如何在颤振中解析复杂的json

Question

我无法从复杂的 json 中获取数据，下面是请求中的 json。

{
   "results":{
      "TotalRecordCount":"1",
      "Records":[
         {
            "code":"PCK_34333338365C93E2D50DB9C",
            "address":"1 AV KHEIREDDINE PACHA Imm Pacha centre BLOC B tunis Tunis 1000",
            "contact_phone":"99608258"
         }
      ],
      "Result":"OK"
   }
}

下面是我制作的模型。

import 'dart:convert';

class Pickup {
  String status;
  List message;
  //Map<String ,dynamic> results;
  Results results;
  Pickup(
    {this.status,
     this.message,
     this.results,
  });
  factory Pickup.fromJson(Map<String, dynamic> json) {
    return Pickup(
             status: json["status"] as String,
             results: Results.fromJson(json["results"]),

           );
  }
}

class Results {
  String TotalRecordCount;
  records Records;

  Results({this.TotalRecordCount,this.Records});

  factory Results.fromJson(Map<String, dynamic> json) {
    return Results(
    TotalRecordCount: json["TotalRecordCount"],
    Records:records.fromJson(json["Records"]),

    );
  }
}

class records{
  String code;
  String address;
  String contact_phone;

  records({
    this.code,
    this.address,
    this.contact_phone
  });

  factory records.fromJson(Map<String, dynamic> json) {
    return records(
      code: json["code"],
      address: json["address"],
      contact_phone: json["contact_phone"],
    );
  }
}

现在我正在尝试解析记录以获取代码或地址并打印出来。

if (response.statusCode == 200) {
  print(response.body);
  final responseJson = json.decode(response.body);
  var da=Pickup.fromJson(responseJson);
  Results dat=da.results;
  records data=dat.Records;
  print(data.address);
}

response.body 工作正常，但是当我尝试解析结果或记录时，我得到 'List' is not a subtype of type 'Map' 错误

Answer 1

{
   "results":{
      "TotalRecordCount":"1",
      "Records":[
         {
            "code":"PCK_34333338365C93E2D50DB9C",
            "address":"1 AV KHEIREDDINE PACHA Imm Pacha centre BLOC B tunis Tunis 1000",
            "contact_phone":"99608258"
         }
      ],
      "Result":"OK"
   }
}

上面的 JSON 文件包含一个“记录”列表，因此它是一个复杂的 JSON。以下是我想对代码做出贡献的修改 -

class Pickup {
  String status;
  List message;
  //Map<String ,dynamic> results;
  Results results;
  Pickup(
    {this.status,
     this.message,
     this.results,
  });
  factory Pickup.fromJson(Map<String, dynamic> json) {
    return Pickup(
             status: json["status"] as String,
             results: Results.fromJson(json["results"]),

           );
  }
}
class Results {
  String TotalRecordCount;
  records Records;

  Results({this.TotalRecordCount,this.Records});



    factory Results.fromJson(Map<String, dynamic> json) {
        //since records is a list hence it is parsed separately by creating a 
        //list out of it.  
    



       listOfRecords = (json["Records"] as List)
      .map((i) => records.fromJson(i))
      .toList();


 
    return Results(
    TotalRecordCount: json["TotalRecordCount"],
    Records:listOfRecords //now the list is used here directly
),

    );
  }
}

class records{
  String code;
  String address;
  String contact_phone;

  records({
    this.code,
    this.address,
    this.contact_phone
  });

  factory records.fromJson(Map<String, dynamic> json) {
    return records(
      code: json["code"],
      address: json["address"],
      contact_phone: json["contact_phone"],
    );
  }

最后，我们可以在任何需要的地方使用它——

if (response.statusCode == 200) {
  print(response.body);
  final responseJson = json.decode(response.body);
  var da = Pickup.fromJson(responseJson);
  Results dat = da.results;
  records data = dat.Records;
  print(data[0].address);  //since records was a list
}

Answer 2

我们将解析一个复杂的 JSON，其中包含一些字段和一个对象数组字段。它看起来像这样：

{
  "title": "Dart Tutorial",
  "description": "Way to parse Json",
  "author": {
    "name": "bezkoder",
    "age": 30
  },
  "tags": [
    {
      "name": "dart",
      "quantity": 12
    },
    {
      "name": "flutter",
      "quantity": 25
    }
  ]
}

