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sql - 如何在 SQL 中查询表的层次结构

问题描述

我有一张桌子，meeting. 在这些meetings 中存在一个层次结构。其中一些是年度会议，大多数只是例会。

所有正则meetings都将与至少一个连接表相关联，meeting_yearly_meeting. Ameeting_yearly_meeting有两列：meeting_id和yearly_meeting_id。

下面是这两个表的样子：

会议：

id SERIAL PRIMARY KEY,
title VARCHAR(255),
mappable BOOLEAN,
phone VARCHAR(255),
email VARCHAR(255),
city VARCHAR(255),
address VARCHAR(255),
zip VARCHAR(255),
latitude NUMERIC,
longitude NUMERIC,
description VARCHAR(255),
worship_time TIME,
state VARCHAR(255),
website VARCHAR(255),
lgbt_affirming BOOLEAN,
created TIMESTAMP default current_timestamp,
updated TIMESTAMP default current_timestamp

会议年度会议：

id SERIAL PRIMARY KEY,
meeting_id SMALLINT,
yearly_meeting_id SMALLINT,
created TIMESTAMP default current_timestamp,
updated TIMESTAMP default current_timestamp

因此，从我的/meetings端点，我想返回所有会议的集合——包括定期会议和年度会议。我想返回会议及其所有列，以及一个附加列：yearly_meeting.

对于meeting具有一个或多个关联meeting_yearly_meeting记录的记录，yearly_meeting将是一个以逗号分隔的记录列表，title该meeting记录被指定为meeting年度会议。对于那些没有任何相关meeting_yearly_meeting记录的会议（因此它们本身就是年度会议），我想要yearly_meeting field to beNULL`。

在追求这个目标的路上，我尝试了这样的事情：

SELECT t1.*, t2.meeting_yearly_meeting AS yearly_meeting
FROM (
  SELECT * FROM meeting
  FULL JOIN meeting_yearly_meeting ON meeting.id = meeting_yearly_meeting.yearly_meeting_id;
) as t1, 
(
  SELECT CASE WHEN (meeting_yearly_meeting.id IS NOT NULL)
    THEN (SELECT title FROM meeting WHERE meeting.id = meeting_yearly_meeting.yearly_meeting_id)
    ELSE NULL
    END
  FROM (
    SELECT meeting_yearly_meeting.* FROM meeting
    FULL JOIN meeting_yearly_meeting ON meeting.id = meeting_yearly_meeting.meeting_id
  ) as meeting_yearly_meeting;
) as t2;

但这会引发语法错误。

我很欣赏其他人可能有的任何见解。如果您需要任何其他上下文或说明，请告诉我！

更新：

样本meeting数据：https ://gist.github.com/micahbales/4013399c3fd23a0caf108124dab827c8

样本meeting_yearly_meeting数据：https ://gist.github.com/micahbales/fcbdeef282bd7bf1014606cee43bfb5e

预期返回值示例：https ://gist.github.com/micahbales/13d2aafdc5d43c4b948dc39c2df51569

标签： sqlpostgresql

解决方案

您可以尝试离开加入年度会议，然后使用string_agg()以获取逗号分隔列表。

SELECT m1.id,
       m1.title,
       m1.mappable,
       m1.phone,
       m1.email,
       m1.city,
       m1.address,
       m1.zip,
       m1.latitude,
       m1.longitude,
       m1.description,
       m1.worship_time,
       m1.state,
       m1.website,
       m1.lgbt_affirming,
       m1.created,
       m1.updated,
       string_agg(m2.title, ', ') yearly_meeting
       FROM meeting m1
            LEFT JOIN meeting_yearly_meeting mym1
                      ON mym1.meeting_id = m1.id
            LEFT JOIN meeting m2
                      ON m2.id = mym1.yearly_meeting_id
       GROUP BY m1.id,
                m1.title,
                m1.mappable,
                m1.phone,
                m1.email,
                m1.city,
                m1.address,
                m1.zip,
                m1.latitude,
                m1.longitude,
                m1.description,
                m1.worship_time,
                m1.state,
                m1.website,
                m1.lgbt_affirming,
                m1.created,
                m1.updated;

编辑：

更“紧凑”的解决方案可能是使用相关子查询。

SELECT m1.*,
       (SELECT string_agg(m2.title, ', ')
               FROM meeting_yearly_meeting mym1
                    LEFT JOIN meeting m2
                              ON m2.id = mym1.yearly_meeting_id
               WHERE mym1.meeting_id = m1.id) yearly_meeting
       FROM meeting m1;

但请注意，虽然它的代码更少，但不一定更快。

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