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在这里，我试图在 webview 中加载深层链接首先在 google 中输入深层链接并将其重定向到应用程序并在 webview 中加载深层链接。如果我把静态 URL 像 var temp = "www.google.com" 然后使用 temp var 作为 webview 中的初始 URL，它将加载谷歌应用内。但我试图将深层链接 var 复制到字符串 var 中，并在 webview 中使用 temp var 作为初始 URL，它不会在 webview 中加载动态链接。甚至我将深层链接 var 转换为字符串（将 URI 类型转换为字符串）

import 'dart:async';
import 'dart:convert';
import 'dart:io';

import 'package:flutter/material.dart';
import 'package:firebase_dynamic_links/firebase_dynamic_links.dart';
import 'package:webview_flutter/webview_flutter.dart';

void main() {
  runApp(MaterialApp(
    title: 'Dynamic Links Example',
    routes: <String, WidgetBuilder>{
      '/': (BuildContext context) => _MainScreen(),
    },
  ));
}

class _MainScreen extends StatefulWidget {
  @override
  State<StatefulWidget> createState() => _MainScreenState();
}

class _MainScreenState extends State<_MainScreen> {

  String temp;

// String temp = "www.google.com"; // 这将作为 webview 中的初始 url 正常工作

  @override
  void initState() {
    initDynamicLinks();
    super.initState();
  }

  Future initDynamicLinks() async {
    // 1. Get the initial dynamic link if the app is opened with a dynamic link
    final PendingDynamicLinkData data =
        await FirebaseDynamicLinks.instance.getInitialLink();

    // 2. handle link that has been retrieved
    _handleDeepLink(data);

    // 3. Register a link callback to fire if the app is opened up from the background
    // using a dynamic link.
    FirebaseDynamicLinks.instance.onLink(
        onSuccess: (PendingDynamicLinkData dynamicLink) async {
      // 3a. handle link that has been retrieved
      _handleDeepLink(dynamicLink);
    }, onError: (OnLinkErrorException e) async {
      print('Link Failed: ${e.message}');
    });
  }

  @override
  Widget build(BuildContext context) {
    return Material(
      child: Scaffold(
        appBar: AppBar(
          title: const Text('Dynamic Links Example'),
        ),
        body: Builder(builder: (BuildContext context) {
          return Center(
            child: WebView(
              initialUrl: temp, // dynamic url copy from deep link var it will not load in web view
              javascriptMode: JavascriptMode.unrestricted,
            ),
          );
        }),
      ),
    );
  }

  void _handleDeepLink(PendingDynamicLinkData data) {
    final Uri deepLink = data?.link;
    if (deepLink != null) {
      temp = deepLink.toString(); // here i had done type casting to uri to string 
      print('_handleDeepLink | deeplink: $deepLink');
    }
  }
}
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