How to select all posts from database? - php

Question

In database in table user I have three columns:
id
name
friends
In column friends are names of people who are friends with person whose name is stored in column name. Column friends looks like this:
friendname1,friendname2,friendname3,friendname4
Each of those friends have it's own row where name is equal to their name.
I also have another table called post where I have four columns:
id
name_posted
post
visible
What I would like now, is to select all posts from table post where name_posted is equal to name of the logged in user or any of his friends which are stored in column friends in table user.
Name of the logged in user is stored in variable $user.
For selecting only posts from logged in user I can use this:
$all_posts = mysqli_query($connect_db, "SELECT * FROM post WHERE name_posted='$user' AND visible='yes'");
but I don't know how to include to select also posts from his friends. Something like Facebook have, when you log in and you see your posts plus your friends posts. I don't know how they created that. Sorry for long post, I just wanted to give you detailed description.

Answer 1

For selecting data based on information across multiple tables I suggest reading up on MySQL Joins.

Answer 2

Maybe with two querys, first select friends something like this:
SELECT * FROM user WHERE name='$user'
You then have all his friends in string like this if I understood correctly :
friend1,friend2,friend3...
Explode $row['friends'] -> explode(',',$row['friends']); to get all friends names in array and then you can do another select in foreach loop to get all posts from friends and display it the way you like or you can even better do IN in query
select * from posts where name_posted IN ($row['friends'])
this is the other way, which would be longer
foreach($friendarray as $k=>$friend){
...
mysqli_query($connect_db,
"SELECT * from post where name_posted='$friend' AND visible='yes'");
...
}
and also the query you already have to get own posts. Don't forget to escape all values and stuff...
You could also join two tables but I cant write that query from my mind , would have to sit down and try it with real data but that would be ultimate solution for you.
Don't know if I hit it right but shout if you need help

Answer 3

You can do it in a single query with something like:
SELECT p.*
FROM user u
join post p
on (u.name = p.name or concat(',',u.friends,',') like concat('%,',p.name,',%')
AND p.visible='yes'
WHERE u.name='$user'
- but the performance is likely to be much poorer than if you had a properly normalised design for the relationship between users and their friends.

Answer 4

You should probably reconsider the design of your DB. From what you've described, you should have three tables: userinfo, friendship, posts.
That way, you can then do a quick union between all three tables to get what you're looking for in one query.

Answer 5

Let me talk about how I will solve that case if its required from me.
I will use the following tables
users - user_id, name, email and whatever I need
relations - relation_id, user_id, fiend_id -- this table will relate one user to other
posts - post_id, user_id, content, visible
Now basically we have everything needed.
For selecting all data from the currently logged user and all of his friend I will use the following query:
SELECT * FROM posts
WHERE Visible = 1
AND (
user_id IN (SELECT friend_id FROM relations WHERE user_id = 1)
OR
user_id = 1)
What I do here, I use nested queries to accomplish that.
The interesting part here is the nested query - which basically return "array" with the ids of my friends. MySQL IN function check the user_id against that "array" After that in the main parentheses I add OR user_id = 1 which is the my user_id.
In that way I should have the content which I want to use to my feed.
However this code I away from fast and optimized but it's good example how to do that.