Question

我遇到了这个问题

***> E/flutter (15532): [错误:flutter/lib/ui/ui_dart_state.cc(186)]

<块引用>

未处理的异常：FormatException：意外字符（在字符 1) E/flutter (15532):
E/flutter (15532): ^ E/flutter (15532): E/flutter (15532): #0 _ChunkedJsonParser.fail (dart:convert-patch/convert_patch.dart:1404:5) E/flutter (15532): #1
_ChunkedJsonParser.parseNumber (dart:convert-patch/convert_patch.dart:1271:9) E/flutter (15532): #2
_ChunkedJsonParser.parse (dart:convert-patch/convert_patch.dart:936:22) E/flutter (15532): #3
_parseJson (dart:convert-patch/convert_patch.dart:40:10) E/flutter (15532): #4 JsonDecoder.convert (dart:convert/json.dart:506:36) E/flutter (15532)：#5 JsonCodec.decode (dart:convert/json.dart:157:41) E/flutter (15532): #6 jsonDecode (dart:convert/json.dart:96:10) E/flutter (15532): #7
_LoginState.login (package:flutter_login/main.dart:35:18) E/flutter (15532): E/flutter (15532):***

这是我的代码

    void main() {
  HttpOverrides.global = new MyHttpOverrides();
  runApp(MaterialApp(
    home: login(),
  ));
}

class login extends StatefulWidget {
  @override
  _LoginState createState() => _LoginState();
}

class _LoginState extends State<login>{
  var email,password;
  final _key = new GlobalKey<FormState>();
  check(){
    final form = _key.currentState;
    if(form.validate()){
      form.save();
      login();
    }
  }

  login() async{
    final response = await http.post(Uri.parse("link"),
        body: {
      "email" : email,
      "password" : password
    });
    final data = jsonDecode(response.body);
    print(data);
  }

  bool _secureText = false;
  showHide(){
    setState(() {
      _secureText = !_secureText;
    });
  }

可以帮我吗？请

Answer 1

在我看来，您的 Uri 解析失败了，因为您已经通过了 "link" 这是一个无效的 uri。

这是来自Uri.parse docs：

<块引用>

如果 uri 字符串作为 URI 或 URI 引用无效，则抛出 FormatException。

Error Flutter (186)] 未处理的异常：FormatException：意外字符（在字符 1 处）

1 个答案: