How can i pass the jquery value to php function - php

Question

Here i want to call the php function with some parameter that will be held in some jquery function.
I have php function that takes one parameter and executes the sql query.
and also i have a jquery function where i am getting different values when select item from dropdown box.
now i want to pass the value of dropdown box to php function, how can i do this?
here is what i have done
$("#product_category").change(function(){
var category = $(this).val(); //getting dropdown value
// here i want to pass this variable value to php function
});
php
function productManagement($procat){
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT product_name,product_category,product_image_url FROM product_list where product_category='$procat'";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo "<tr><td>". $row["product_name"]."</td><td><a href=''><span class='glyphicon glyphicon-remove'></span>Remove</a></td><td><a href='edit-product.php'><span class='glyphicon glyphicon-edit'></span>Edit</a></td></tr>";
}
} else {
echo "No results found";
}
$conn->close();
}
How can i do this?

Answer 1

Another way is to use Ajax like :
Jquery code:
$("#product_category").change(function(){
var category = $(this).val(); //getting dropdown value
// here i want to pass this variable value to php function
-----------------------------------------------------
$.ajax({
url: 'newFunctionFile.php',
type: 'POST',
data: 'category='+category,
success: function(data) {
//you can add the code to show success message
},
error: function(e) {
//called when there is an error
//console.log(e.message);
}
});
});
and you have to create a file for php function let say file name is newFunctionFile.php as i have mention in ajax call:
if(isset($_POST['category'])) {
productManagement($_POST['category']);
}
function productManagement($procat){
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT product_name,product_category,product_image_url FROM product_list where product_category='$procat'";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo "<tr><td>". $row["product_name"]."</td><td><a href=''><span class='glyphicon glyphicon-remove'></span>Remove</a></td><td><a href='edit-product.php'><span class='glyphicon glyphicon-edit'></span>Edit</a></td></tr>";
}
} else {
echo "No results found";
}
$conn->close();
}

Answer 2

How about something along these lines.
In your javascript:
$("#product_category").change(function(){
var category = $(this).val(); //getting dropdown value
location.href = location.href + "?val="+category; //reload the page with a parameter
});
And then in your PHP, add this code as your function call
if(isset($_GET['val'])) { //if 'val' parameter is set (i.e. is in the url)
productManagement($_GET['val']); //call function, with that parameter
}