PHP displaying data from four tables in one query (ie: LEFT JOIN) - php

Question

I have three tables:
I want to display 'Event Details' which shows attending employee details (listo f employee ids from the 'attending_employees table > corresponding employee details form 'employee' table), what team they belong to (from the club_teams table) and the event details from the 'club_events' table).
Currently I am using multiple mysqli queries to display this information however cannot get my head around pulling the data from the database in one query (ie: LEFT JOIN). Your assistance would be greatly appreciated!
Below are the queries i am currently using:
$query = msqli_query($con, "SELECT * FROM attending_employees")or die(mysqli_error($con));
if(mysqli_num_rows($query) > 0){
while($attending = mysqli_fetch_array($query)){
foreach($attending['club_event']){
$eventid = $attending['club_event'];
$query = msqli_query($con, "SELECT * FROM club_events WHERE club_event_id = '$eventid'")or die(mysqli_error($con));
while($event_details = mysqli_fetch_array($query)){
// Echo event details
}
}foreach($attending['employee']){
$empid = $attending['employee'];
$query = msqli_query($con, "SELECT * FROM employees WHERE employee_id = '$empid'")or die(mysqli_error($con));
while($event_employees = mysqli_fetch_array($query)){
// Echo employee details
}
}foreach($attending['team']){
$teamid = $attending['team'];
$query = msqli_query($con, "SELECT * FROM club_teams WHERE clb_team_id = '$teamid'")or die(mysqli_error($con));
while($event_team = mysqli_fetch_array($query)){
// Echo team details
}
}
}
}
This method is highly inefficient and wasteful since its retrieving duplicate data (ie: all repeated 'club_event_id's in the 'attending_employees' table.)

Answer 1

Try this:
$query = msqli_query(
$con,
"SELECT"
. " attending_employees.*"
. ", club_events.*"
. ", employees.*"
. ", club_teams.*"
. " FROM"
. " attending_employees"
. " LEFT JOIN club_events ON club_events.club_event_id = attending_employees.club_event"
. " LEFT JOIN employees ON employees.employee_id = attending_employees.employee"
. " LEFT JOIN club_teams ON club_teams.clb_team_id = attending_employees.team"
) or die(mysqli_error($con));