Flutter - 如何在 firebase 动态链接中为应用邀请功能传递自定义 arguments？

Question

你能告诉我如何使用 flutter 插件 firebase_dynamic_links[ https://pub.dev/packages/firebase_dynamic_links#-readme-发送和获取 arguments？

我想在深层链接/动态链接中传递用户名和密码等参数，如下所示：

uriPrefix: 'https://test.page.link/appinvite?username=Test&password=123456',
link: Uri.parse('https://test.page.link/appinvite?username=Test&password=123456'),

这是传递数据的正确方法吗？

之后我使用下面的代码来获取数据，

await FirebaseDynamicLinks.instance.getInitialLink();
final Uri deepLink = data?.link;

But, it is providing me the dummy web URL i have added in the firebase console for firebase dynamic links in Deep Link URL

目前我可以从 firebase 动态链接打开应用程序，但无法获取自定义参数。

任何帮助将不胜感激。

更新代码：

发送邀请：

_generateAndShareDynamicLink() async {
    final DynamicLinkParameters parameters = DynamicLinkParameters(
      uriPrefix: 'https://test.page.link/groupinvite',
      link: Uri.parse('https://test.page.link/groupinvite'),
      androidParameters: AndroidParameters(
        packageName: 'com.test.flutter_authentication',
        minimumVersion: 0,
      ),
      dynamicLinkParametersOptions: DynamicLinkParametersOptions(
        shortDynamicLinkPathLength: ShortDynamicLinkPathLength.unguessable,
      ),
    );

    Uri url = await parameters.buildUrl();

    shareURL(Uri.https(url.authority, url.path, {"username": "Test"}));
}

在 initState() 中获取数据：

FirebaseDynamicLinks.instance.onLink(
        onSuccess: (PendingDynamicLinkData dynamicLink) async {
          final Uri deepLink = dynamicLink?.link;
          _scaffoldKey.currentState.showSnackBar(SnackBar(content: Text("deepLink2 : ${deepLink}",)));
          if (deepLink != null) {
            Map sharedListId = deepLink.queryParameters;
            print("sharedListId : ${sharedListId}");

            String username=sharedListId["username"];
            print("username : ${username}");

            Navigator.pushNamed(context, deepLink.path);
          }
        }, onError: (OnLinkErrorException e) async {
      print('onLinkError');
      print(e.message);
    });

仍然检索到的数据是 null。

有人可以帮忙吗？

Answer 1

使用动态链接的最佳方式是，

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

class TestPage extends StatefulWidget {
  @override
  _TestPageState createState() => _TestPageState();
}

class _TestPageState extends State<TestPage> {

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

  void fetchLinkData() async {
    // FirebaseDynamicLinks.getInitialLInk does a call to firebase to get us the real link because we have shortened it.
    var link = await FirebaseDynamicLinks.instance.getInitialLink();

    // This link may exist if the app was opened fresh so we'll want to handle it the same way onLink will.
    handleLinkData(link);

    // This will handle incoming links if the application is already opened
    FirebaseDynamicLinks.instance.onLink(onSuccess: (PendingDynamicLinkData dynamicLink) async {
      handleLinkData(dynamicLink);
    });
  }

  void handleLinkData(PendingDynamicLinkData data) {
    final Uri uri = data?.link;
    if(uri != null) {
      final queryParams = uri.queryParameters;
      if(queryParams.length > 0) {
        String userName = queryParams["username"];
        // verify the username is parsed correctly
        print("My users username is: $userName");
      }
    }
  }

  @override
  Widget build(BuildContext context) {
    return Scaffold(
      appBar: AppBar(
        title: Text("Sample"),
      ),
      body: Center(
        child: Text("Test"),
      ),
      floatingActionButton: FloatingActionButton(
        onPressed: () async {
          var dynamicLink = await createDynamicLink(userName: "Test");
          // dynamicLink has been generated. share it with others to use it accordingly.
          print("Dynamic Link: $dynamicLink");
        },
        child: Icon(
          Icons.add,
          color: Colors.white,
        ),
      ),
    );
  }

  Future<Uri> createDynamicLink({@required String userName}) async {
    final DynamicLinkParameters parameters = DynamicLinkParameters(
      // This should match firebase but without the username query param
      uriPrefix: 'https://test.page.link',
      // This can be whatever you want for the uri, https://yourapp.com/groupinvite?username=$userName
      link: Uri.parse('https://test.page.link/groupinvite?username=$userName'),
      androidParameters: AndroidParameters(
        packageName: 'com.test.demo',
        minimumVersion: 1,
      ),
      iosParameters: IosParameters(
        bundleId: 'com.test.demo',
        minimumVersion: '1',
        appStoreId: '',
      ),
    );
    final link = await parameters.buildUrl();
    final ShortDynamicLink shortenedLink = await DynamicLinkParameters.shortenUrl(
      link,
      DynamicLinkParametersOptions(shortDynamicLinkPathLength: ShortDynamicLinkPathLength.unguessable),
    );
    return shortenedLink.shortUrl;
  }
}

你完成了。

Answer 2

这对我有用

发送方：

Future<String> generateLink() async {
final DynamicLinkParameters parameters = DynamicLinkParameters(
      uriPrefix: 'https://<your-domain-name>.page.link',
      link: Uri.parse(
          'https://<your-domain-name>.page.link/<your-route>/?id=acb&name=me'), // <- your paramaters
      dynamicLinkParametersOptions: DynamicLinkParametersOptions(
          shortDynamicLinkPathLength: ShortDynamicLinkPathLength.unguessable),
      androidParameters: AndroidParameters(
        packageName: '<package-name>',
        minimumVersion: 0,
      ),
      socialMetaTagParameters: SocialMetaTagParameters(
        title: "click the link",
      ),
    );
    final Uri dynamicUrl = await parameters.buildUrl();
    final ShortDynamicLink shortenedLink =
        await DynamicLinkParameters.shortenUrl(
      dynamicUrl,
      DynamicLinkParametersOptions(
          shortDynamicLinkPathLength: ShortDynamicLinkPathLength.unguessable),
    );
    final Uri shortUrl = shortenedLink.shortUrl;
    return "https://<your-domain-name>.page.link" + shortUrl.path;
  }

接收方：

FirebaseDynamicLinks.instance.onLink(
    onSuccess: (PendingDynamicLinkData dynamicLink) async {
      final Uri deepLink = dynamicLink?.link;
      if (deepLink != null) {
        print(deepLink.queryParameters['id']); // <- prints 'abc'
      }
    }, onError: (OnLinkErrorException e) async {
    print('onLinkError');
    print(e.message);
  }
);

Answer 3

来自diidu的一个可能的解决方案 - Firebase 动态链接支持自定义参数？