我想获取来自其他Web服务,这样的城市的名单:
Future<List<City>> fetchCities() async {
final response =
await http.get('https://my-url/city',
headers: {HttpHeaders.acceptHeader: "application/json"});
if (response.statusCode == 200) {
// If the call to the server was successful, parse the JSON
return compute(parseCities, response.body);
} else {
// If that call was not successful, throw an error.
throw Exception('Failed to load cities');
}
}
然后解析:
List<City> parseCities(String responseBody) {
final parsed = json.decode(responseBody)['data']['children'].cast<Map<String, dynamic>>();
return parsed.map<City>((json) => City.fromJson(json['data'])).toList();
}
这是City
类的定义:
class City {
final String id;
final String name;
City({this.id, this.name});
factory City.fromJson(Map<String, dynamic> json) {
return City(
id: json['id'],
name: json['name'],
);
}
}
我的例子responseBody
是:
[{\"id\":\"599\",\"name\":\"Bia\u0142ystok-dev\",\"location\":{\"long\":\"23.15\",\"lat\":\"53.13\"}}]
(现在,我想ommit location
,只取ID和名称)。我json.decode
抛出异常:
类型“字符串”不是“索引”的类型“INT”的子类型
如何解决呢?有任何想法吗?
你已经发布的JSON字符串不包含data
或children
所以解析JSON你需要改变你的解析方法,下面的按键:
List<City> parseCities(String responseBody) {
final parsed = json.decode(responseBody).cast<Map<String, dynamic>>();
return parsed.map<City>((json) => City.fromJson(json)).toList();
}
虽然我认为这是使用qazxsw好的做法POI代替qazxsw POI
List.from