user_fk,
             row_number() over (partition by ug.user_fk order by g.name) as seqnum_g,
             row_number() over (partition by ug.user_rk order by r.name) as seqnum_r
      from user_group ug join
           group g join
           on g.id = ug.group_fk join
           group_role gr
           on g.id = gr.group_fk join
           role r
           on gr.role_fk = r.id 
     ) ugr
     on ugr.user_fk = u.id
group by u.last_name, u.first_name, u.user_name; 

Answer 4

试试这样的：

SELECT
user.last_name,
user.first_name,
user.user_name,
listagg(group.name || ' - ' ||role.name, ', ') within group (order by group.name, role.name)
FROM user 
INNER JOIN user_group ON user_group.user_fk = user.id
INNER JOIN group ON group.id = user_group.group_fk
INNER JOIN group_role ON group.id = group_role.group_fk
INNER JOIN role ON group_role.role_fk = role.id
group by user.last_name,
user.first_name,
user.user_name

你需要两件事：

1. 对您的结果进行分组，它是按子句分组完成的
2. 连接每个组的角色，由 listagg 函数完成