How do i get selected results to be displayed on another page?

Question

The output of the code below has a list with check boxes, i want only the results with checked check boxes to be be displayed on a different page, please assist...

<?php
mysql_connect("localhost", "root","") or die(mysql_error()); //Connect to server
mysql_select_db("first_db") or die("Cannot connect to database"); //connect to database

$search = $_POST['search'];

$query = mysql_query("SELECT * FROM medical WHERE conditions LIKE '%$search%'"); // SQL Query

while($row = mysql_fetch_array($query))
{
Print "<tr>";

Print '<td align="center" >'.'<input type="checkbox" name="gender" value="male" >'. $row['conditions'] . "</td>";
Print '<td align="center">'. $row['icd_10'] . "</td>";
Print '<td align="center">'. $row['drugs']. "</td>";
Print '<td>'."<input type='number'/>"."</td>";
Print '<td>'."<input type='number'/>"."</td>";
Print '<td>'."<input type='number'/>"."</td>";
Print '<td>'."<input type='text'/>"."</td>";
Print '<td>'."<select name='Dosage'>
<option value='space'> </option>
<option value='one'>1 tablet a day</option>
<option value='two'>1 tablet nocte</option>
<option value='three'>1 tablet twice a day</option>
<option value='four'>1 tablet 3 times ad day</option>
<option value='five'>2 tablet twice a day</option>
<option value='six'>2 tablet 3 times a day</option>
<option value='seven'>1 spoon twice a day</option>
<option value='eight'>1 spoon three times a day</option>
<option value='nine'>2 spoons twice a day</option>
</select>"."</td>";
Print '<td>'."<input type='text'/>"."</td>";
Print '<td>'."<form method='post'>
<select name='Dosage'>
<option value='space'> </option>
<option value='one'>Capsules</option>
<option value='two'>Ointment/Cream</option>
<option value='three'>Injectible</option>

Answer 1

give to all your inputs names and take values by $_POST['name of your input']