我们修改了 Tutorial 类，使它有一个新的 List<Tag> 标签字段。

class Tutorial {
  String title;
  String description;
  User author;
  List<Tag> tags;

  Tutorial(this.title, this.description, this.author, [this.tags]);

  factory Tutorial.fromJson(dynamic json) {
    if (json['tags'] != null) {
      var tagObjsJson = json['tags'] as List;
      List<Tag> _tags = tagObjsJson.map((tagJson) => Tag.fromJson(tagJson)).toList();

      return Tutorial(
        json['title'] as String,
        json['description'] as String,
        User.fromJson(json['author']),
        _tags
      );
    } else {
      return Tutorial(
        json['title'] as String,
        json['description'] as String,
        User.fromJson(json['author'])
      );
    }
  }

  @override
  String toString() {
    return '{ ${this.title}, ${this.description}, ${this.author}, ${this.tags} }';
  }
}

– 在构造方法中，我们使用方括号[this.tags]来指定这个tags数组是一个选项字段。

– 就像我们将 JSON 对象数组解析成 Dart List 的方式一样，我们使用 map() 和 toList() 来获取 List<Tag>。

List<Tag> _tags = tagObjsJson.map((tagJson) => Tag.fromJson(tagJson)).toList();

main() 函数现在看起来像这样。

import 'dart:convert';

main() {
  String complexText =
      '{"title": "Dart Tutorial", "description": "Way to parse Json", "author": {"name": "bezkoder", "age": 30}, "tags": [{"name": "dart", "quantity": 12}, {"name": "flutter", "quantity": 25}]}';

  Tutorial complexTutorial = Tutorial.fromJson(jsonDecode(complexText));

  print(complexTutorial);
}

运行上面的代码后，你可以看到带有标题、描述、作者、标签数组的结果。

{ Dart Tutorial, Way to parse Json, { bezkoder, 30 }, [{ dart, 12 }, { flutter, 25 }] }

基于：parse complex JSON

Answer 3

我一定会向你推荐这个网站 json to dart App Quicktype 只是不要忘记在右侧选择 Dart。

你只需将你的 json 放在 la left 中，in 就会给你这样的东西：

// To parse this JSON data, do
//
//     final pickUp = pickUpFromJson(jsonString);

import 'dart:convert';

PickUp pickUpFromJson(String str) => PickUp.fromJson(json.decode(str));

String pickUpToJson(PickUp data) => json.encode(data.toJson());

class PickUp {
    Results results;

    PickUp({
        this.results,
    });

    factory PickUp.fromJson(Map<String, dynamic> json) => new PickUp(
        results: Results.fromJson(json["results"]),
    );

    Map<String, dynamic> toJson() => {
        "results": results.toJson(),
    };
}

class Results {
    String totalRecordCount;
    List<Record> records;
    String result;

    Results({
        this.totalRecordCount,
        this.records,
        this.result,
    });

    factory Results.fromJson(Map<String, dynamic> json) => new Results(
        totalRecordCount: json["TotalRecordCount"],
        records: new List<Record>.from(json["Records"].map((x) => Record.fromJson(x))),
        result: json["Result"],
    );

    Map<String, dynamic> toJson() => {
        "TotalRecordCount": totalRecordCount,
        "Records": new List<dynamic>.from(records.map((x) => x.toJson())),
        "Result": result,
    };
}

class Record {
    String code;
    String address;
    String contactPhone;

    Record({
        this.code,
        this.address,
        this.contactPhone,
    });

    factory Record.fromJson(Map<String, dynamic> json) => new Record(
        code: json["code"],
        address: json["address"],
        contactPhone: json["contact_phone"],
    );

    Map<String, dynamic> toJson() => {
        "code": code,
        "address": address,
        "contact_phone": contactPhone,
    };
}

一开始它会告诉你如何使用它。

// To parse this JSON data, do
//
//     final pickUp = pickUpFromJson(jsonString);

所以当你在你的代码中调用它时，它会是这样的。

  Future<Pickup> getPickup() async {
    var response = await http.get(url);
    return pickUpFromJson(response.body);
  }

例如，此代码可以调用 FutureBuilder 或您将代码设置为等待未来的任何位置。

Answer 4

试试这个：https://flutter.dev/docs/development/data-and-backend/json#serializing-json-using-code-generation-libraries

具体使用json_serializable的部分