getting a result of a table Mysqli left join - php

Question

So what i want is the following. I have 3 tables: first i will explain in english what i want for final result:
let me explain all in english:
table 1 is REGISTERED_MEMBERS
table 2 is OPEN CLASSES
table 3 i want who appplied and to which class they applied and show that in a table when date is selected from a dropdown list!
This is how it should show at the end:
table = razpisani_tecaji with (ID_TECAJA, DATE, STATUS, ST_ODPRTIH_MEST)
table = registrirani_clani wtih (ID_TECAJNIKA,IME,PRIIMEK, EMAIL)
table = prijave_na_tecaj with (ID_TECAJNIKA, ID_TECAJA, PLACILO)
now in HTML i have a dropdown list populated with dates (this works ok) from table 1.
<form>
<select name="dates" onchange="showUser(this.value)">
<option value="" selected="selected">Izberi datum za pregled</option>;
<?php
$con = mysqli_connect('localhost','root','','viverius_education');
if (!$con) {
die('Could not connect: ' . mysqli_error($con));
}
$sql = mysqli_query($con, "SELECT ID_TECAJA, DATUM FROM razpisani_tecaji");
while ($row = $sql->fetch_assoc()){
echo "<option value='" . $row['ID_TECAJA'] . "'>" . $row['DATUM'] . "</option>";
}
?>
Here a ID is saved to a variable q which is then send to php via JS. This all works. Now what i would like is for user to select a date and get a results from 3. table only to show IME from linked to ID_TECAJNIKA and PLACILO linked to ID_TECAJNIKA.
THis is what i have in PHP so far:
<?php
$q = intval($_GET['q']);
$con = mysqli_connect('localhost','root','','viverius_education');
if (!$con) {
die('Could not connect: ' . mysqli_error($con));
}
$sql="SELECT razpisani_tecaji.ID_TECAJA, registrirani_clani._ID_TECAJNIKA FROM prijave_na_tecaj
LEFT JOIN razpisani_tecaji ON prijave_na_tecaj.ID_TECAJA = razpisani_tecaji.ID_TECAJA
LEFT JOIN registrirani_clani ON prijave_na_tecaj.ID_TECAJNIKA = registrirani_clani.ID_TECAJNIKA";
$result = mysqli_query($con,$sql);
echo "<table>
<tr>
<th>Ime</th>
<th>Placilo</th>
</tr>";
while($row = mysqli_fetch_array($result)) {
echo "<tr>";
echo "<td>" . $row['ID_TECAJNIKA'] . "</td>";
echo "<td>" . $row['PLACILO'] . "</td>";
echo "</tr>";
}
echo "</table>";
mysqli_close($con);
?>

Answer 1

$sql="select registrirani_clani., prijave_na_tecaj.,razpisani_tecaji.*
from registrirani_clani
left join prijave_na_tecaj on registrirani_clani.ID_TECAJNIKA = prijave_na_tecaj.ID_TECAJNIKA
left join razpisani_tecaji ON razpisani_tecaji.ID_TECAJA = prijave_na_tecaj.ID_TECAJA
where razpisani_tecaji.ID_TECAJA = '".$q."'";