Flutter：将 JSON 映射到对象列表中返回 null

Posted 2023-02-23

【中文标题】Flutter：将 JSON 映射到对象列表中返回 null

【英文标题】：Flutter: Mapping JSON into List of Objects returns null

【发布时间】：2020-12-20 19:17:18

【问题描述】：

我有一个 json 结构

  "message": "",
  "status" : "complete",
  "results": [
      
        "id": 66,
        "user": 
            "id": 80,
            "email": "[e-mail]"
        ,
          "order": 
              "id": 1,
              "customer": 2,
              "receiver_name": "Something",
              "receiver_phone_number": "464"
        ,
          "order": 
              "id": 1,
              "customer": 2,
              "receiver_name": "Something",
              "receiver_phone_number": "464"
          ,
          "pickedup_time": "sksjkns",
          "delivered_time": "hkvjsfsf"
      
    ]
  

我正在以这种方式解析 json，并尝试将其映射到对象列表中

Future<void> get() async
  res = await http.get(url);
  var json = jsonDecode(res.body);
  objectsJson = json['results'] as List;
  objects = List<Object>.from(objectsJson.map((i) => Object.fromMap(i))).toList();

其中 res,objectsJson,objects 是声明为的全局变量

var res;
List<Object> objects;
var objectsJson;

这是我使用 fromMap 方法创建类的 Object 类

class Object
  int id, customer;
  String 
      receiverName,
      receiverPhoneNumber,
      pickedUpTime,
      deliveredTime;

  Object(
        
        int id,
          int customer,
          String receiverName,
          String receiverPhoneNumber,
          String pickedUpTime,
          String deliveredTime
        
        );


  factory Object.fromMap(Map<String, dynamic> json) 

    print(json['order']['id'].toString()+
        json['order']['customer'].toString()+
        json['order']['receiver_name'].toString()+
        json['order']['receiver_phone_number'].toString()+
        json['pickedup_time'].toString()+
        json['delivered_time'].toString()
        );

    return new Object(
        id: json['order']['id'],
        customer: json['order']['customer'],
        receiverName: json['order']['receiver_name'],
        receiverPhoneNumber: json['order']['receiver_phone_number'],
        pickedUpTime: json['pickedup_time'],
        deliveredTime: json['delivered_time']
        );
  

如您所见，我在创建对象之前打印了这些值，并且它显示正确。 但是当我尝试从 futurebuilder 的对象列表中访问对象时，我在整个过程中都得到了空值。

FutureBuilder(
  future: get(),
  build: (BuildContext context, AsyncSnapshot snapshot)
    if(snapshot.connectionState == ConnectionState.done)
      print(objects[0].id);
      print(objects[0].customer); 
      print(objects[0].receiverName);
      print(objects[0].receiverPhoneNumber);
      print(objects[0].pickedUpTime);
      print(objects[0].receivdeliveredTimeerName);
      return Container();
    
    return Container();
  

我已经打印出 json 并且它是完整的，当我在返回类中的对象之前打印出值时，可以通过不获取 null 而是实际值来支持它。

我尝试在映射后打印出对象，这就是我得到的

['Instance of Object']

我还尝试将 .toString() 添加到返回对象的字符串中。例如：

receiverName: json['order']['receiver_name'].toString()

我不知道 null 是从哪里来的。

【问题讨论】：

不要将你的类命名为 'Object' 。对象类是每个类都隐式继承的***类

我没有把它命名为对象。我只是将其更改为反对在此处发布。

【参考方案1】：

很有趣，我刚刚找到了解决方案。 我将我的 fromJson 函数更改为使用 this.variableName 来分配值

Object.fromJson(Map json) 
      this.id=json['order']['id'];
      this.customer=json['order']['customer'];
      this.receiverName=json['order']['receiver_name'].toString();
      this.receiverPhoneNumber=json['order']['receiver_phone_number'].toString();
      this.pickedUpTime=json['pickedup_time'].toString();
      this.deliveredTime=json['delivered_time'].toString();
  

【参考方案2】：

试试

在你的State

Future<void> get() async
  res = await http.get(url);
  var json = jsonDecode(res.body);
  objectsJson = json['results'] as List;
  setState(()
    objects = List<Object>.from(objectsJson.map((i) => Object.fromMap(i))).toList(););


List<Object> objects = [];


@override
initState()
 get();

那么您可以使用Builder 代替FutureBuilder

Builder(
 builder: (context)
   if(objects.isEmpty)
      return Container();
   
   print(objects[0].id);
   print(objects[0].customer); 
   print(objects[0].receiverName);
   print(objects[0].receiverPhoneNumber);
   print(objects[0].pickedUpTime);
   print(objects[0].receivdeliveredTimeerName);
   return Container();
 
)

【讨论】：

我也试过这个。这与更新 UI 无关，而是对象中的实际值，