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首页 > 解决方案 > 将数据从 Flutter 传递到 SwiftUI 主屏幕 WidgetExtension？访问 SQLite DB 本身还是使用方法通道？

flutter - 将数据从 Flutter 传递到 SwiftUI 主屏幕 WidgetExtension？访问 SQLite DB 本身还是使用方法通道？

问题描述

我有一个使用 SQFLite 来持久化数据的颤振应用程序。我有兴趣尝试做一个 iOS 主屏幕小部件，据我所知，它必须在 swiftui 中构建。我想知道是否可以从 Swiftui 中的本机小部件访问我的颤振 sqlite 数据库？

更新：尝试使用以下建议但遇到问题。由于如果我将 appdelegate 模块添加为共享目标，小部件扩展似乎会导致 xcode 构建失败......我试图将它跳到一个快速文件调用然后共享它。但我遇到了错误。

我的 Flutter 助手（它目前只是试图从列表中发送一个整数或多或少作为概念证明）

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

import 'db.dart';


final channelName = 'widgetDataCheckItOff';

Future<void> _handleMethod(MethodCall call) async {
  final String utterance = call.arguments;
  switch(call.method) {
    case "getDataFromFlutter":
      var db = new DB();
      dynamic result = await db.widgetQuery();
      return result;
  }
  final methodChannel = MethodChannel(channelName);
  methodChannel.setMethodCallHandler(_handleMethod);

}

应用委托

import UIKit
import Flutter

@UIApplicationMain
@objc class AppDelegate: FlutterAppDelegate {
    var controller : FlutterViewController!;
    var widgetDataCheckItOff : FlutterMethodChannel!;
    class func shared() -> AppDelegate
    {
        return UIApplication.shared.delegate as! AppDelegate
    }
    
    override func application(
        _ application: UIApplication,
        didFinishLaunchingWithOptions launchOptions: [UIApplication.LaunchOptionsKey: Any]?
    ) -> Bool {
        controller = window?.rootViewController as? FlutterViewController;
        widgetDataCheckItOff = FlutterMethodChannel (
            name : "widgetDataCheckItOff",
            binaryMessenger : controller as! FlutterBinaryMessenger
        )
        
        GeneratedPluginRegistrant.register(with: self)
        if #available(iOS 10.0, *) { UNUserNotificationCenter.current().delegate = self as? UNUserNotificationCenterDelegate }
        return super.application(application, didFinishLaunchingWithOptions: launchOptions)
    }
    @available(iOS 9.0, *)
    override func application(_ application: UIApplication, performActionFor shortcutItem: UIApplicationShortcutItem, completionHandler: @escaping (Bool) -> Void) {
        let controller = window.rootViewController as? FlutterViewController
        
        let channel = FlutterMethodChannel(name: "plugins.flutter.io/quick_actions", binaryMessenger: controller! as! FlutterBinaryMessenger)
        channel.invokeMethod("launch", arguments: shortcutItem.type)
    }
    public func getTasks ( args: String, completion:@escaping (Int) -> Int) {
        var x = -1
        widgetDataCheckItOff.setMethodCallHandler { (call : FlutterMethodCall, result: @escaping FlutterResult) in
            //        print("set method call handler")
            if(call.method == "getDataFromFlutter") {
                self.widgetDataCheckItOff.invokeMethod ("getDataFromFlutter", arguments: args, result: {(r:Any?) -> () in
                    x=r as! Int
                });
            }
            else{
                result(FlutterMethodNotImplemented)
                x = -1// to handle undeclared methods
            }
            let result = x
            completion(result)
        }
    }
    
}

跳文件我称为widgetSwift

import Foundation

@objc class WidgetData: NSObject{
    public func getWidgetData(result: Int)-> Int {
        var x = -1
        let a = UIApplication.shared.delegate as! AppDelegate
        a.getTasks(args: "getDataFromFlutter",  completion: {(result) -> Int in
            x = result
        })
        return x
//        let x = AppDelegate.getTasks(<#T##self: AppDelegate##AppDelegate#>);
        
    }
}

和我的小部件扩展部分调用它

struct widget_extension_widgetEntryView : View {
    var entry: Provider.Entry

       var body: some View {
        let w = WidgetData()
        let x: Int =         w.getWidgetData(result: <#Int#>)
        
        let str = "Today's Tasks:  \(x)"
        Text("", tableName: str)
       }
}

扩展给我函数调用的错误：源文件中的编辑器占位符。

widgetSwift 在 getTasks 调用时给了我错误：在范围内找不到“AppDelegate”声明的闭包结果“（）”与上下文类型“Int”不兼容

标签： flutterswiftuisqflite

解决方案

Flutter 提供了一种使用MethodChannel.

