找到最受欢迎的人的追随者。一个人的追随者越多
他们很受欢迎。
我需要SQL查询来选择最受欢迎的追随者。
我的桌子-(追随者)
id | person_id | follower_person_id
1 1 2
2 1 3
3 2 1
4 2 4
5 3 1
6 3 2
7 3 4
8 4 3
个人id 1共有2个跟随者(个人id 2,个人id 3),个人id
2有2个追随者(1号人物,4号人物),3号人物
共有3名追随者(个人id 1、个人id 2、个人id 4)和个人id
4人共有1名追随者(3人)。
因此,person_id 3是person_id 1最受欢迎的追随者,
person_id 1是person_id 2、person_id 1(或
person_id 2)是person_id 3和person_id 3最受欢迎的追随者
最受4号人物欢迎。
这是查询。。。
SELECT t1.person_id, t1.follower_person_id, t2.cnt
FROM followers AS t1
JOIN (
SELECT person_id, COUNT(*) AS cnt
FROM followers
GROUP BY person_id
) AS t2 ON t1.follower_person_id = t2.person_id
WHERE t1.person_id = 1
ORDER BY t2.cnt DESC LIMIT 1
上面的查询输出是
person_id, follower_person_id, cnt
-----------------------------------
1, 3, 3
Here is explanation of above query
此查询仅适用于查找特定人员的常用人员,但
我想找一对他们最“受欢迎”的追随者为所有人。
所以输出应该是
person_id, follower_person_id, cnt
-----------------------------------
1, 3, 3
2, 1, 2
3, 1, 2
4, 3, 3
现在我又有了一个
person table
id | name
1 John
2 Ali
3 Rohn
4 Veronica
现在我想把这个id转换成人名。
最终输出应该是
person_name, follower_person_name, cnt
--------------------------------------
John, Rohn, 3
Ali, John, 2
Rohn, John, 2
Veronica, Rohn, 3
我需要sql查询来获取这些数据。
最佳答案
您可以使用以下查询:
SELECT person_name, follower_name, cnt
FROM (
SELECT person_name, follower_name, cnt,
@rn := IF(@pname = person_name, @rn + 1,
IF(@pname := person_name, 1, 1)) AS rn
FROM (
SELECT t3.name AS person_name, t4.name AS follower_name, t2.cnt
FROM followers AS t1
JOIN (
SELECT person_id, COUNT(*) AS cnt
FROM followers
GROUP BY person_id
) AS t2 ON t1.follower_person_id = t2.person_id
JOIN person AS t3 ON t1.person_id = t3.id
JOIN person AS t4 ON t1.follower_person_id = t4.id
) AS x
CROSS JOIN (SELECT @rn := 0, @pname := '') AS vars
ORDER BY person_name, cnt DESC) AS v
wHERE v.rn = 1;
输出:
person_name follower_name cnt
--------------------------------
John Rohn 3
Veronica Rohn 3
Ali John 2
Rohn Ali 2
查询使用变量以获得最大的每个组记录。
Demo here
关于mysql - 返回所有用户的列表,以及其“最受欢迎”的关注者。某人的追随者越多,他们越“受欢迎”,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/43285818/