通过方法通道颤振传递自定义对象

Question

我正在尝试将类型为 User 的自定义对象从原生平台传递给 Flutter。 User 类是库的一部分，不能直接访问以进行编辑。这是我的 android 和 iOS 代码实现。问题是我无法找到如何通过方法通道传递此对象的解决方案，以便我可以轻松地在 Dart 代码中解析它。

Android 部分：

private fun loginUser(uid: String, apiKey: String, result: MethodChannel.Result) {
        MyChat.login(uid, apiKey, object : MyChat.CallbackListener<User>() {
            override fun onSuccess(user: User) {
                Log.e(TAG, user.toString())
                result.success(hashMapOf("RESULT" to true, "AVATAR" to user.avatar,
                        "CREDITS" to user.credits,
                        "EMAIL" to user.email,
                        "LAST_ACTIVE" to user.lastActiveAt,
                        "NAME" to user.name,
                        "ROLE" to user.role,
                        "STATUS" to user.status,
                        "STATUS_MESSAGE" to user.statusMessage).toString())
            }

            override fun onError(p0: MyChatException?) {
                Log.e(TAG, p0?.message)
                result.error("FAILED", "Unable to create login", null)
            }
        })
    } 

iOS 实现：

func loginUser(result: @escaping FlutterResult, uid: String, apiKey: String){
        MyChat.login(UID: uid, apiKey: apiKey, onSuccess: { (user) in
            // Login Successful
            let data: [String: Any] = ["RESULT":true,
                                       "AVATAR":user.avatar!,
                                       "CREDITS": user.credits,
                                       "EMAIL": user.email!,
                                       "LAST_ACTIVE":String(user.lastActiveAt),
                                       "NAME":user.name!,
                                       "ROLE":user.role!,
                                       "STATUS":user.status.rawValue,
                                       "STATUS_MESSAGE":user.statusMessage]
            let jsonData =  try? JSONSerialization.data(withJSONObject: data, options: [.prettyPrinted])
            result(String(data: jsonData!, encoding: .ascii))
        }) { (error) in
            // Login error
            result(FlutterError(code: "FAILED", message:"Login failed with exception: " + error.errorDescription, details: nil))

        }
    }

我的飞镖代码：

    Future<String> isUserLoggedIn() async {
    String status = "";
    try {
      final String result = await platform
          .invokeMethod('loginUser', {"UID": UID, "API_KEY": API_KEY});
      print(result); //How to parse?
    status = "Hello";
    } on PlatformException catch (e) {
      print("Exception");
      status = e.message;
    }
    return status;
  }

Answer 1

您可以使用invokeMapMethod，它是invokeMethod 的一个实现，可以返回类型化的地图。 像这样：

final result = await platform.invokeMapMethod('loginUser', ...);

或者您可以像这样将 json 对象作为字符串传递：

在安卓中

platform.success(
    "{\"CREDITS\":\"${user.credits}\",\"EMAIL\":\"${user.email}\",\"LAST_ACTIVE\":\"${user.lastActiveAt}\"}"
)

在颤抖中

var result = await methodChannel.invokeMethod('loginUser' , '');
var json = json.decode(result);

var credit = json['CREDITS'];
var email = json['EMAIL'];
var lastActive = json['LAST_ACTIVE'];

Answer 2

您可以在哈希映射中传递数据。

在 Android 中：

result.success(hashMapOf(
    "CREDITS" to user.credits,
    "EMAIL" to user.email,
    ...
))

在 iOS 中：

let data: [String: Any] = [...]
result(data)

在 Flutter 中：

final result = await platform.invokeMethod<Map<String, dynamic>>('loginUser', ...);
final credits = result['CREDITS'] as String;
final email = result['EMAIL'] as String;
...