Question

需要帮助解决此问题，我试图学习颤振并在运行测试apk时被卡住。它只是显示错误未处理的异常：类型'String'不是类型'int'的子类型。我尝试过但是没有运气这是我的代码

class Login {
  final String email;
  final String password;

  Login(this.email,this.password);

  Future<bool> getData(BuildContext context) async {
    bool isSuccessed;
    int id;
    int rule;
    String image;


Response response = await get(InfixApi.login(email, password));
if (response.statusCode == 200) {
  Map<String, dynamic> user = jsonDecode(response.body) as Map;
  isSuccessed = user['success'];
  id = user['data']['user']['id'];
  rule = user['data']['user']['role_id'];
  image = user['data']['userDetails']['staff_photo'];

  if (isSuccessed) {
    saveBooleanValue('isLogged', isSuccessed);
    saveStringValue('email', email);
    saveStringValue('password', password);
    saveStringValue('id','$id');
    saveStringValue('rule', '$rule');
    saveStringValue('image', '$image');
    saveStringValue('lang', 'en');
    AppFunction.getFunctions(context,rule.toString());
  }

}
return isSuccessed;


 }



 Future<bool> saveBooleanValue(String key ,bool value) async {
    final SharedPreferences prefs = await SharedPreferences.getInstance();

    return prefs.setBool(key, value);
  }

  Future<bool> saveStringValue(String key ,String value) async {
    final SharedPreferences prefs = await SharedPreferences.getInstance();

    return prefs.setString(key, value);
  }

}

Answer 1

我认为问题来了，因为您试图将ID和规则作为字符串存储在int变量中，请尝试替换这两行：

id = user['data']['user']['id'];
rule = user['data']['user']['role_id'];

id = int.parse(user['data']['user']['id']);
rule = int.parse(user['data']['user']['role_id']);

希望有帮助

未处理的异常：类型'String'不是类型'int'的子类型

1 个答案: