Multiple SQL fetch and print html-code if two tables is true - php

Question

I'm trying to achieve a PHP, SQL and HTML solution. This is probably commonly done.
I have coded a project overview system from scratch. The state right now is that I have a SQL table for USERS and another table for PROJECTS. It works great to sign in with each user and also to list all projects for all the users.
But to be able to track which users that are connected to each project I created a third SQL table which is called PROJECTMEMBERS. It contains the rows USERS_ID and PROJECTS_ID which are supposed to hold the ID from the user and the project to be able to track the projectmembers. I'm not sure if this is the best solution but found it in another thread and thought it made sense. But now..
I have issues to figure out how to access different SQL tables in the same mysqli_fetch() query. How is this possibly done to present the USER IMAGE in each project? The current fetch is used to present all the projects but I still want to show all projects, but with an image of the active users on the project on each.
I hope I was clear enough for anyone to understand. Looking forward to any help out there.
SQL STRUCTURE EXAMPLE:
tbl_USERS tbl_PROJECTS
| id | username | USER_IMG | etc | | id | PROJECTNAME | DATE | etc |
---------------------------------- ----------------------------------
| 1 | XXXX | XXXXXXXX | 1 | | 1 | AAAA | AAAA | 1 |
| 2 | YYYY | YYYYYYYY | 2 | | 2 | BBBB | BBBB | 3 |
| 3 | ZZZZ | ZZZZZZZZ | 1 | | 3 | CCCC | CCCC | 4 |
tbl_PROJECTMEMBERS
| USERS_ID | PROJECTS_ID |
-------------------------
| 1 | 1 |
| 1 | 5 |
| 2 | 1 |
How it should be presented where USER_IMG is fetched from the users table:
-------------------------------- --------------------------------
| PROJECTNAME | | PROJECTNAME |
| DATE | | DATE |
| | | |
| | | |
| USER_IMG, USER_IMG | | USER_IMG, USER_IMG |
-------------------------------- --------------------------------
--------------------------------
| PROJECTNAME |
| DATE |
| |
| |
| USER_IMG |
--------------------------------

Answer 1

As you have 3 table users and projects and user_projects and you are fetching only projects from projects table then you can add join the first query where you fetch the projects it could some like this
SELECT p.,user. FROM projects as p
INNER JOIN user_projects as up
ON up.project_id = p.id
INNER JOIN users
ON users.id = up.user_id
Above I have used multijoin where project table connected with user_project and user_project conned with users table and in the select query we have fetch project and users table.
In above query you might get query if same attribute get repate into the both table like (ambiguity) so you can just skip the error just by alise it

Answer 2

SELECT user.id AS 'User ID', project.id as 'Project ID', project.name as 'Project Name' from user tbl_USERS , project tbl_PROJECTS, userprojects tbl_PROJECTMEMBERS where user.id = userprojects.USERS_ID AND project.id = userprojects.PROJECT_ID
This would be your mysql query, it would fetch the user id as User ID, project Id as Project ID, and project name as Project Name.
You can access those variables in plain php and output anyway you like.

Answer 3

Something like this?
$query = "SELECT a.*, b.* FROM USERS a INNER JOIN PROJECTS b on a.userID = b.userID
RIGHT JOIN USER_PROJECTS c ON a.userID = c.userID
WHERE a.userID = $variable";
$result = $db->Execute($query)
foreach($result as $data){
if($data['userID'] == $data['projectID']){
//do stuff here
}
}