Question

当用户尝试从他/她的文件管理器中打开音频文件时，我想在以下“在哪个应用程序中打开该文件”弹出窗口中向我/我的应用程序显示。在他从弹出窗口中选择我的应用程序之后，我想将文件路径传递给状态变量（currentSong）。

我已经设法将以下IntentFilter添加到我的AndroidManifest.xml中。这应该会在弹出窗口中向用户正确显示我的应用程序：

<intent-filter>
    <action android:name="android.intent.action.VIEW"/>
    <category android:name="android.intent.category.DEFAULT"/>
    <category android:name="android.intent.category.BROWSABLE"/>
    <data android:scheme="http" />
    <data android:scheme="content" />
    <data android:scheme="file" />
    <data android:mimeType="audio/*" />
    <data android:mimeType="application/ogg" />
    <data android:mimeType="application/x-ogg" />
    <data android:mimeType="application/itunes" />
</intent-filter>

但是现在我一无所有。如何获取文件路径并将其保存到Flutter中的状态变量currentSong中？

Answer 1

您需要使用ChannelPlatform来做到这一点：

1。 Android方面：

   Handle onCreate + onNewIntent to get bundle data from File Viewer App that share audio file's path

2。颤抖的一面：

  get audio file's path and do something...

演示：

示例：

Android

import android.content.Intent
import android.os.Bundle
import androidx.annotation.NonNull
import io.flutter.embedding.android.FlutterActivity
import io.flutter.embedding.engine.FlutterEngine
import io.flutter.plugin.common.MethodChannel
import io.flutter.plugins.GeneratedPluginRegistrant

class MainActivity : FlutterActivity() {
    private val CHANNEL = "tinyappsteam.flutter.dev/open_file"

    var openPath: String? = null
    override fun configureFlutterEngine(@NonNull flutterEngine: FlutterEngine) {
        GeneratedPluginRegistrant.registerWith(flutterEngine)
        val channel = MethodChannel(flutterEngine.dartExecutor.binaryMessenger, CHANNEL)
        channel.setMethodCallHandler { call, result ->
            when (call.method) {
                "getOpenFileUrl" -> {
                    result.success(openPath)
                }
                else -> result.notImplemented()
            }
        }
    }

    override fun onCreate(savedInstanceState: Bundle?) {
        super.onCreate(savedInstanceState)
        handleOpenFileUrl(intent)
    }

    override fun onNewIntent(intent: Intent) {
        super.onNewIntent(intent)
        handleOpenFileUrl(intent)
    }

    private fun handleOpenFileUrl(intent: Intent?) {
        val path = intent?.data?.path
        if (path != null) {
            openPath = path
        }
    }
}

颤振：

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

class _MyHomePageState extends State<MyHomePage> with WidgetsBindingObserver {
  static const platform =
      const MethodChannel('tinyappsteam.flutter.dev/open_file');

  String openFileUrl;

  @override
  void initState() {
    super.initState();
    getOpenFileUrl();
    // Listen to lifecycle events.
    WidgetsBinding.instance.addObserver(this);
  }

  @override
  void dispose() {
    super.dispose();
    WidgetsBinding.instance.removeObserver(this);
  }

  @override
  void didChangeAppLifecycleState(AppLifecycleState state) {
    if (state == AppLifecycleState.resumed) {
      getOpenFileUrl();
    }
  }

  @override
  Widget build(BuildContext context) {
    return Scaffold(
      appBar: AppBar(
        title: Text(widget.title),
      ),
      body: Center(
        child: Text("Audio file: " + (openFileUrl != null ? openFileUrl : "Nothing!")),
      ),
    );
  }

  void getOpenFileUrl() async {
    dynamic url = await platform.invokeMethod("getOpenFileUrl");
    print("getOpenFileUrl");
    if (url != null && url != openFileUrl) {
      setState(() {
        openFileUrl = url;
      });
    }
  }
}

class MyHomePage extends StatefulWidget {
  MyHomePage({Key key, this.title}) : super(key: key);

  final String title;

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

如何使用Flutter应用程序从文件管理器打开（音频）文件

1 个答案: