HTML复选框，用于在多对多查找表上添加/删除条目（PHP / MySQL）

Question

I have the following entities:

EMPLOYEE (id, fname, lname, ...)

PROJECT (id, ...)

EMPLOYEE_PROJECT (employee_id, project_id, ...)

An employee can work on more than one project, and a project can be worked on by more than one employee, so the EMPLOYEE_PROJECT table is used to relate them.

I would like to generate in my application (PHP/MySQL) a form for a given PROJECT ID with a checkbox next to each employee name, that will display as checked if an entity for that employee id exists for that project id in the EMPLOYEE_PROJECT table, and that will allow the user to check or uncheck any employee and then click an "update" button to add/remove the selected/unselected employees from the project by adding/removing the entries on the EMPLOYEE_PROJECT table.

I am able to use an IF statement to display "checked" on any entity that exists in the EMPLOYEE_PROJECT table. My problems are:

1. Properly forming the query to get the list. If I get the list of employees by a JOIN of EMPLOYEE ON EMPLOYEE_PROJECT, it only shows employees who have been involved in a project, not all employees. I can do a LEFT JOIN to show all, but that would repeat names of employees who have multiple projects, and I only want to show each name once. If I add a GROUP BY employee.id, I can get a proper list of only one of each employee, however the results don't necessarily match the correct project_id. If I add a HAVING with the project_id, I lose employees who are not on the project (keep in mind I want to list ALL employees, but only show a check next to the ones that are currently on the project).

2. Assuming I figure out #1 and can list all employees and display a check next to the ones which exist as an EMPLOYEE_PROJECT entity for the specified project, how can I allow the check boxes to be modified by the user, and then submitted and have the one submit action add/remove the relevant entities from the EMPLOYEE_PROJECT table?

Thanks so much!!

Answer 1

@zzarbi is correct for #2. For #1 you need a sub-query:

SELECT employee.*, (SELECT count(id) FROM employee_project 
   WHERE project_id=<project_id> AND employee_id=employee.id) 
   AS in_project 
   FROM employee

Answer 2

For #1 "list ALL employees, but only show a check next to the ones that are currently on the project". (That work only if you show only one project at the time)

SELECT * from employee as e
LEFT JOIN EMPLOYEE_PROJECT as ep on ep.employee_id = e.id
WHERE ep.project_id = <project id>

in php :

if("employee_id" IS NOT NULL/empty){
    echo "check";
}

For #2, usually my middle table project_employee has nothing else than two id (employeeId, projectId). So i generally just delete every employee for this projectId and re-add them. If you have other data in the same table, this might not work. And you will have to compare old data store to new data to be stored.