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LEFT JOIN中的COUNT返回重复值

[英]COUNT in LEFT JOIN returning duplicated value

I've the following tables (example): 我有以下表格(示例):

users : 用户

id | user | photo   | joined | country
1  | Igor | abc.jpg | 2015   | Brazil
2  | John | cga.png | 2014   | USA
3  | Lucas| hes.jpg | 2016   | Japan

posts (see that there are two lines with author = Igor and ft = 2 and one line with author = Igor and ft = 3 and Igor have three posts): (看到有两行与author = Igorft = 2 ,并用一条线author = Igorft = 3和Igor具有三个柱):

id | author | content | date | ft (2 = photos and 3 = videos)
1  | Igor   | hi      | 2016 | 2
2  | Igor   | hello   | 2016 | 3
3  | John   | hehehe  | 2016 | 2
4  | Igor   | huhuhuh | 2016 | 2
5  | Lucas  | lol     | 2016 | 3

friendship (when status = 2 means that they are friends): 友谊status = 2表示他们是朋友时):

id | friend1 | friend2 | status
1  | Igor    | Lucas   | 2
2  | Lucas   | John    | 2
3  | John    | Igor    | 2

And I want to do a COUNT of posts with ft = 2 and a COUNT of friends ( status = 2 ) according to the currently logged user (Igor, in this case). 我想根据当前登录的用户(在这种情况下为Igor)执行ft = 2COUNT 个帖子COUNT个朋友( status = 2 )的帖子

So, I do (assuming that the current user logged in is Igor ): 因此,我做了(假设当前登录的用户是Igor ):

SELECT photo, joined, country, sum(CASE WHEN ft = 2 THEN 1 ELSE 0 END) AS numPhotos, sum(CASE WHEN ft = 3 THEN 1 ELSE 0 END) AS numVideos
FROM users
LEFT JOIN posts
ON users.user = posts.author
WHERE users.user = 'Igor'
GROUP BY users.user
LIMIT 1

And when I check on a foreach , the data is correct: numPhotos = 2 and numVideos = 1 . 当我检查foreach ,数据是正确的: numPhotos = 2numVideos = 1

But, I want to select too the number of friends, so, I do: 但是,我也想选择朋友数,因此,我这样做:

SELECT photo, joined, country, sum(CASE WHEN ft = 2 THEN 1 ELSE 0 END) AS numPhotos, sum(CASE WHEN ft = 3 THEN 1 ELSE 0 END) AS numVideos, count(friendship.status) AS numFriends
FROM users
LEFT JOIN posts
ON users.user = posts.author
LEFT JOIN friendship
ON (users.user = friend1 OR users.user = friend2) AND friendship.status = 2
WHERE users.user = 'Igor'
GROUP BY users.user
LIMIT 1

But, the output is: numPhotos = 4 , numVideos = 2 and numFriends = 6 . 但是,输出为: numPhotos = 4numVideos = 2numFriends = 6

In other words, he is duplicating all results but in numFriends he's taking the total of posts of Igor (3) and duplicating the value too. 换句话说,他正在复制所有结果,但在numFriends他正在获取Igor(3)的帖子总数,并且也在复制值。 And if I change count(friendship.status) to sum(friendship.status) the output is: numPhotos = 4 , numVideos = 2 and numFriends = 18 (triples the numFriends ). 如果我将count(friendship.status)更改为sum(friendship.status)则输出为: numPhotos = 4numVideos = 2numFriends = 18 (将numFriends三倍)。

I tried too with count(distinct friendship.status) and the result is: numPhotos = 4 , numVideos = 2 and numFriends = 1 (duplicates the values again as well as return the wrong value 1 for numFriends that should be 2 knowing he has two friends). 我也尝试过使用count(distinct friendship.status) numPhotos = 4 count(distinct friendship.status) ,结果是: numPhotos = 4numVideos = 2numFriends = 1 (再次复制值以及为numFriends返回错误的值1,因为他知道2有2朋友)。

So, how I can do this? 那么,我该怎么做呢? (I'm using MySQL) (我正在使用MySQL)

EDIT : 编辑

I changed the count(distinct friendship.status) to count(distinct friendship.id) and it worked to select the number of friends. 我改变了count(distinct friendship.status)count(distinct friendship.id)和它的工作选择的朋友的数量。 But the rest of values ( numPhotos and numVideos ) continue duplicated. 但是其余值( numPhotosnumVideos )将继续重复。

I discovered that the problem is in ON (users.user = friend1 OR users.user = friend2) , because if I leave only ON (users.user = friend1) or ON (users.user = friend2) the output isn't duplicated. 我发现问题出在ON (users.user = friend1 OR users.user = friend2) ,因为如果我仅保留ON (users.user = friend1)ON (users.user = friend2)则输出不会重复。 I tried too with ON 'Igor' IN (friend1, friend2) but the result is the same ( numPhotos and numVideos` continue duplicated). 我也尝试使用ON 'Igor' IN (friend1, friend2) but the result is the same ( numPhotos and numVideos`继续重复)。

I think the left join may be joining on a one-to-many relationship, which is causing inflated counts. 我认为左联接可能正在一对多关系上联接,这会导致计数虚高。 Since you are only retrieving the counts for 1 user, I suggest using a subquery to retrieve the friendship counts (for retrieving the counts for multiple users, a derived table may be faster than a subquery): 由于您只检索1个用户的计数,因此我建议使用子查询来检索友谊计数(要检索多个用户的计数,派生表可能比子查询更快):

SELECT 
    sum(ft = 2) AS numPhotos, 
    sum(ft = 3) AS numVideos,
    (select count(*) from friendships f
    where (friend1 = users.user 
    or friend2 = users.user)
    and status = 2) as friendship_count
FROM users
LEFT JOIN posts
ON users.user = posts.author
WHERE users.user = 'Igor'

Note that I removed the group by because users.user is already in the where clause, which means there is only 1 group. 请注意,我删除了group by原因是因为users.user已经在where子句中,这意味着只有1个组。

Instead of count(distinct friendship.status) , try using count(distinct friendship.id) . 代替count(distinct friendship.status) ,请尝试使用count(distinct friendship.id) That should give you the number of unique friends. 那应该给您带来许多独特的朋友。 Counting distinct statuses doesn't work because all the statuses will be 2 by definition, so there is only one distinct value. 计数不同的状态是行不通的,因为根据定义,所有状态将为2,因此只有一个不同的值。

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