Question

我有3张桌子。 user_sticker保存用户之间发送的所有贴纸。加载配置文件视图屏幕时，我需要显示哪个是给用户最常见的标签。

用户

id_user
名称

贴纸

id_sticker
名称

user_sticker

id_user_sticker
id_sticker
id_user_from
id_user_to

因此，如果user_sticker拥有此信息：

1,3,254,205
2,2,362,205
3,2,519,205
4,3,945,205
5,3,199,205

（这意味着用户254,362,519,945,199向用户205发送了贴纸）。结果必须在同一记录中返回两者用户205信息（名称）和最常见的标签ID（在本例中为＃3）。

Answer 1

这将是选择：

   select id_user_to, count(id_sticker) quant from user_sticker group by id_sticker order by quant desc limit 1;

现在有了用户的信息：JOIN

select us.id_user_to, u.name, count(us.id_sticker) quant 
from user_sticker us
join user u on u.id_user=us.id_user_to
group by us.id_sticker 
order by quant desc limit 1;

Answer 2

好的，让我们看看

原则上，您需要使用user_sticker表。所以：

SELECT *
FROM user_sticker us

然后，匹配user_to的用户表

SELECT *
FROM user_sticker us JOIN user u ON us.id_user_to = u.id_user

然后，您需要要收到的用户的名称和贴纸

SELECT us.id_sticker, u.name
FROM user_sticker us JOIN user u ON us.id_user_to = u.id_user
WHERE u.id_user = "yourwanteduser"

现在，选择第一个

SELECT TOP 1 us.id_sticker, u.name
FROM user_sticker us JOIN user u ON us.id_user_to = u.id_user
WHERE u.id_user = "yourwanteduser"
ORDER BY (SELECT COUNT(*)
          FROM user_sticker us2
          WHERE us2.id_user_to=u.id_user AND us2.id_sticker=us.id_sticker)

ORDER BY用于搜索向该用户发送了多少贴纸

抱歉我的英语不好！我希望这有帮助！

Answer 3

select test.id_sticker, test.id_user_to, name from 
(select s.id_sticker, s.id_user_to
from user_sticker s, user u
group by id_sticker, id_user_to
order by count(*) desc
limit 1) 
as test, user
where test.id_user_to=user.id_user

请参阅sql fiddle链接：http://sqlfiddle.com/#!2/e02ca/6/0

返回用户的最常见链接ID

3 个答案: