Call a database variable with SQL in PHP - php

Question

I have a database with the tables program and class. The class table has a foreign key from program, the id_program. In the PHP code below, I fetch the data from the class table and works great.
<?php
$A_apotelesma = NULL;
$query = "SELECT * FROM class";
$result = mysqli_query($con,$query) or die('Error, query failed');
$i=-1;
if ($result) {
while ($row = mysqli_fetch_array($result)) {
$i++;
$A_apotelesma[$i]['id_program'] = $row['id_program'];
$A_apotelesma[$i]['id_class'] = $row['id_class'];
$A_apotelesma[$i]['onoma'] = $row['onoma'];
$A_apotelesma[$i]['imera'] = $row['imera'];
$A_apotelesma[$i]['time_start'] = $row['time_start'];
$A_apotelesma[$i]['time_end'] = $row['time_end'];
}
}
echo "<table border=\"1\" bordercolor=\"#ff8533\">";
if(is_array($A_apotelesma)){
echo " <tr>
<td colspan='7' align='center'><strong><br>Στοιχεία υπάρχοντων τμημάτων:</strong><br><br></td>
</tr>
<tr>
<td align='left'>ID</td>
<td align='left'>Πρόγραμμα</td>
<td align='left'>Όνομα</td>
<td align='left'>Ημέρα</td>
<td align='left'>Ώρα έναρξης</td>
<td align='left'>Ώρα λήξης</td>
</tr>";
foreach($A_apotelesma as $in => $value){
echo "<tr>
<th width='16%'>".$value['id_class']."</th>
<th width='16%'>".$value['id_program']."</th>
<th width='16%'>".$value['onoma']."</th>
<th width='16%'>".$value['imera']."</th>
<th width='16%'>".$value['time_start']."</th>
<th width='16%'>".$value['time_end']."</th>
<td>
Διαγραφή τμήματος<br>
<img src=\"images/del.jpg\" onclick='delete_class(".$value['id_class'].")' onmouseover=\"this.style.cursor='pointer'\" />
</td>
</tr>";
}
}
else
echo "<tr><th>Δεν βρέθηκαν τμήματα.</th></tr>";
echo "</table>";
?>
I just would like in the foreach command the ".$value['id_program']." not to be the id_program value, but another value, called onoma from the "program" table. I tried:
SELECT onoma FROM program LEFT JOIN class ON program.id_program=class.id_program
but no luck. Any help would be really appreciated.

Answer 1

Something like this should work. You cannot overwite the id_program, you have to call it something else in the select statement
SELECT Program.Onoma as Onoma
FROM Program
INNER JOIN Class
ON Class.id_program = Program.id;

Answer 2

Need to join two tables rather than selecting values from single table.Replace your query:
$query = "SELECT c.id_class,c.imera,c.time_start,c.time_end,p.program_name,p.onoma FROM class c INNER JOIN program p ON c.id_program=p.id_program";
See program_name name is used instead of program_id to display program name from program table.