PHP Group PDO或JSON结果按3个字段和输出

如何解决PHP Group PDO或JSON结果按3个字段和输出

我正在尝试按3个字段delivery_date，van_no和drop_no将结果分组到一个数组中，这是我到目前为止的结果：

// Gets all orders and sorts by delivery date,van and drop number
    $getorders = $conn->prepare("SELECT * FROM orders INNER JOIN customers ON orders.customer_id = customers.customer_id INNER JOIN addresses on orders.address_id = addresses.address_id ORDER BY delivery_date,van_no,drop_no");
    $getorders->execute();

    $result = $getorders->fetchAll(\PDO::FETCH_ASSOC);    

    var_dump($result);

这给了我一个输出，如：

array (size=6)
  0 => 
    array (size=8)
      'orders_id' => string '6' (length=1)
      'customer_id' => string '1' (length=1)
      'address_id' => string '1' (length=1)
      'van_no' => string '1' (length=1)
      'drop_no' => string '0' (length=1)
      'delivery_date' => string '2020-08-25' (length=10)
      'customer_name' => string 'One' (length=3)
      'address_postcode' => string 'b1' (length=2)
  1 => 
    array (size=8)
      'orders_id' => string '1' (length=1)
      'customer_id' => string '1' (length=1)
      'address_id' => string '1' (length=1)
      'van_no' => string '1' (length=1)
      'drop_no' => string '1' (length=1)
      'delivery_date' => string '2020-08-25' (length=10)
      'customer_name' => string 'One' (length=3)
      'address_postcode' => string 'b1' (length=2)
  2 => 
    array (size=8)
      'orders_id' => string '3' (length=1)
      'customer_id' => string '2' (length=1)
      'address_id' => string '2' (length=1)
      'van_no' => string '1' (length=1)
      'drop_no' => string '2' (length=1)
      'delivery_date' => string '2020-08-25' (length=10)
      'customer_name' => string 'Two' (length=3)
      'address_postcode' => string 'b2' (length=2)

如果我将其更改为\ PDO :: FETCH_GROUP，它似乎只是将每个数组的每个字段加倍，并且如果我在查询中使用GROUP BY，则只会给我每个结果一个。 我已经尝试过：

 foreach($data->delivery_date as $values)
 {
      echo $values->van_no . "\n";
 }

但只是得到注意：尝试获取非对象的属性“ delivery_date”，我尝试将其编码为JSON，但卡在同一位置，试图进行分组。

现在基本上我已经完全迷失了，而我在Google上围绕PDO数组或JSON数组进行分组的所有事情最终都无济于事或引发了上述错误。

编辑：

Id至少使输出看起来像这样：

[
    "delivery_date": "2020-08-25"
    {
        "van_no": "1"
            {
                "orders_id": "6","customer_id": "1","address_id": "1","van_no": "1","drop_no": "0","delivery_date": "2020-08-25","customer_name": "One","address_postcode": "b1"
            },{
                "orders_id": "1","drop_no": "1",{
                "orders_id": "3","customer_id": "2","address_id": "2","drop_no": "2","customer_name": "Two","address_postcode": "b2"
            }

    }

解决方法

输出...

"delivery_date": "2020-08-25"
    {

是无法实现的，因此此代码将创建输出，其中包含送货和货车清单...

$output = [];
while( $result = $getorders->fetch(\PDO::FETCH_ASSOC))  {
    if ( !isset($output[$result['delivery_date']]) )    {
        $output[$result['delivery_date']]['date'] = $result['delivery_date'];
    }
    if ( !isset($output[$result['delivery_date']][$result['van_no']]) ) {
        $output[$result['delivery_date']][$result['van_no']]['van'] 
            = $result['van_no'];
    }
    $output[$result['delivery_date']]
            [$result['van_no']][] = $result;
}
echo json_encode(array_values($output),JSON_PRETTY_PRINT);

这将给您类似...

[
    {
        "date": "2020-08-25","1": {
            "van": 1,"0": {
                "delivery_date": "2020-08-25","van_no": 1,