Missing the result printed from database - php

Question

My purpose is search book type from input keyboard and count it. I have error: counting the number of book is right but when print, it miss some result. For example, I put 'f' it show 7 result( right) however it print only 2 two result. Here is my code:
<form action="" method="POST">
Enter type of book here:<input type="text" size="20" name="sbt"> <br />
<input type="submit" name="sb" value="Search">
</form>
<table align="center" border="1" width="600">
<thead><tr align="center">
<tr align="center">
<td><b>Book ID</b></td>
<td><b>Book Title</b></td>
<td><b>Book Author</b></td>
<td><b>Pulished Year</b></td>
<td><b>Book Type</b></td>
<td><b>Status</b></td>
</tr>
<?php
if (isset($_POST['sb'])) {
$s="";
if ($_POST['sbt'] == null) {
echo "Please re-enter <br>";
} else
{
$s = $_POST['sbt'];
}
$q = "SELECT * FROM book WHERE book_type LIKE '%$s%' ";
$r= mysqli_query($conn,$q);
while($row = mysqli_fetch_array($r)) {
?>
<tr align="center">
<td><?php echo $row['book_no'];?></td>
<td><?php echo $row['book_title'];?></td>
<td><?php echo $row['book_author'];?></td>
<td><?php echo $row['book_year'];?></td>
<td><?php echo $row['book_type'];?></td>
<td>
<?php
if ($row['book_quantity'] == 1) {
echo "Available";
}
else {
echo "Not available";
}
?>
</td>
<?php
$c=" SELECT COUNT(DISTINCT(book_no)) AS totaltype FROM book WHERE book_type LIKE '%$s%'";
$s= mysqli_query($conn, $c);
if (mysqli_num_rows($s)> 0 )
{
$row2= mysqli_fetch_array($s);
echo " Total book type result:"."{$row2['totaltype']}";
}
}
?>
<?php } ?>
</table>

Answer 1

Remove "%" before $s and try.
for example:
$query = mysql_query("SELECT * FROM table WHERE the_number LIKE '$yourPHPVAR%'");