Php Laravel中的通配符Undot数组

在Laravel中，我执行了一个JSON映射，其中我需要将JSON转换为不同的模式.为此，我遵循以下步骤:

1. 用Arr::dot helper点array.这样我就得到了这个结果，例如:

array(11) {
  ["0.identificador"]=>
  int(1)
  ["0.nome_completo"]=>
  string(8) "John Doe"
  ["0.email"]=>
  string(11) "j@email.com"
  ["0.empresa.razao_social"]=>
  string(11) "ABC Company"
  ["0.empresa.endereco"]=>
  string(11) "123 Main St"
  ["0.pedidos.0.SKU"]=>
  int(1)
  ["0.pedidos.0.descricao"]=>
  string(5) "Shoes"
  ["0.pedidos.0.qtd"]=>
  int(2)
  ["0.pedidos.1.SKU"]=>
  int(2)
  ["0.pedidos.1.descricao"]=>
  string(5) "Shirt"
  ["0.pedidos.1.qtd"]=>
  int(1)
}

2. 迭代通过点数组，并与我的键映射进行比较.如果它们匹配，我用映射键设置值.

我的映射看起来像这样:

array(8) {
    ["*.identificador"]=> string(4) "*.id"
    ["*.nome_completo"]=> string(6) "*.name"
    ["*.empresa"]=> string(9) "*.company"
    ["*.empresa.razao_social"]=> string(14) "*.company.name"
    ["*.empresa.endereco"]=> string(17) "*.company.address"
    ["*.pedidos.*.SKU"]=> string(13) "*.orders.*.id"
    ["*.pedidos.*.descricao"]=> string(18) "*.orders.*.product"
    ["*.pedidos.*.qtd"]=> string(19) "*.orders.*.quantity"
  }

这就是我的代码到目前为止的样子:

$payload = [
    [
        "identificador" => 1,
        "nome_completo" => "John Doe",
        "email" => "j@email.com",
        "empresa" => [
            "razao_social" => "ABC Company",
            "endereco" => "123 Main St"
        ],
        "pedidos" => [
            [
                "SKU" => 1,
                "descricao" => "Shoes",
                "qtd" => 2
            ],
            [
                "SKU" => 2,
                "descricao" => "Shirt",
                "qtd" => 1
            ]
        ]
    ]
];

$dottedPayload = Arr::dot($payload);

$transformedPayload = [];
foreach ($dottedPayload as $key => $value) {
    $newKey = preg_replace('/\d+./', '*.', $key);
    $newKey = preg_replace('/\.\d+\./', '.*.', $newKey);
    $newKey = preg_replace('/\.\d+$/', '.*', $newKey);

    $mappedKey = $mappingConfig[$newKey] ?? $key;

    data_fill($transformedPayload, $mappedKey, $value);
}

我的输出是这样的:

array(2) {
  ["*"]=>
  array(4) {
    ["id"]=>
    int(1)
    ["name"]=>
    string(8) "John Doe"
    ["company"]=>
    array(2) {
      ["name"]=>
      string(11) "ABC Company"
      ["address"]=>
      string(11) "123 Main St"
    }
    ["orders"]=>
    array(1) {
      ["*"]=>
      array(3) {
        ["id"]=>
        int(2)
        ["product"]=>
        string(5) "Shirt"
        ["quantity"]=>
        int(1)
      }
    }
  }
  [0]=>
  array(1) {
    ["email"]=>
    string(11) "j@email.com"
  }
}

但这是预期的:

$expectedResult = [
    [
        "id" => 1,
        "name" => "John Doe",
        "email" => "j@email.com",
        "company" => [
            "name" => "ABC Company",
            "address" => "123 Main St"
        ],
        "orders" => [
            [
                "id" => 1,
                "product" => "Shoes",
                "quantity" => 2
            ],
            [
                "id" => 2,
                "product" => "Shirt",
                "quantity" => 1
            ]
        ]
    ]
];

如何在映射后正确取消数组的点，使用通配符，正确设置索引？任何其他建议来执行这个JSON映射也是受欢迎的.