Using variable with multiple values in MySQL LIKE clause - php

Question

can someone help me on the proper way of using a variable with multiple values in MySQL LIKE clause?
by the way, I have two tables teacher and student.
Teacher table contains. (fname,lname,subject,yr,sec) columns and for the student table (lrn,yr,sec)
what I'm trying to achieve is to count the total number of students of the teacher from all sections.
Here's my query for getting teacher's sections:
<?php
$result= mysqli_query($con, "select *from teacher where lname
= 'Uy' and subject = 'Science & Technology 7'" );
while($rowx = mysqli_fetch_array($result)) {
$section = $rowx['sec'];
}
?>
echoing $section output this: Our Lady of FatimaOur Lady of Guadalupe
Heres my query to count the students.
<?php
$ratequery = mysqli_query($con, "SELECT *,
(SELECT COUNT(*) FROM student WHERE
sec LIKE '%$section%') AS responseCount FROM student");
$rateresult = mysqli_fetch_array($ratequery);
?>
Output: the count works but only the first value [Our Lady of Fatima] of $section is getting recognized by LIKE clause, so only 1 section is getting counted.
by the way, sorry for my bad English and explanation.

Answer 1

Use Section as a input for second query:
<?php
$final_result = array();
$result= mysqli_query($con, "select *from teacher where lname
= 'Uy' and subject = 'Science & Technology 7'" );
while($rowx = mysqli_fetch_array($result))
{
$section = $rowx['sec'];
$ratequery = mysqli_query($con, "SELECT *, (SELECT COUNT(*) FROM student WHERE
sec =$section) AS responseCount FROM student");
$rateresult = mysqli_fetch_array($ratequery);
array_push($final_result,array("section"=>$section,"rateresult"=>$rateresult));
}
var_dump($final_result); // Display section wise data
?>

Answer 2

somehow I have managed to achieve my goal it but in other way . I used WHERE IN clause instead of LIKE clause. stored my query results into an array.. then used implode to add seperator.
Thanks for the help mr. #ashnu
<?php
$query="$con, select *from teacher where lname = 'Uy' and subject = 'Science & Technology 7'";
$result = mysqli_query($query) or die;
$sections = array();
while($row = mysqli_fetch_assoc($result))
{
$sections[] = $row['sec'];
}
$string = implode("','",$sections)
?>
<?php
$ratequery = mysqli_query($con, "SELECT *,
(SELECT COUNT(*) FROM student WHERE sec IN ('$string')) AS responseCount FROM student");
$rateresult = mysqli_fetch_array($ratequery);
?>