Question

这是JSON响应，我需要将数据检索到混乱状态。 在投射详细信息时出错。

{code": 200,
    "message": "Success",
    "data": [
        {

           "fname": "LG",
            "lname": "Electronics",
            "email": "lg@gmail.com",
            "address": "Girish Cold Drinks, Chimanlal Girdharlal Road, Mithakhali, Navrangpura, Ahmedabad, Gujarat, India",

        }
    }

这是客户类别

class Customer {
  int code;
  String message;
  List<Data> data;
  Customer({this.code, this.message, this.data});
  Customer.fromJson(Map<String, dynamic> json) {
    code = json['code'];
    message = json['message'];
    if (json['data'] != null) {
      data = new List<Data>();
      json['data'].forEach((v) {
        data.add(new Data.fromJson(v));
      });
    }
  }
  Map<String, dynamic> toJson() {
    final Map<String, dynamic> data = new Map<String, dynamic>();
    data['code'] = this.code;
    data['message'] = this.message;
    if (this.data != null) {
      data['data'] = this.data.map((v) => v.toJson()).toList();
    }
    return data;
  }
}



  class Data {
      String fname;
      String lname;
      String email;
      String address;

    Data(
          {
            this.fname,
            this.lname,
            this.email,
            this.address,
            });





 Data.fromJson(Map<String, dynamic> json) {
        fname = json['fname'];
        lname = json['lname'];
        email = json['email'];
        address = json['address'];
      }

   Map<String, dynamic> toJson() {
    final Map<String, dynamic> data = new Map<String, dynamic>();

    data['fname'] = this.fname;
    data['lname'] = this.lname;
    data['email'] = this.email;
    data['address'] = this.address;

    return data;
  }
}

我在这里遇到投放错误

final jsonResponse = json.decode(response.body).cast<Map<String, dynamic>>();
      List<Customer> listOfUsers = jsonResponse.map<Customer>((json) async {
        return  Customer.fromJson(json);
      }).toList()

错误：Flutter：InternalLinkedHashMap'没有具有匹配参数的实例方法'cast'

Answer 1

您可以使用Customer.fromJson(json.decode(response.body))。

对于任何复杂的JSON解析，请使用quicktype.io。

更新：您的pojo课堂将是这样的：

import 'dart:convert';
Customer customerFromJson(String str) => Customer.fromJson(json.decode(str));
String customerToJson(Customer data) => json.encode(data.toJson());
class Customer {
    int code;
    String message;
    List<Data> data;
    Customer({
        this.code,
        this.message,
        this.data,
    });
    factory Customer.fromJson(Map<String, dynamic> json) => Customer(
        code: json["code"],
        message: json["message"],
        data: List<Data>.from(json["data"].map((x) => Data.fromJson(x))),
    );
    Map<String, dynamic> toJson() => {
        "code": code,
        "message": message,
        "data": List<dynamic>.from(data.map((x) => x.toJson())),
    };
}
class Data {
    String fname;
    String lname;
    String email;
    String address;
    Data({
        this.fname,
        this.lname,
        this.email,
        this.address,
    });
    factory Data.fromJson(Map<String, dynamic> json) => Data(
        fname: json["fname"],
        lname: json["lname"],
        email: json["email"],
        address: json["address"],
    );
    Map<String, dynamic> toJson() => {
        "fname": fname,
        "lname": lname,
        "email": email,
        "address": address,
    };
}

Answer 2

在Customer中尝试此操作，

   class Customer {
      int code;
      String message;
      List<Data> data;
      Customer({this.code, this.message, this.data});
      factory  Customer.fromJson(Map<String, dynamic> json) {
        code = json['code'];
        message = json['message'];
         var list = json['data'] as List;
        List<Data> dataList =
            list != null ? list.map((i) => Data.fromJson(i)).toList() : [];

      retrun Customer(code:code,message,data:dataList)
      }


      Map<String, dynamic> toJson() {
        final Map<String, dynamic> data = new Map<String, dynamic>();
        data['code'] = this.code;
        data['message'] = this.message;
        if (this.data != null) {
          data['data'] = this.data;
        }
        return data;
      }
}

Data正在关注

class Data {
      String fname;
      String lname;
      String email;
      String address;

    Data({
           this.fname,
            this.lname,
            this.email,
            this.address,
            });

 factory Data.fromJson(Map<String, dynamic> json) {
        fname = json['fname'];
        lname = json['lname'];
        email = json['email'];
        address = json['address'];
        return Data(fname:fname,lname:lname,email:email,address:address);
      }

   Map<String, dynamic> toJson() {
    final Map<String, dynamic> data = new Map<String, dynamic>();

    data['fname'] = this.fname;
    data['lname'] = this.lname;
    data['email'] = this.email;
    data['address'] = this.address;

    return data;
  }
}

Flutter：InternalLinkedHashMap'没有具有匹配参数的实例方法“ cast”

2 个答案: