什么查询会给我这个结果？ JSON 和 PHP

Question

{
    "cars": {
        "toyota": [
            {"model":"*****", "doors":"*","color":"*****"},
            {"model":"*****", "doors":"*","color":"*****"}
                  ],
        "Ford": [
            {"model":"*****", "doors":"*","color":"*****"},
            {"model":"*****", "doors":"*","color":"*****"}
        ]
    }
}

什么查询会给我以下结果？

它是一个 JSON 对象（汽车），其中包含多个数组，每个数组用于不同型号的汽车。在每个数组中都会有其他类型的数据，例如门、颜色、年份……等等

我试过这段代码：

<?php 

require 'conn_pdo.php';

$conn -> setAttribute(PDO::ATTR_DEFAULT_FETCH_MODE,PDO::FETCH_ASSOC); 

$sql = "select * from test_cars where cars in (select cars from test_cars GROUP BY cars) ORDER BY RAND()";

$stmt = $conn -> prepare($sql);
$stmt -> execute();
$row = $stmt -> fetchAll();
$json['cars'] = $row;
echo json_encode($json);

?>

但结果不是我希望的

{ "cars": 
         [ {"model":"*****", "doors":*","color":*****"},  
           {"model":"*****","doors":"*","color":*****"} ,
           {"model":"*****", "doors":"*","color":*****"}, 
           {"model":"*****", "doors":"*","color":*****"},
           .......
         ] 
}

我得到了汽车对象，其中包含 一个数组 用于所有不同型号的汽车！！

Answer 1

这是一种方法。我删除了 GROUP BY 方法，因为问题中的方法没有正确使用它 - 它通过使用 IN 查询将分组丢弃。无论如何，我认为用 PHP 很容易做到。

// It would be good to check to ensure you have a valid connection,
// and that the prepare/execute calls are successful
$sql = "SELECT * from test_cars";
$stmt = $conn->prepare($sql);
$stmt->execute();
$rows = $stmt->fetchAll();

$out = [];
foreach ($rows as $row)
{
    // Let's process by brand (since you have not shown your
    // schema, you will need to adjust column names to suit)
    $brand = $row['brand'];

    // Get other data
    $model = $row['model'];
    $doors = $row['doors'];
    $colour = $row['colour'];

    // Check to ensure the brand is an array, so that we don't
    // get an error when we push
    if (!isset($out[$brand]))
    {
        $out[$brand] = [];
    }

    // Now push the data on to the end (using [])
    $out[$brand][] = [
        'model' => $model,
        'doors' => $doors,
        'colour' => $colour,
    ];
}

$json['cars'] = $out;
echo json_encode($json);

这是未经测试的，但应该给你一个大致的想法。您应该期望必须对其进行一些调整，就像 Stack Overflow 上的所有答案一样。我使用了新的数组语法（从 PHP 5.4 开始），但如果你将它换成 array()，它将适用于早期版本。