Question

我不知道我需要做什么，我试图更改我的 URL。但我仍然收到此错误。

这里是错误。

Exception has occurred.
FormatException (FormatException: Unexpected character (at character 1)
<!DOCTYPE html>
^
)

下面是我请求数据的代码

    Map<dynamic, dynamic> venuesData;

  Future<void> fetchAndSetVenues() async {
    try {
      final response = await http.get(Uri.parse(venuesURL));
      final extractedData = json.decode(response.body) as Map<dynamic, dynamic>;
      venuesData = extractedData;
      print(venuesData);
    } catch (error) {
      throw error;
    }
  }

  // events
  Map<dynamic, dynamic> eventsData;

  Future<void> fetchAndSetEvents() async {
    try {
      final response = await http.get(Uri.parse(eventsURL));
      final extractedData = json.decode(response.body) as Map<dynamic, dynamic>;
      eventsData = extractedData;
    } catch (error) {
      throw error;
    }
  }

这是我的网址

String eventsURL = 'http://medeni.art/api/v1/events?page=1'

String venuesURL = 'http://medeni.art/api/v1/venues'

Answer 1

此错误发生在 json.decode 方法中。无论出于何种原因，如果 http 请求出错，则不应调用 json.decode。您可以检查请求响应状态代码。

import 'package:http/http.dart' as http;

var url = Uri.parse('https://example.com/whatsit/create');
var response = await http.post(url, body: {'name': 'doodle', 'color': 'blue'});
if(response.statusCode ==200){
  final result = json.decode(response.body);
}

从http请求数据时出错

1 个答案: