我对这个特定的查询一直在争论。我正在制作一个排行榜'已向其指南添加或收到最多位置的用户。
我的数据模型如下:
Users
uuid id
string first_name
bool blogger
Guests
uuid id
Tips (join model)
uuid id
uuid guide_id
uuid place_id
integer status
uuid owner_id
string owner_type
Guides
uuid id
uuid user_id
string name
Place
uuid id
string name
用户可以向他/她的指南添加地点(通过提示)。该用户还可以从其他用户接收他/她的指南的地点(提示)。可以接受这些提示tips.status = 1
。
我想要的是以下内容:
所有地点的所有地点的first_name和数量,以及所有指南的提示,但没有提供给users.blogger = 1
分组的其他用户的提示。
示例:
Guest = true
you 40
user1 30
user2 25
Guest = false
user3 20
user4 15
user5 5
这是我到目前为止所做的:
SELECT tips.owner_id, tips.owner_type, count(tips.owner_id) AS places_count
FROM "tips"
LEFT JOIN users on (owner_type ='User' AND users.id = owner_id)
GROUP BY "tips"."owner_id", "tips"."owner_type"
ORDER BY places_count DESC
LIMIT 16
此查询会返回计数,但不会收到有关的提示,并且还会向其他用户计算给定的提示。我有一种预感,我需要使用子查询,首先从给定用户中选择所有指南ID,然后再简单地选择'选择guide_id = selected_guide_ids
和tips.status = 1
所有提示的计数。最后按users.blogger = 1
但我怎么写这个?
我已使用其他Guest
表格更新了原始问题(这就是我使用owner_type and owner_id instead
table_id
的原因。我已经使用博客更新了用户表(bool )我想对结果进行分组。
示例数据:
Users
id first_name blogger
user1 Daniel true
user2 Quassnoi false
user3 vkp true
Guests
id
guest_1
guest_2
Guides
id user_id name
guide_1 user_1 Bugers
guide_2 user_1 Cool places
guide_3 user_2 Amsterdam
Tips
id guide_id place_id status owner_id owner_type
tip1 guide_1 place_1 1 user_1 User # user_1 added place_1 to his own guide guide_1 (accepted)
tip2 guide_1 place_2 1 guest_1 Guest # guest_1 suggested place_2 to user_1's guide guide_1 (accepted)
tip2 guide_1 place_2 0 guest_1 Guest # guest_1 suggested place_2 to user_1's guide guide_1 (rejected)
tip_3 guide_2 place_1 1 user_2 User # user_2 added place_1 to his own guide guide_3 (accepted)
tip_4 guide_2 place_2 1 user_2 User # user_2 added place_2 to his own guide guide_3 (accepted)
tip_5 guide_2 place_3 1 user_1 User # user_1 added place_3 to user_2's guide guide_2 (accepted)
Places
id name
place1 burgerbar
place2 burgermeester
place_3 bbq shack
我期望的结果是什么:
请注意,提供给其他用户的提示不适用于小费提供者。
first_name tips_count blogger
Quassnoi 3 false (2 added by himself, 1 received from user_1)
Daniel 2 true (1 added by himself, 1 received from guest1. Note that the rejected tip does not count)
vkp 0 false
我已经改变了Quassnoi对此的回答:
SELECT *
FROM users u
LEFT JOIN
(
SELECT g.user_id, COUNT(*) tips_count
FROM guides g
JOIN tips t
ON t.guide_id = g.id
AND (t.owner_id = g.user_id AND t.status = 1)
GROUP BY g.user_id
) g
ON g.user_id = u.id
ORDER BY tips_count DESC
然而,这将首先返回tips_count为NULL的所有记录。我希望那些是0而不是NULL。如何将NULL tips_count转换为0?
我已更新了查询,因此它只计算了guide_id等于给定用户的指南ID的提示。
SELECT *
FROM users u
LEFT JOIN
(
SELECT g.user_id, COUNT(*) tips_count
FROM guides g
JOIN tips t
ON t.guide_id = g.id
AND (t.guide_id = g.id AND t.status = 1)
GROUP BY g.user_id
) g
ON g.user_id = u.id
ORDER BY tips_count DESC
答案 0 :(得分:1)
现在设置架构的方式,无法确定哪个指南属于哪个用户。
假设有guide.owner
您忘了提及(或忘记添加),那就是:
SELECT *
FROM user u
LEFT JOIN
(
SELECT g.owner, COUNT(*) cnt
guide g
JOIN tips t
ON t.guide_id = g.id
AND (t.owner_id = g.owner OR t.status = 1)
GROUP BY
g.owner
) g
ON g.owner = u.id
答案 1 :(得分:1)
似乎添加/建议一个地方并不重要。一旦被接受(状态1),它就属于指南,因此属于指南的用户。因此:
select u.first_name, u.blogger, count(t.id)
from users u
left join guides g on g.user_id = u.id
left join tips t on t.guide_id = g.id and t.status = 1
group by u.id
order by count(t.id) desc;