MySQL Query Multiple ID's - php

Question

I have built a membership application that allows users to assemble projects whose contents are contained across 2 tables ('projects' and 'notes'). Each member can create, update or delete as many projects as they want.
Good so far...
I'd like the members to be able to share their projects with other members they choose. At this point I have built a function that allows Member A to type in an email address in order to share a project (with say, Member B). If that email exists in the DB it updates a third table 'sharing' with the project owner's ID (Member A), the "shared_with" member's ID (Member B) and the project ID. (Perhaps I have gone bullheaded in the wrong direction?)
The problem: How do I query the DB to show Member B all of their own projects + any projects that have been shared with them? The query below illustrates the direction I've been which has been useless. I am trying to say, "Select all from projects where user_id = (me) AND all corresponding projects where my ID is in the 'sharing' table under the 'shared_with' column. ...Oh yeah, and grab that project_id in order to know which project while you're at it."
My brain is mush. Any direction would be sincerely appreciated.
function find_all_projects($id) {
global $db;
$sql = "
SELECT *
FROM projects
LEFT
JOIN sharing
on projects.id = sharing.project_id
WHERE user_id = '" . db_escape($db, $id) . "'
OR sharing.shared_with = '" . db_escape($db, $id) . "'
ORDER
BY project_name
";
$result = mysqli_query($db, $sql);
confirm_result_set($result);
return $result;
}

Answer 1

Current Table Structure
From your question I believe your current table structure to be something like the following:
TABLE: user TABLE: project TABLE: shared
id | email | | id | user_id | content | | id | user_id | project_id
---+-------------------- ---+---------+------------------------------ ---+---------+------------
1 | james#website.com | | 1 | 1 | Project for James | | 9 | 1 | 5
2 | hannah#website.com | | 2 | 1 | Some other project for James | | 10 | 3 | 5
3 | lucy#website.com | | 3 | 2 | Project for Hannah | | 11 | 1 | 8
| | | 4 | 2 | A new project for hannah | | 12 | 2 | 8
| | | 5 | 2 | Hannah's pride and Joy | |
| | | 6 | 3 | Lucy cracking down | |
| | | 7 | 3 | Lucy's second project | |
| | | 8 | 3 | Lucy's public stuff | |
SQL
Example: https://www.db-fiddle.com/f/6KnEsGUmy5PS42usmzyTEX/0
SELECT project.id, project.user_id AS owner_id, shared.user_id AS shared_id, project.content
FROM project
LEFT JOIN shared
ON project.id = shared.project_id
AND project.user_id <> ?
WHERE project.user_id = ?
OR shared.user_id = ?;
N.B.
The key difference between this SQL statement and the one in your question is
AND project.user_id <> ?
Without that condition in the ON clause you will get duplicate records for every shared project for that user. I.e. if the user has shared the project with 20 users then there will be 20 duplicates.
This is expected behaviour as explained here: PHP while statement echoes duplicates
PHP
$sql = "
SELECT project.id, project.user_id AS owner_id, shared.user_id AS shared_id, project.content
FROM project
LEFT JOIN shared
ON project.id = shared.project_id
AND project.user_id <> ?
WHERE project.user_id = ?
OR shared.user_id = ?
";
$query = $mysqli->prepare($sql);
$query->bind_param("iii", $user_id, $user_id, $user_id);
$query->execute();
Alternate Table Structure
I suggest updating your table structure so that you have three tables (effectively: users, projects, and project_users). The project_user table then acts as a conduit between the two entities (users and projects). In this case storing the relationship between the two (i.e. owner vs shared with).
TABLE: user TABLE: project TABLE: project_user
id | email | | id | content | | id | user_id | project_id | role
---+-------------------- ---+------------------------------ ---+---------+------------+-----
1 | james#website.com | | 1 | Project for James | | 1 | 1 | 1 | 1
2 | hannah#website.com | | 2 | Some other project for James | | 2 | 1 | 2 | 1
3 | lucy#website.com | | 3 | Project for Hannah | | 3 | 2 | 3 | 1
| | | 4 | A new project for hannah | | 4 | 2 | 4 | 1
| | | 5 | Hannah's pride and Joy | | 5 | 2 | 5 | 1
| | | 6 | Lucy cracking down | | 6 | 3 | 6 | 1
| | | 7 | Lucy's second project | | 7 | 3 | 7 | 1
| | | 8 | Lucy's public stuff | | 8 | 3 | 8 | 1
| | | | | | 9 | 1 | 5 | 2
| | | | | | 10 | 3 | 5 | 2
| | | | | | 11 | 1 | 8 | 2
| | | | | | 12 | 2 | 8 | 2
SQL
Example: https://www.db-fiddle.com/f/imQZ6cvEEff4VgRQ4v22Qo/0
SELECT project.id, project_user.user_id, project_user.role, project.content
FROM project
JOIN project_user
ON project_user.project_id = project.id
WHERE project_user.user_id = ?;
PHP
$sql = "
SELECT project.id, project_user.user_id, project_user.role, project.content
FROM project
JOIN project_user
ON project_user.project_id = project.id
WHERE project_user.user_id = ?
";
$query = $mysqli->prepare($sql);
$query->bind_param("i", $user_id);
$query->execute();

Answer 2

You can use another relationship between members and projects with a table like this :
CREATE TABLE `project_members` (
`project_id` INT NOT NULL,
`member_id` INT NOT NULL,
`is_owner` TINYINT NOT NULL DEFAULT 0,
PRIMARY KEY (`project_id`, `member_id`));
This table allows you to have many members linked to many projects.
The column is_owner is a boolean to easily see if the member is the owner or if the project has been shared to him.
Also it would be good to add foreign keys to project_id and member_id.