Siri - 类似于 Skype 音频通话的联系人搜索行为

Siri - contact searching behaviour similar to skype for audio call

提问人:makboney 提问时间:1/4/2017 最后编辑:Paul Roubmakboney 更新时间:1/6/2017 访问量:1013

问:

我正在实施一个 VoIP 应用程序,我需要通过 Siri 发起呼叫。我能够通过 Siri 发起呼叫。但问题是 - 每次启动应用程序时,尽管联系人不在应用程序的联系人列表中。

我不确定如何处理以及在哪里处理。我的意思是,如果应用程序没有像Skype那样的联系人,则不要启动该应用程序。Skype的回复如下:

嗯,Skype没有找到<用户>。

你想给谁打电话?

下面是扩展处理程序的代码片段:

- (id)handlerForIntent:(INIntent *)intent {
    // This is the default implementation.  If you want different objects to handle different intents,
    // you can override this and return the handler you want for that particular intent.
    return self;
}

#pragma mark - INStartAudioCallIntentHandling

- (void)resolveContactsForStartAudioCall:(INStartAudioCallIntent *)intent
                          withCompletion:(void (^)(NSArray<INPersonResolutionResult *> *resolutionResults))completion{
    NSArray<INPerson *> *recipients = intent.contacts;
    NSMutableArray<INPersonResolutionResult *> *resolutionResults = [NSMutableArray array];
    if (recipients.count == 0) {
        completion(@[[INPersonResolutionResult needsValue]]);
        return;
    }else if(recipients.count==1){
        [resolutionResults addObject:[INPersonResolutionResult successWithResolvedPerson:recipients.firstObject]];
    }else if(recipients.count>1){
        [resolutionResults addObject:[INPersonResolutionResult disambiguationWithPeopleToDisambiguate:recipients]];
    }else{
        [resolutionResults addObject:[INPersonResolutionResult unsupported]];
    }
    completion(resolutionResults);
}

- (void)confirmStartAudioCall:(INStartAudioCallIntent *)intent
                   completion:(void (^)(INStartAudioCallIntentResponse *response))completion{
    NSUserActivity *userActivity = [[NSUserActivity alloc] initWithActivityType:NSStringFromClass([INStartAudioCallIntent class])];
    INStartAudioCallIntentResponse *response = [[INStartAudioCallIntentResponse alloc] initWithCode:INStartAudioCallIntentResponseCodeReady userActivity:userActivity];
    completion(response);
}

- (void)handleStartAudioCall:(INStartAudioCallIntent *)intent
                  completion:(void (^)(INStartAudioCallIntentResponse *response))completion{
    NSUserActivity *userActivity = [[NSUserActivity alloc] initWithActivityType:NSStringFromClass([INStartAudioCallIntent class])];
    INStartAudioCallIntentResponse *response = [[INStartAudioCallIntentResponse alloc] initWithCode:INStartAudioCallIntentResponseCodeContinueInApp userActivity:userActivity];
    completion(response);
}
iOS Objective-C Sirikit

评论

10赞 Thomas Weller 1/5/2017
在 Meta 上讨论。我认为我们可以停止投反对票。
0赞 makboney 1/6/2017
@ThomasWeller感谢您的回复。我试图在元讨论中说清楚。错误是我的,我为此道歉。但仍然投反对票。

答:

3赞 makboney 1/4/2017 #1

您可以在方法中处理该方法,检查您获得该意图的人是否包含在您的应用程序联系人列表中。resolveContactsForStartAudioCall

- (void)resolveContactsForStartAudioCall:(INStartAudioCallIntent *)intent
                              withCompletion:(void (^)(NSArray<INPersonResolutionResult *> *resolutionResults))completion{
        NSArray<INPerson *> *recipients = intent.contacts;
        NSMutableArray<INPersonResolutionResult *> *resolutionResults = [NSMutableArray array];
        if (recipients.count == 0) {
            completion(@[[INPersonResolutionResult needsValue]]);
            return;
        }else if(recipients.count==1){
            if([self containContact:recipients.firstObject.displayName]){
               [resolutionResults addObject:[INPersonResolutionResult successWithResolvedPerson:recipients.firstObject]];
            }else [resolutionResults addObject:[INPersonResolutionResult unsupported]];
        }else if(recipients.count>1){
            [resolutionResults addObject:[INPersonResolutionResult disambiguationWithPeopleToDisambiguate:recipients]];
        }else{
            [resolutionResults addObject:[INPersonResolutionResult unsupported]];
        }
        completion(resolutionResults);
}
- (BOOL)containContact:(NSString *)displayName {
           //fetch contacts and check, if exist retun YES else NO
}

请注意,如果要将应用程序中的联系人共享到任何扩展,则可能需要启用应用组支持。以下是一些准则:

  1. Apple 文档
  2. Stack Overflow 链接