Issue in php script for android app

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;
}

Nigel Ren · Accepted Answer · 2018-08-07 07:24:27Z

2

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;
}

answered Aug 7, 2018 at 7:24

Nigel Ren

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