如何计算Flutter中的触摸坐标？

Question

我想在触摸坐标处显示一个弹出窗口。我正在使用 Stack 和 Positioned 小部件来放置弹出窗口。

Answer 1

您可以为您的侦听器添加一个 GestureDetector as parent of stack, and register onTapDownDetails listener. This should call your listener on every tapdown event, with global offset of the tap in TapDownDetails 参数。

下面是演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Flutter中的触摸坐标？

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