Question

我希望在触摸坐标时显示弹出窗口。我正在使用Stack和Positioned小部件来放置弹出窗口。

Answer 1

您可以添加GestureDetector作为堆栈的父级，并注册onTapDownDetails侦听器。这应该在每个tapdown事件中调用您的侦听器，并在侦听器的TapDownDetails参数中使用tap的全局偏移量。

以下是展示相同内容的示例代码。

import 'package:flutter/material.dart';

void main() {
  runApp(new MyApp());
}

class MyApp extends StatelessWidget {
  @override
  Widget build(BuildContext context) {
    return new MaterialApp(
      title: 'Flutter Demo',
      home: new MyHomePage(),
    );
  }
}

class MyHomePage extends StatefulWidget {
  @override
  MyHomePageState createState() => new MyHomePageState();
}

class MyHomePageState extends State<MyHomePage> {
  @override
  Widget build(BuildContext context) {
    return new Scaffold(
        appBar: new AppBar(
          title: new Text('Popup Demo'),
        ),
        body: new MyWidget());
  }
}

class MyWidget extends StatefulWidget {
  @override
  State<StatefulWidget> createState() {
    return new MyWidgetState();
  }
}

class MyWidgetState extends State<MyWidget> {
  double posx = 100.0;
  double posy = 100.0;

  void onTapDown(BuildContext context, TapDownDetails details) {
    print('${details.globalPosition}');
    final RenderBox box = context.findRenderObject();
    final Offset localOffset = box.globalToLocal(details.globalPosition);
    setState(() {
      posx = localOffset.dx;
      posy = localOffset.dy;
    });
  }

  @override
  Widget build(BuildContext context) {
    return new GestureDetector(
      onTapDown: (TapDownDetails details) => onTapDown(context, details),
      child: new Stack(fit: StackFit.expand, children: <Widget>[
        // Hack to expand stack to fill all the space. There must be a better
        // way to do it.
        new Container(color: Colors.white),
        new Positioned(
          child: new Text('hello'),
          left: posx,
          top: posy,
        )
      ]),
    );
  }
}

如何计算颤动中的触摸坐标？

1 个答案: