如何从没有匹配列的表中获取数据？

Question

I've been bumping my head over trying to find the perfect query that will allow me to display the first and last names of all people involved in a conversation about a specific project.

conversations table

convo_id | project_id | toEmployee_id | fromEmployee_id | message

employees

employee_id | first_name | last_name |

myinbox.php

SELECT * FROM projects as p
JOIN employeeprojects AS ep
ON p.project_id = ep.project_id
JOIN employees AS e
ON ep.assigned_by = e.employee_id
JOIN clients AS c
ON p.client_id = c.id
WHERE ep.employee_id='$session_myemployeeid'

<a data-toggle="tooltip" title="view conversation"  href='conversation_feed.php?viewproject=conversation&emprojectid=<?=$employeeproject['project_id'];?>View Conversation</a>

displayconversation.php

$projectconvoid =  $_GET['emprojectid'] ;
SELECT * FROM employeeprojects_conversation AS epc
JOIN projects AS p
ON epc.project_id=p.project_id
JOIN employeeprojects AS ep
ON p.project_id=ep.project_id
WHERE epc.project_id='$projectconvoid'

While it all work great at displaying specific project conversation with the people involved, I would like to be able to display their name rather their employee_id .

How can I do that?

Answer 1

join employees table twice, one for toEmployee_id and one for fromEmployee_id like so:

SELECT
    *
FROM
    employeeprojects_conversation AS epc
JOIN projects AS p
    ON epc.project_id=p.project_id
JOIN employeeprojects AS ep
    ON p.project_id=ep.project_id
JOIN employees AS emp
    ON emp.employee_id = epc.toEmployee_id
JOIN employees AS emp2
    ON emp2.employee_id = epc.fromEmployee_id
WHERE
    epc.project_id='$projectconvoid'