如何使用laravel/php手动映射集合并获取总量

我目前有API返回的以下json响应。我能用拉威尔的口才把它还给你。有几个用户，每个用户都有几个收据。收据具有类型和状态。我想尝试获取与类型和状态相关的每个收据的总金额。我能够使用

$this->user->with'receipts'->has'receipts'->get['id'，name']

我尝试过使用多种laravel collections方法，但仍然无法获得所需的响应

[
  {
    "id": 1,
    "name": "kent",
    "receipts": [
      {
        "id": 1,
        "user_id": 1,
        "type_id": 1,
        "status": 0,
        "amount": 100
      },
      {
        "id": 2,
        "user_id": 1,
        "type_id": 1,
        "status": 0,
        "amount": 100
      },
      {
        "id": 3,
        "user_id": 1,
        "type_id": 2,
        "status": 1,
        "amount": 50
      },
      {
        "id": 4,
        "user_id": 1,
        "type_id": 2,
        "status": 0,
        "amount": 30
      },
      {
        "id": 5,
        "user_id": 1,
        "type_id": 2,
        "status": 0,
        "amount": 30
      },
      {
        "id": 6,
        "user_id": 1,
        "type_id": 1,
        "status": 0,
        "amount": 20
      },
      {
        "id": 7,
        "user_id": 1,
        "type_id": 1,
        "status": 1,
        "amount": 10
      }
    ]
  },
  {
    "id": 2,
    "name": "allison",
    "receipts": [
      {
        "id": 9,
        "user_id": 2,
        "type_id": 1,
        "status": 0,
        "amount": 20
      }
    ]
  }
]

我希望得到这个：

[
  {
    "id": 1,
    "name": "kent",
    "receipts": [
      {
        "performance and deleted": 220,
        "performance and not deleted": 10,
        "project and deleted": 60,
        "project and deleted": 50
      }
    ]
  },
  {
    "id": 2,
    "name": "allison",
    "receipts": [
      {
        "performance and deleted": 20,
        "performance and not deleted": 0,
        "project and deleted": 0,
        "project and not deleted": 0
      }
    ]
  }
]

---

我主要关心的是使用laravel收集方法和易于阅读的代码来获得预期的结果

我认为这会有所帮助

$user->receipts->sum('amount');

---

通过这种方式，我相信您可以通过可读性好的代码获得预期的结果

我假设您有$users变量，该变量包含集合中的用户列表，并且可能已经加载了收据

// You need to name the keys as you desire. (I can't figure out the labels by the type_id and status numbers).
$statusPair = collect([
    'performance and deleted' => [
        'type_id' => 1,
        'status'  => 0,
    ],
    'something'               => [
        'type_id' => 1,
        'status'  => 1,
    ],
    'something 2'             => [
        'type_id' => 2,
        'status'  => 0,
    ],
    'something 3'             => [
        'type_id' => 2,
        'status'  => 1,
    ],
]);

$data = $users->map(function ($user) use ($statusPair) {
    $receipts = $user->receipts;

    $sums = $statusPair->mapWithKeys(function ($pair, $statusLabel) use ($receipts) {
        $filtered = $receipts;
        foreach ($pair as $key => $value) {
            $filtered = $filtered->where($key, $value);
        }
        $sum = $filtered->sum('amount');

        return [
            $statusLabel => $sum,
        ];
    });

    return [
        'id'       => $user->id,
        'name'     => $user->name,
        'receipts' => $sums->toArray(),
    ];
});

---

请提供一些数据的数据库结构。我建议用户使用原始查询，而不是编写复杂的雄辩的方法，在那里您最终将使用用户原始！db数据是我上面的数据，高于预期值answer@VishalTarkar我怎样才能得到想要的结果？谢谢！foreach可以用each方法替换吗？什么是$pair和$statusLabel？foreach可以是每个，它们基本上是相同的。对于收据标签，如performance and deleted或performance and not deleted，您的要求并不明确，这就是$statusLabel在我的代码中的含义，因此我假设它们是由type_id和status对决定的。如果不是您想要的，您可以修改这些对。