包含 List<Object> 的对象的映射用于 sqlite [英] Map of object containing a List&lt;Object&gt; for sqlite

问题描述

我正在设置我的模型类以确认 sqflite 的文档，其中建议包含一个命名构造函数来转换到/从 Maps 以更好地处理类和数据库之间的数据.我能找到的每个例子都很简单，类属性都是简单的数据类型.

I am setting up my model classes to confirm to the docs for sqflite which suggest including a named constructor to convert to/from Maps to better handling of data between the classes and the DB. Every example I can find is very simple, with class properties all being simple data types.

使用下面显示的构造函数和方法，在处理这样的类时，转换到/从 Map 非常简单.

Using the constructor and method shown below, converting to/from Map is quite simple when dealing with a class such as this.

class Human{
  final String name;
  final String height;
  Final String weight;

  Human({this.name, this.height, this.weight});
}

然而，当你有一个字段有点复杂的类时，我不明白如何在命名构造函数和 xxx 方法中构造事物以返回我相信"我应该得到的数据映射.

However, when you have a class where one of the fields is a bit more complex, I do not understand how to structure things within the named constructor and xxx method to return the map of data that I 'believe' I should get.

class Human{
      final String name;
      final String height;
      Final String weight;
      List<Child> children = [];

      Human({this.name, this.height, this.weight, this.children});
    }

Human({this.name, this.height, this.weight, this.children});

  Human.fromMap(Map<String, dynamic> map)
    : name = map['name'],
      height = map['height'],
      weight = map['weight'],
      children = map['children'];

  Map<String, dynamic> toMap() {
   return {
     'name': name,
     'height': height,
     'weight': weight,
     'children': children,
   }; 
  }

列表孩子是我正在努力解决的部分.我相信您必须将每个 Child 对象也转换为父地图中的地图，但我在这里输了.

The List children is the part I am struggling with. I believe you have to get each Child object ALSO converted to a map within the parent map, but am losing the battle here.

我的方法在这儿吗?我应该使用其他方法来完成此操作吗?

Is my approach way off here? Is there some other method I should be using to accomplish this?

如有任何帮助，我们将不胜感激.

Any assistance would be much appreciated.

推荐答案

这里我解释一下

1. 如何将模型对象转换为 Map 以与 sqlite 一起使用
2. 如何将 Map 对象从 sqlite 转换为模型类.
3. 如何在 Flutter 中正确解析 JSON 响应
4. 如何将模型对象转换为 JSON

以上所有问题都有相同的答案.Dart 对这些操作有很大的支持.这里我会用一个详细的例子来说明.

All of the above questions has same answer. Dart has great support for these operations. Here I am going to illustrate it with a detailed example.

class DoctorList{
  final List<Doctor> doctorList;

  DoctorList({this.doctorList});

  factory DoctorList.fromMap(Map<String, dynamic> json) {
    return DoctorList(
      doctorList: json['doctorList'] != null
          ? (json['doctorList'] as List).map((i) => Doctor.fromJson(i)).toList()
          : null,
    );
  }

  Map<String, dynamic> toMap() {
    final Map<String, dynamic> data = Map<String, dynamic>();
    if (this.doctorList != null) {
      data['doctorList'] = this.doctorList.map((v) => v.toMap()).toList();
    }
    return data;
  }
}

上面的 DoctorList 类有一个成员，它包含一个 'Doctor' 对象列表..

The above DoctorList class has a member which holds a list of 'Doctor' objects..

看看我是如何解析医生列表的.

And see how I parsed the doctorList.

 doctorList: json['doctorList'] != null
      ? (json['doctorList'] as List).map((i) => Doctor.fromMap(i)).toList()
      : null,

您可能想知道 Doctor 类会是什么样子.给你

You may wonder, how the Doctor class may look like. Here you go

class Doctor {
  final String doCode;
  final String doctorName;

  Doctor({this.doCode, this.doctorName});

  factory Doctor.fromMap(Map<String, dynamic> json) {
    return Doctor(
      doCode: json['doCode'],
      doctorName: json['doctorName'],
    );
  }

  Map<String, dynamic> toMap() {
    final Map<String, dynamic> data = Map<String, dynamic>();
    data['doCode'] = this.doCode;
    data['doctorName'] = this.doctorName;
    return data;
  }

}

<小时>

仅此而已.希望你有这个想法.干杯！

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That's all. Hope you got the idea. Cheers!

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