尝试我的第一个 Flutter 插件时，我尝试在 iOS 和 Android 世界中调用一个方法。我成功地调用了这样一个没有任何参数的方法。

但现在我想调用一个有参数的方法。

对于 iOS，由于某种原因我无法让它工作。 (也许这只是我一直在监督的自动完成的事情，因为 VSCode 不会自动完成我的 Swift 代码)。但也许是另外一回事。请对此提供任何帮助。

这是我的代码:

我的库(F​​lutter-world)是这样的:

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

class SomeName {
  static const MethodChannel _channel =
      const MethodChannel('myTestMethod');

  static Future<String> get sendParamsTest async {
    final String version = await _channel.invokeMethod('sendParams',<String, dynamic>{
        'someInfo1': "test123",
        'someInfo2': "hello",
      });
    return version;
  }
}

.

我的 swift 插件 (iOS-world) 看起来像这样:

import Flutter
import UIKit

public class SwiftSomeNamePlugin: NSObject, FlutterPlugin {

  public static func register(with registrar: FlutterPluginRegistrar) {
    let channel = FlutterMethodChannel(name: "myTestMethod", binaryMessenger: registrar.messenger())
    let instance = SwiftSomeNamePlugin()
    registrar.addMethodCallDelegate(instance, channel: channel)
  }

  public func handle(_ call: FlutterMethodCall, result: @escaping FlutterResult) {

    // flutter cmds dispatched on iOS device :
    if call.method == "sendParams" {

      guard let args = call.arguments else {
        result("iOS could not recognize flutter arguments in method: (sendParams)") 
      }
      String someInfo1 = args["someInfo1"]
      String someInfo2 = args["someInfo2"]
      print(someInfo1)
      print(someInfo2)
      result("Params received on iOS = \(someInfo1), \(someInfo2)")
    } else {
      result("Flutter method not implemented on iOS")
    }
  }
}

错误消息说:

注意:在类型后面加参数，构造一个类型的值 String someInfo1 = args["someInfo1"]

注意:在类型后面加参数，构造一个类型的值 String someInfo2 = args["someInfo2"]

注意:使用'.self'来引用类型对象 String someInfo1 = args["someInfo1"]

注意:使用'.self'来引用类型对象 String someInfo2 = args["someInfo2"]

警告:“String.Type”类型的表达式未使用 String someInfo1 = args["someInfo1"]

警告:“String.Type”类型的表达式未使用 String someInfo2 = args["someInfo2"]

最佳答案

在miguelpruivo的帮助下，我找到了解决方案。

这是工作代码:

Dart 中的 Flutter-world 是正确的:

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

class SomeName {
  static const MethodChannel _channel =
      const MethodChannel('myTestMethod');

  static Future<String> get sendParamsTest async {
    final String version = await _channel.invokeMethod('sendParams',<String, dynamic>{
        'someInfo1': "test123",
        'someInfo2': 3.22,
      });
    return version;
  }
}

.

下面是 Swift 中的 iOS 世界——现在也能正常工作了……

(Dart的dynamic对应Swift的Any)

(方法参数是 [String:Any] 类型的字典 - 有点像 Swift 经常使用的 userInfo - 因此你需要在接收处理程序中转换)...

import Flutter
import UIKit

public class SwiftSomeNamePlugin: NSObject, FlutterPlugin {
  public static func register(with registrar: FlutterPluginRegistrar) {
    let channel = FlutterMethodChannel(name: "myTestMethod", binaryMessenger: registrar.messenger())
    let instance = SwiftSomeNamePlugin()
    registrar.addMethodCallDelegate(instance, channel: channel)
  }

  public func handle(_ call: FlutterMethodCall, result: @escaping FlutterResult) {

    // flutter cmds dispatched on iOS device :
    if call.method == "sendParams" {

      guard let args = call.arguments else {
        return
      }
      if let myArgs = args as? [String: Any],
         let someInfo1 = myArgs["someInfo1"] as? String,
         let someInfo2 = myArgs["someInfo2"] as? Double {
        result("Params received on iOS = \(someInfo1), \(someInfo2)")
      } else {
        result(FlutterError(code: "-1", message: "iOS could not extract " + 
           "flutter arguments in method: (sendParams)", details: nil))
      } 
    } else if call.method == "getPlatformVersion" {
      result("Running on: iOS " + UIDevice.current.systemVersion)
    } else {
      result(FlutterMethodNotImplemented)
    }
  }
}

关于android - flutter 插件 : invoking iOS and Android method including parameters not working，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/53898765/