PHP - 比较来自两个MySQL表的Connections / Feed

Question

I am working on a small social networking project that will allow users to connect with each other, similar to the "friends" idea of Facebook. The site has it's own unique features that set it apart from others, so please hold the comments about the social networking market being saturated.

Each time a user connects with a friend, it is added to the 'Connections' table in the MySQL DB. Each time a user sends a gift, invite or anything else, it is added to the 'Feed' table.

I need to be able to display only the activity of friends from the 'Feed' table, based on the connections in the 'Connections' table.

My first thought was to collect all of the connection for the currently logged in user from the 'Connections' table and place the IDs in an array. Then loop through that array, checking each ID against the 'Feed' table for activity. This, however, just doesn't seem like a reasonable way to do this. It will also group all of the activity of each user together, instead of in the order it happened.

Any ideas?

Answer 1

Something along the lines of a straight forward join:

SELECT feed.*
FROM connections
JOIN feed
  ON connections.friend_id = feed.user_id
WHERE connections.user_id = <insert id of user>

Answer 2

You can use IN clause in mysql, try this..

select * 
from feeds f 
where user_id IN (
    select friends_id 
    from connections 
    where user_id = $user_id
)

OR

select * 
from feeds f join connections c 
on c.friends_id = f.user_id 
where c.user_id = $user_id

Joins are faster than sub-queries, also don't forget to add indexes on related columns(friends_id, user_id).