Dart -如何搜索JSON？

Question

我有这样的JSON代码：

final String response = await rootBundle.loadString('assets/Schools.json');
List data = await json.decode(response);
print(data);

/* output:
[{ "city": "ISTANBUL", "districty": "Kagithane", "name": "Kagithane anadolu lisesi"}, { "city": "ISTANBUL", "districty": "Sisli", "name": "Aziz Sancar anadolu lisesi"}, { "city": "IZMIR", "districty": "Adalar", "name": "Kemal Sunal anadolu lisesi"}, { "city": "ISTANBUL", "districty": "Bagcilar", "name": "Bagcilar Fen lisesi"}, { "city": "ISTANBUL", "districty": "Kagithane", "name": "Kagithane Meslek lisesi"}]
*/

艾还写了这样一个模型：

List <School> schools = [];

List<School> allSchools() {
  return schools;
}

class School {
  String city;
  String districty;
  String name;
  School({required this.city, required this.name, required this.districty});
}

如何将数据从JSON导出到list？所以我想这样传递：

List <School> schools = [
  School(city: "ISTANBUL", districty: "Kagithane", name: "Kagithane anadolu lisesi"),
  School(city: "ISTANBUL", districty: "Sisli", name: "Aziz Sancar anadolu lisesi"),
  School(city: "IZMIR", districty: "ADALAR", name: "Kemal Sunal anadolu lisesi"),
  School(city: "ISTANBUL", districty: "BAGCILAR", name: "Bagcilar Fen lisesi"),
  School(city: "ISTANBUL", districty: "Kagithane", name: "Kagithane Meslek lisesi")
];

我感谢你提前帮忙，谢谢。

Answer 1

正如@Spanching所说，工厂使用工厂是最好的方式，因为它是简单和有效的使用。

但是，由于您的问题包含了如何搜索特定的学校，我将在这里添加一些额外的提示。

首先，创建一个工厂构造函数：

class School {
  String city;
  String districty;
  String name;
  School({required this.city, required this.name, required this.districty});

  factory School.fromJson(Map<String, dynamic> json) {
    return School(
      city: json['city'],
      name: json['name'],
      districty: json['district'],
    );
  }
}

现在创建另一所班级来轻松地处理数据：

class Schools {
  List<School> list;
  Schools({required this.list});

  factory Schools.fromJson(List<dynamic> json) {
    return Schools(
      list: json.map((e) => School.fromJson(e)).toList(),
    );
  }
}

添加getter以获取类的实例：

static Future<Schools> get instance async {
    final String response = await rootBundle.loadString('assets/Schools.json');
    List data = await json.decode(response);
    return Schools.fromJson(data);
  }

根据需要添加方法。例如，如果您想按城市搜索一所学校，可以添加如下方法：

List<School> searchByCity(String city) {
    return list.where((element) => element.city == city).toList();
  }

现在，您的主程序甚至可以阅读如下：

  final schools = await Schools.instance;
  final schoolList = schools.searchByCity('ISTANBUL');
  print(schoolList);

那是很多代码，但我希望它能帮到你。如果您想了解整个代码，可以使用在这里查一下。

Answer 2

您可以使用List.map()方法并传递一个函数，该函数从列表中的每个JSON对象创建一个School：

List data = json.decode(response);
schools = data.map((schoolData) => School(
  city: schoolData['city'],
  districty: schoolData['districty'],
  name: schoolData['name']
)).toList();

Answer 3

你可以在你学校的班级里用这样的工厂：

factory School.fromJson(Map<String, dynamic> json) {
  return School(city: json["city"], districty: json["districty"], name: json["name"]);
}

然后从您的json中映射列表：

var schools = data.map((school) => School.fromJson(school)).toList();