Issue in php script for android app - php

Question

I have a showProduct.php file from where i want to call a function showProduct() in another file. In showProduct() i want to extract all rows from database and to showProduct.php file. the issue is that when i return the array only last row is showing. I want to show all the rows.
The showProduct.php is:
<?php
require_once '../includes/DbOperations.php';
$response = array();
$result = array();
if($_SERVER['REQUEST_METHOD'] == 'POST')
{
$db = new DbOperations();
$result = $db->showProduct();
if(!empty($result))
{
$response["prod_name"] = $result["prod_name"];
$response["prod_desc"] = $result["prod_desc"];
$response["prod_image"] = $result["prod_image"];
}
else
{
$response["error"] = true;
$response["message"] = "products are not shown";
}
}
echo json_encode($response);
?>
and showProduct() function is:
public function showProduct(){
$menu = array();
$query = mysqli_query($this->con,"SELECT * FROM `products` WHERE 1");
while ($row = mysqli_fetch_array($query)) {
$menu['prod_name'] = $row['prod_name'] ;
$menu['prod_desc'] = $row['prod_desc'] ;
$menu['prod_image'] = $row['prod_image'];
}
return $menu;
}

Answer 1

In your function, you are just overwriting the last data each time, you need to build this data up. Create an array with the new data and use $menu[] to add this new data to the list of menus...
public function showProduct(){
$menu = array();
$query = mysqli_query($this->con,"SELECT * FROM `products` WHERE 1");
while ($row = mysqli_fetch_array($query)) {
$newMenu = []; // Clear array to ensure no details left over
$newMenu['prod_name'] = $row['prod_name'] ;
$newMenu['prod_desc'] = $row['prod_desc'] ;
$newMenu['prod_image'] = $row['prod_image'];
$menu[] = $newMenu;
}
return $menu;
}