从颤振MethodChannel文档（这里）：

Flutter 使用灵活的系统，允许您调用特定于平台的 API，无论是 Android 上的 Kotlin 或 Java 代码，还是 iOS 上的 Swift 或 Objective-C 代码。

注意：如果需要，方法调用也可以反向发送，平台充当 Dart 中实现的方法的客户端。一个具体的例子是quick_actions插件。

MethodChannel从颤振调用本机平台代码上的函数的用法：

我们在 Flutter、iOS 和 Android 中创建一个同名的方法通道对象。Android 和 iOS 对象将设置一个方法调用处理程序来接收来自 Flutter 的调用。然后 Flutter 代码可以调用 invokeMethod 来调用原生对象上的处理程序

扑：

static final channelName = 'com.example.widget/database';
final methodChannel = MethodChannel(channelName);

await this.methodChannel.invokeMethod("getDataFromDatabase");

安卓：

import androidx.annotation.NonNull;
import io.flutter.embedding.android.FlutterActivity;
import io.flutter.embedding.engine.FlutterEngine;
import io.flutter.plugin.common.MethodChannel;

public class MainActivity extends FlutterActivity {
  private static final String CHANNEL = "com.example.widget/database";

  @Override
  public void configureFlutterEngine(@NonNull FlutterEngine flutterEngine) {
  super.configureFlutterEngine(flutterEngine);
    new MethodChannel(flutterEngine.getDartExecutor().getBinaryMessenger(), CHANNEL)
        .setMethodCallHandler(
          (call, result) -> {
            if (call.method.equals("getDataFromDatabase")) {
              //TODO: Perform method to query database here
              String data;

              if (data != null) {
                result.success(data);
              } else {
                result.error("UNAVAILABLE", "Data not available", null);
              }
            } else {
              result.notImplemented();
            }
          }
        );
  }
}

IOS：

@UIApplicationMain
@objc class AppDelegate: FlutterAppDelegate {
  override func application(
    _ application: UIApplication,
    didFinishLaunchingWithOptions launchOptions: [UIApplication.LaunchOptionsKey: Any]?) -> Bool {

    let controller : FlutterViewController = window?.rootViewController as! FlutterViewController
    let batteryChannel = FlutterMethodChannel(name: "com.example.widget/database",
                                              binaryMessenger: controller.binaryMessenger)
    batteryChannel.setMethodCallHandler({
      (call: FlutterMethodCall, result: @escaping FlutterResult) -> Void in
      // Note: this method is invoked on the UI thread.
      guard call.method == "getDataFromDatabase" else {
         result(FlutterMethodNotImplemented)
         return
      }

      //TODO: Perform method to query database here
      let data;

      if data == nil {
         result(FlutterError(code: "UNAVAILABLE",
                        message: "Data not available",
                        details: nil))
      } else {
         result(data)
      }     
    })

    GeneratedPluginRegistrant.register(with: self)
    return super.application(application, didFinishLaunchingWithOptions: launchOptions)
  }
}

逆向示例——从 Native 调用 Flutter 如果您需要从 Native 代码调用 Flutter 中定义的方法，那么您调用之前的相同方法，但这次您setMethodCallHandler在 Flutter 端和invokeMethodNative 端调用。

扑：

final channelName = 'dataFromFlutterChannel';

final methodChannel = MethodChannel(channelName);
methodChannel.setMethodCallHandler(this._handleMethod);

Future<void> _handleMethod(MethodCall call) async {
   final String utterance = call.arguments; 
   switch(call.method) {
     case "getDataFromFlutter":
        //TODO: Query database here and return the data to native side
   }
}

安卓：

val channelName = 'dataFromFlutterChannel'
val methodChannel = MethodChannel(flutterView, channelName)
    
methodChannel.invokeMethod("getDataFromFlutter", utterance)

IOS：

let rootViewController : FlutterViewController = window?.rootViewController as! FlutterViewController
let channelName = "dataFromFlutterChannel"
let methodChannel = FlutterEventChannel(name: channelName, binaryMessenger: rootViewController)
    
methodChannel.invokeMethod("getDataFromFlutter", utterance)

本文可以帮助您进行更详细的解释。

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