无法使用静态访问访问 Flutter 实例成员“{0}" [英] Flutter Instance member ‘{0}’ can’t be accessed using static access

问题描述

我正在将变量从一个活动传递到另一个活动，但收到错误无法使用静态访问访问实例成员‘纬度’"，我需要在该块中对其进行转换，以便我可以将其分配给静态 URL.

I am passing variables from one activity to another in flutter but getting the error "Instance member ‘latitude’ can’t be accessed using static access" I need it converted in that block so I can assign it to a static URL.

class Xsecond extends StatefulWidget {
  final double latitude;
  final double longitude;
  Xsecond(this.latitude, this.longitude, {Key key}): super(key: key);

  @override
  _Xsecond createState() => _Xsecond();
}

class _Xsecond extends State<Xsecond> {
  static String lat = Xsecond.latitude.toString(); // Error: Instance member ‘latitude’ can’t be accessed using static access
  ...

关注

  ...
  String url = 'https://maps.googleapis.com/maps/api/place/nearbysearch/json?location=${lat},$lng&radius=$radius&type=restaurant&key=$api';
  ...

推荐答案

在您的代码中，纬度和经度都被定义为非静态的，即实例变量.这意味着它们只能使用类实例调用.

In your code both latitude and longitude are defined as non-static i.e. are instance variables. Which means they can only be called using a class instance.

class _Xsecond extends State<Xsecond> {
      final xsecond = Xsecond();
      static String lat = xsecond.latitude.toString();
      ...

请阅读任何面向对象编程语言的基础知识，例如Dart，Java，C++

Please read the basics of any Object Oriented Programming language e.g. Dart, java, C++

但是，在您的上下文中，第一类是您的 StatefullWidget.因此，您可以通过状态类的 widget 字段访问它.

However, in your context the first class is your StatefullWidget. So you can access that by the widget field of your state class.

修正:

class _Xsecond extends State<Xsecond> {
          static String lat = widget.latitude.toString();
          ...

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