Question

我有两个mysql表order和orderitems 现在我想在一个单独的监视器上使用angularjs显示这些数据我的桌子结构：
表格order包含id, table_id, date_time
表格orderitems包含order_id, food_item, item_qty, dalivery

现在我想在单独的html div中显示这些数据。 table_id为标题，orderitems为内容。

为此，我想要一个像这样的数据结构

[{
 table_id,
 date_time,
 items: [
    {
   food_item,
   item_qty,
   delivery
},
{
   food_item,
   item_qty,
   delivery
}
]
}]

那么如何使用codeigniter或MySQL进行查询呢？

Answer 1

您可以选择包含订单数据的所有订单商品

这是一个Mysqli解决方案

$mysqli = new mysqli("example.com", "user", "password", "database");

$query = " SELECT * FROM order LEFT JOIN  orderitems ON order.id = orderitems.order_id ";
$stmt = $mysqli->prepare($query);
!$stmt ? die('error') : "";
$stmt->execute();
$result = $stmt->get_result();


while ($row = $result->fetch_array(MYSQLI_ASSOC)) {
    if($lastOrderId != $row['id']){
        if($order){
            $finalArray[] = $order;
        }
        $lastOrderId = $row['id'];
        $order = new \stdClass();
        //$order->id = $row['id'];
        $order->table_id = $row['table_id'];
        $order->date_time = $row['date_time'];
        $order->items = [];
    }
    if($row['food_item']){
        $orderItem = new \stdClass();
        $orderItem->food_item = $row['food_item'];
        $orderItem->item_qty = $row['item_qty'];
        $orderItem->dalivery = $row['dalivery'];
        $order->items[] = $orderItem;
    }
}
//last order in loop
if($order){
    $finalArray[] = $order;
}
print_r($finalArray) ;
exit;

尝试现场演示（https://eval.in/835997）

此演示依赖于虚假数据进行测试

$row1 = ["id"=>1,"table_id"=>5,"date_time"=>"2017-12-31","food_item"=>1,"item_qty"=>1,"dalivery"=>"yes"];
$row2 = ["id"=>1,"table_id"=>5,"date_time"=>"2017-12-31","food_item"=>2,"item_qty"=>2,"dalivery"=>"no"];
$row3 = ["id"=>2,"table_id"=>5,"date_time"=>"2017-12-31","food_item"=>3,"item_qty"=>3,"dalivery"=>"yes"];
$row4 = ["id"=>2,"table_id"=>5,"date_time"=>"2017-12-31","food_item"=>4,"item_qty"=>4,"dalivery"=>"no"];

$rows = [$row1, $row2, $row3, $row4];

此输出

Array
(
    [0] => stdClass Object
        (
            [table_id] => 5
            [date_time] => 2017-12-31
            [items] => Array
                (
                    [0] => stdClass Object
                        (
                            [food_item] => 1
                            [item_qty] => 1
                            [dalivery] => yes
                        )
                    [1] => stdClass Object
                        (
                            [food_item] => 2
                            [item_qty] => 2
                            [dalivery] => no
                        )
                )
        )
    [1] => stdClass Object
        (
            [table_id] => 5
            [date_time] => 2017-12-31
            [items] => Array
                (
                    [0] => stdClass Object
                        (
                            [food_item] => 3
                            [item_qty] => 3
                            [dalivery] => yes
                        )
                    [1] => stdClass Object
                        (
                            [food_item] => 4
                            [item_qty] => 4
                            [dalivery] => no
                        )
                )
        )
)

Answer 2

使用Codeigniter。对于PDO可能相同。

$data = array();
        $tables = $this->db->select('table_data.*, order.id As order_id, order.date_time')
                            ->join('order', 'order.table_id = table_data.id')
                            ->where('order.status', 'Pending')
                            ->get('table_data')->result_array();
        foreach ($tables as $table) {
            $items = $this->db->select('orderitems.food_item, orderitems.item_qty')
                                ->where('order_id', $table['order_id'])
                                ->get('orderitems')->result_array();
            array_push($data, array('details' => $table, 'items' => $items));
        }
        $this->response($data);

检索数据以便以特定格式显示

2 个答案: