这是两个表:

1.user
user_id  |  full_name   | username
1              A             A_1
2              B             B_2
3              C             C_3
4              D             D_4
5              E             E_5

2.user_follower
user_id  |  follower_id  |  follow_dtm
2              4            2018-10-09 10:10:10
2              3            2018-01-09 11:10:10
1              5            2018-11-09 07:10:10
4              2            2018-10-09 06:10:10
4              5            2018-10-09 00:10:10


查找用户的关注者:2
输出应为:

user_id: 4  fullname: D username: D_4  f_total_flwr: 2 (num of flwr of id-4)  following: yes
user_id: 3  fullname: C username: C_3  f_total_flwr: 0 (num of flwr of id-3)  following: no


我需要一个mysql查询来找到特定用户的所有关注者,并具有user表中关注者的详细信息,并且需要知道每个关注者具有的关注者数量,并且我还需要特定用户是否也关注该关注者。这是我尝试过的:

SELECT u.user_id
     , u.full_name
     , u.username
     , COUNT(DISTINCT uf.follower_id) f_total_flwr
     , case when b.user_id is null then 'no' else 'yes' end following
  FROM user_follower
  JOIN user u
    ON user_follower.follower_id = u.user_id
  LEFT
  JOIN user_follower uf
    ON u.user_id = uf.user_id
  LEFT
  JOIN user_follower b
    ON b.user_id = user_follower.follower_id
   and b.user_id = u.user_id
 WHERE user_follower.user_id=2
 GROUP
    BY u.user_id
 ORDER
    BY uf.follow_dtm DESC
 LIMIT 30


我知道我要靠近了;)。问题是,即使用户没有跟进,我仍在使用following值获取yes这是另一个奇怪的东西-not allsome of them显示yes应该是no。谢谢!

最佳答案

尝试这个:

select u2.*,b.fcount, case when uf3.user_id is null then 'no' else 'yes' end as connected from user u2 inner join
(
select u.user_id,count(distinct(uf2.follower_id)) fcount
from user u
inner join user_follower uf1 on u.user_id=uf1.follower_id and uf1.user_id=1
left join user_follower uf2 on uf2.user_id=uf1.follower_id
group by u.user_id
) b on u2.user_id=b.user_id
left join user_follower uf3 on u2.user_id=uf3.user_id and uf3.follower_id=1


我使用以下数据集进行了尝试:

用户

1,a,abcd

2,b,bcde

3,c,cdef

user_follower

1,2,10

1,3,11

2,3,10

3,1,13

并得到了预期的结果:

2,b,bcde,1,no

3,c,cdef,1,是

关于mysql - MySQL查询查找关注者并关注,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/53609175/

10-15 21:06