php- check radio button then pass the value to mysql query - php

Question

Good day every one, i am trying to check if a radio button were clicked, and iwant the value of that clicked radio button to pass on a variable, i will used that variable to compare records in database with the same values from it. and display all records on list box??
but when i try to run this code nothings happen
<td><input type="radio" name="1stChoice" value="TESDA" ></td><br>
<td align = "center">
<select name="course1">
<?php
include('dbconnection.php');
if(isset($_POST['1stChoice'])) {
if($_POST['1stChoice'] == 'TESDA') {
$choose='TESDA';
} elseif($_POST['1stChoice'] == 'CHED') {
$choose='CHED';
}
}
$mysql_select=mysql_query("select * from courses where Institution = '$choose' ",$mysql);
while($row=mysql_fetch_array($mysql_select))
{
?>
<option><?php $row['Program']; ?></option></td>
</select>
<?php } ?>
</tr>
<td width="20%">2nd choice:</td>
<td><input type="radio" name="2ndChoice" value="CHED" ></td>
<td><input type="radio" name="2ndChoice" value="TESDA" ></td><br>
<td align = "center">
<select name="course2">
<?php
include('dbconnection.php');
if(isset($_POST['2ndChoice'])) {
if($_POST['2ndChoice'] == 'TESDA') {
$choose='TESDA';
} elseif($_POST['2ndChoice'] == 'CHED') {
$choose='CHED';
}
}
$mysql_select=mysql_query("select * from courses where Institution = '$choose' ",$mysql);
while($row=mysql_fetch_array($mysql_select))
{
?>
<option><?php $row['Program']; ?></option></td>
</select>
<?php } ?>

Answer 1

I tested your code and as far as I can see, you're not echoing your <?php $row['Program']; ?> which should read as <?php echo $row['Program']; ?>
This is for both your <option> tags.
I also didn't notice any form tags <form></form>, so you will need to add those if you're not presently using them.
Using a submit button could be useful also. Although I'm not sure if you're using JS/jQuery with your code.
<input type="submit" name="submit" value="Submit">
Here is what I used to test it with, along with a few additions/modifications:
(I added form tags, a submit button and the echo for the <select> tags)
<form action="" method="post">
<td><input type="radio" name="1stChoice" value="TESDA" ></td><br>
<input type="submit" name="submit" value="Submit">
<td align = "center">
<select name="course1">
<?php
include('dbconnection.php');
if(isset($_POST['1stChoice'])) {
if($_POST['1stChoice'] == 'TESDA') {
$choose='TESDA';
} elseif($_POST['1stChoice'] == 'CHED') {
$choose='CHED';
}
}
$mysql_select=mysql_query("select * from courses where Institution = '$choose' ",$mysql);
while($row=mysql_fetch_array($mysql_select))
{
?>
<option><?php echo $row['Program']; ?></option></td>
</select>
<?php } ?>
</tr>
<td width="20%">2nd choice:</td>
<td><input type="radio" name="2ndChoice" value="CHED" ></td>
<td><input type="radio" name="2ndChoice" value="TESDA" ></td><br>
<td align = "center">
<select name="course2">
<?php
include('dbconnection.php');
if(isset($_POST['2ndChoice'])) {
if($_POST['2ndChoice'] == 'TESDA') {
$choose='TESDA';
} elseif($_POST['2ndChoice'] == 'CHED') {
$choose='CHED';
}
}
$mysql_select=mysql_query("select * from courses where Institution = '$choose' ",$mysql);
while($row=mysql_fetch_array($mysql_select))
{
?>
<option><?php echo $row['Program']; ?></option></td>
</select>
</form>
<?php } ?>

Answer 2

This is not a php work. IF your html is good, your browser will check the checked radio button. See! In your code, name must start with a letter or a _(underscore).
So, replace all name='2ndChoice' with
<td><input type="radio" name="ck_2ndChoice" value="CHED" ></td>
<td><input type="radio" name="ck_2ndChoice" value="TESDA" ></td><br>
and all name='1stChoice' with name='ck_1stChoice'