How to output the name of their foreign key id - php

Question

I have a problem, I want that each id in the foreign key can output the name instead of their id. Here is the image.
Here's my code :
<table class="altrowstable" data-responsive="table" >
<thead >
<tr>
<th> IDno</th>
<th> Lastname </th>
<th> Firstname </th>
<th> Department </th>
<th> Program </th>
<th> Action</th>
</tr>
</thead>
<tbody>
<div style="text-align:center; line-height:50px;">
<?php
include('../connection/connect.php');
$YearNow=Date('Y');
$result = $db->prepare("SELECT * FROM student,school_year where user_type =3 AND student.syearid = school_year.syearid AND school_year.from_year like $YearNow ");
$result->execute();
for($i=0; $row = $result->fetch(); $i++){
?>
<tr class="record">
<td><?php echo $row['idno']; ?></td>
<td><?php echo $row['lastname']; ?></td>
<td><?php echo $row['firstname']; ?></td>
//name belong's to their id's
<td><?php echo $row['dept_id']; ?></td>
<td><?php echo $row['progid']; ?></td>
<td><a style="border:1px solid grey; background:grey; border-radius:10%; padding:7px 12px; color:white; text-decoration:none; " href="addcandidates.php?idno=<?php echo $row['idno']; ?>" > Running</a></div></td>
</tr>
<?php
}
?>
</tbody>
</table>
Thanks guys need a help

Answer 1

Just replace this line
<td><?php echo $row['progid']; ?></td>
by this one
<td><?php echo $row['prog_name']; ?></td>
To use the field you want
You also need to adapt your query to select program infos if not already in the table school_year or student (with a join) :
$result = $db->prepare("SELECT * FROM student
INNER JOIN school_year ON student.syearid = school_year.syearid
INNER JOIN program ON student.progid = program.progid
WHERE user_type =3
AND school_year.from_year like $YearNow ");
Assuming the program table is called "program"

Answer 2

Change
<td><?php echo $row['progid']; ?></td>
to
<td><?php echo $row['prog_name']; ?></td>
And I would recommend to study both php and sql

Answer 3

Change
<td><?php echo $row['progid']; ?></td>
by
<td><?php echo $row['prog_name']; ?></td>

Answer 4

inside the loop:
for($i=0; $row = $result->fetch(); $i++){
$program = $db->prepare("SELECT * FROM program where progid = :progid");
$program->execute(array('progid' => $row['progid']));
$p = $program->fetch();
now you can use it as follow: $p['prog_name']