是否可以在 Flutter/Dart 中解析缺少属性的 json？

Question

我有来自 API 的 json 形式的对象列表。这些对象有一些属性，问题是一些对象缺少属性。那么如何正确解析其他属性的数据呢？

例如我有这个缺少paymentType的属性

"user":{
"id": 6,
"name": "User",
"email": "",
"phone": "123",
"credit": null,
"fireBaseId": null,
"created_at": "2021-10-28T10:28:34.000000Z",
"updated_at": "2021-10-29T22:10:09.000000Z"
},
"paymentType": null,
"booking_type":{
"id": 1,
"name": "Mobile",
"price": 0,
"created_at": null,
"updated_at": null
}

但这所有属性都存在：

"user":{
"id": 6,
"name": "User",
"email": "",
"phone": "",
"credit": null,
"fireBaseId": null,
"created_at": "2021-10-28T10:28:34.000000Z",
"updated_at": "2021-10-29T22:10:09.000000Z"
},
"paymentType":{
"id": 1,
"name": "Cash",
"active": 1,
"created_at": "2021-10-29T00:36:06.000000Z",
"updated_at": "2021-10-29T17:24:42.000000Z"
},
"booking_type":{
"id": 2,
"name": "Parking",
"price": 10,
"created_at": null,
"updated_at": null
}

Answer 1

如果您觉得某些属性可能为空，请为其添加空安全，然后使用 if() 条件将映射中的值解析为对象的属性。

这是一种可能的方式。它可能不是理想的，但效果很好！

class User {
  int? id;
  String? user;
  String? email;
  String? phone;
  String? credit;

  User.fromMap(Map<String, dynamic> map) {
    if (map.containsKey('id')) {
      this.id = int.parse(map['id']);
    }
    if (map.containsKey('user')) {
      this.user = "${map['user']}";
    }
    if (map.containsKey('email')) {
      this.email = "${map['email']}";
    }
    if (map.containsKey('phone')) {
      this.phone = "${map['phone']}";
    }
    if (map.containsKey('credit')) {
      this.credit = "${map['credit']}";
    }
  }
}

Answer 2

您可以为 paymentType 创建特殊的对象类。试试看吧。

创建 payment.dart

class PaymentType {
  int id;
  String name;      
  int active;
  DateTime createdAt;
  DateTime updatedAt;

  PaymentType({
    this.id,
    this.name,
    this.active,
    this.createdAt,
    this.updatedAt,
  }) ;

  factory PaymentType.fromJson(Map<String, dynamic> json) =>PaymentType(
        id: json["id"],
        name: json["name"],
        active: json["active"],
        createdAt: DateTime.parse(json["created_at"]),
        updatedAt: DateTime.parse(json["updated_at"]),
      );
}

并在解析时这样使用它：

......
"paymentType" : json["payment_type"] != null ? PaymentType.fromJson(json["payment_type"]) : null;
......

如果您有任何疑问，请随意:)