用于查找好友签入的SQL查询

我在PostgreSQL中有以下表格：

Categories | Locations   | Checkins     | Users  | Friendships
id         | id          | id           | id     | user_id
name       | category_id | location_id  | gender | friend_id
icon       | name        | user_id      |        |

现在，我想检索以下关于场地的信息

• 一个位置有多少女性和男性用户
• 类别名称和图标
• 地点名称
• 在一个位置（从给定的用户id）签入的朋友数

除了最后一点，我解决了。但我很难从给定的用户id计算朋友数。我尝试了以下查询：

SELECT distinct locations.id,
 max(locations.name) as name,
 max(locations.location) as location,
 max(categories.name) as cat,
 max(categories.icon) as caticon,
 SUM(CASE WHEN users.gender = 'm' THEN 1 ELSE 0 END) AS male,
 SUM(CASE WHEN users.gender = 'f' THEN 1 ELSE 0 END) AS female,
 SUM(CASE WHEN friendships.user_id = 1 OR friendships.friend_id=1 THEN 1 ELSE 0 END) AS friends
 FROM locations
 INNER JOIN checkins ON checkins.location_id = locations.id 
 INNER JOIN users ON users.id = checkins.user_id 
 INNER JOIN categories ON categories.id = locations.category_id
 LEFT JOIN friendships ON friendships.user_id = users.id OR friendships.friend_id = users.id
 WHERE locations.id=7
 GROUP BY locations.id

但我对女性用户的计数是错误的。知道我做错了什么吗？我想我需要一个用于friendships表的左连接，因为如果用户没有朋友（或者没有用户），那么它应该只返回0作为朋友计数

希望我说清楚，

---

thx，tux

在第一行中删除

distinct

。对于同一字段，您已经有了一个

group by

子句。如果有帮助，请告诉我。

从各个位置，我相信应该是这样。另外，当friendships.user\u id=1或friendships.user\u id=1时，
可能应该是
当friendships.user\u id=1或friendships.friends\u id=1时
。我有一个问题：如果两个用户是彼此的朋友，那么在
友谊中会有多少记录，1还是2？两张唱片应该产生一个或两个友谊吗？谢谢你的回答（我修正了，现在是凌晨；））。当两个用户是彼此的朋友时，友谊表中正好有一条记录，请求友谊的用户存储在
user\u id
中，另一条存储在
friend\u id
SELECT
  L.id,
  L.name,
  c.name AS cat,
  c.icon AS caticon,
  COUNT(CASE u.gender WHEN 'm' THEN 1 END) AS male,
  COUNT(CASE u.gender WHEN 'f' THEN 1 END) AS female,
  COUNT(f.user_id) AS friends
FROM Locations L
  INNER JOIN Categories c ON c.id = L.category_id
  INNER JOIN Checkins ch ON ch.location_id = L.id
  INNER JOIN Users u ON u.id = ch.user_id
  LEFT JOIN Friendships f ON f.user_id = @user_id AND f.friend_id = ch.user_id
                          OR f.user_id = ch.user_id AND f.friend_id = @user_id
WHERE L.id = @location_id
GROUP BY L.id, L.name, c.name, c.icon