How to tell php to echo something acquired to the selected option - php

Question

I have two tags, the first one is to select the brand and the second one is to select the product.
My question/problem is that I want to tell php to echo inside the second select tag. The echoed option needs to have the same brand_name as the selected option from the first tag.
What I need is that if I choose Adidas as the brand, php would show every product that has the brand_name Adidas.
I have two different database tables, the first one called brand:
brand_id | brand_name
1 | ADIDAS
2 | NIKE
The second one called product:
product_id | brand_name | product_name | product_image | amount | sell | buy
1 | ADIDAS | T-Shirt | none | 50 | 30 | 28
2 | NIKE | Shoes | none | 20 | 130 | 120
Here is my code:
<select>
<option disabled selected>----SELECT----</option>
<?php
require_once 'connections/dbc.php';
$query = "SELECT `brand_name` FROM `brands`";
$response = #mysqli_query($conn, $query);
if (!$response) {
$_SESSION['errortitle'] ='Error loading the brands';
$_SESSION['errormessage'] = 'Something wrong happend while loading the brands.<br/>Please contact The developer';
header("location: error.php");
exit();
} else {
while($row = mysqli_fetch_array($response)){
echo '<option name="brand_name" value='.$row['brand_name'].'>'.$row['brand_name'].'</option>';
}
}
?>
</select>
<select>
<option disabled selected>----SELECT----</option>
<?php
require_once 'connections/dbc.php';
###########HERE WHERE I NEED MY CODE##############
$query = "SELECT `product_name` FROM `product WHERE brand_name = "##### The selected option from the first select tag" ";
$response = #mysqli_query($conn, $query);
if (!$response) {
$_SESSION['errortitle'] = "Error loading the prouducts";
$_SESSION['errormessage'] = 'Something wrong happend while loading the proudcts.<br/>Please contact The developer';
header("location: error.php");
exit();
} else {
while($row = mysqli_fetch_array($response)){
echo '<option name="product_name" value='.$row['product_name'].'>'.$row['product_name'].'</option>';
}
}
?>
</select>

Answer 1

There are two options:
First: Download all contents of 'products' table, and make script that will add items with selected 'brand' attribute to box.
Second: (Recommended) When user selects a brand, make an AJAX query, and get all items with selected brand.
Second one is much better, because you will need to download less data, and page will load faster.
Imagine you have 10^9 products, and 10^7 brands(average 100 products per brand)
Using first solution, you will need to download all 10^9 products, and then use JavaScript(which is not very fast) to select 100 of them.
If you use second solution two things are going to be much better:
1. Download 100 items instead of 10^9.
2. To select items use fast SQL engine instead of slow JavaScript
It is not possible to do it with php, because php is server-side. Server is making page code with php, and then sending it to user. It's impossible to make callback to php without AJAX

Answer 2

The easiest way to do this would be to use a single SELECT with LEFT JOIN, to get all of the columns from both tables at the same time:
SELECT brands.brand_name, products.product_name FROM brands LEFT JOIN products on brands.brand_name = products.brand_name GROUP BY brands.brand_name
And build both of your <select> nodes from that single query. To do this, build your first <select> node from the query, the same as you are doing now, then in the lower part of the code reset the $response with $response->data_seek(0). Then you can build the second <select> as though you had done another query.
It will look something like the following code. There are only two minor changes to your original. Look for some comments in the top section, and some in the bottom section. I don't have an easy mechanism to run PHP, so I can't test it, and it might not run perfectly as it is now. But I think it will give you an idea of what to do.
<select>
<option disabled selected>----SELECT----</option>
<?php
require_once 'connections/dbc.php';
// Notice that the query includes LEFT JOIN, to get the
// columns from both tables in a single query.
$query = "SELECT brands.brand_name, products.product_name
FROM brands left join products on brands.brand_name = products.brand_name"
$response = #mysqli_query($conn, $query);
if (!$response) {
$_SESSION['errortitle'] ='Error loading the brands';
$_SESSION['errormessage'] = 'Something wrong happend while loading the brands.<br/>Please contact The developer';
header("location: error.php");
exit();
} else {
while($row = mysqli_fetch_array($response)){
echo '<option name="brand_name" value='.$row['brand_name'].'>'.$row['brand_name'].'</option>';
}
}
?>
</select>
<select>
<option disabled selected>----SELECT----</option>
<?php
require_once 'connections/dbc.php';
// Here, you don't need another query. You simply reset
// $response, and then fetch as you would from a
// normal query.
$response->data_seek(0);
while($row = mysqli_fetch_array($response)){
echo '<option name="product_name" value='.$row['product_name'].'>'.$row['product_name'].'</option>';
}
?>
</select>

Answer 3

You can simply use Ajax to do this. I'll try to explain step by step.
I assume jQuery is loaded.
Your product table should use brand_id not brand_name.
product_id|brand_name | product_name | product_image | amount | sell | buy
1 |brand_name_1| T-Shirt | none | 50 | 30 | 28
2 |brand_name_2| Shoes | none | 20 | 130 | 120
I hope this will help you.
Step - Create your first and second select box by using name or id attributes. I will be using the name to define them.
<select name="brand_name">
<?php //call brand_name options from your database
//Sample : <option value="brand_name_1">Brand_Name_1</option>?>
</select>
<select name="product_name" disabled >
<?php //dont fill this select box.
//Leave it empty, contents will be created after our ajax call
?>
</select>
Step
Add jQuery script to your page as below.
<script>
$('[name="brand_name"]').change(function(event){
var brand_name = $(this).val();
$.get(
"ajax.php",
{ brand_name: brand_name },
function(data) {
var opts = $.parseJSON(data);
$.each(opts, function(i, d) {
$('[name="product_name"]').append($('<option>', {
value: d.product_name,
text : d.product_name
}));
});
//Enable the product_name select box
$('[name="product_name"]').prop('disabled', false);
//Refresh the product_name select box
$('[name="product_name"]').selectpicker('refresh');
}
);
});
</script>
Step - The ajax.php must look something like below. I have used your php query code here.
<?php
require_once 'connections/dbc.php';
$getBrandName = $_REQUEST['brand_name']; // This is the id of the selected brand name
$query = "SELECT product_name FROM product WHERE brand_id = '$getBrandName' ";
$response = #mysqli_query($conn, $query);
if (!$response) {
$_SESSION['errortitle'] = "Error loading the prouducts";
$_SESSION['errormessage'] = 'Something wrong happend while loading the proudcts.<br/>Please contact The developer';
//Below Header method will not work cause headers are already been set. You will not be able to redirect page. So just delete it. instead you can return an error by using the json.
header("location: error.php");
exit();
} else {
while($row = mysqli_fetch_array($response)){
$prodyctsArray[]['name'] = $row['product_name'];
}
echo json_encode($brandName);
}
?>
For more information I suggest you to visit references below:
jQuery.ajax()
jQuery.get()
Introducing JSON