siri

【腾讯Bugly干货分享】iOS10 SiriKit QQ适配详解

我怕爱的太早我们不能终老 提交于 2019-12-06 08:51:04
本文来自于 腾讯bugly开发者社区 ,非经作者同意,请勿转载,原文地址: http://dev.qq.com/topic/57ece0331288fb4d31137da6 1. 概述 苹果在iOS10开放了siriKit接口给第三方应用。目前,QQ已经率先适配了Siri的发消息和打电话功能。这意味着在iOS10中你可以直接告诉Siri让它帮你发QQ消息和打QQ电话了,听起来是不是很酷炫? 那么第三方应用使用Siri的体验究竟如何?哪些应用可以接入SiriKit?接入SiriKit又需要做哪些工作呢?这篇文章会为你一一解答这些疑惑。 图1 用Siri发QQ消息效果展示 2. SiriKit简介 我们都知道Siri是iphone手机中的智能语音助手,那么什么是SiriKit呢?SiriKit是苹果为第三方应用支持Siri提供的开发框架。在官方文档中,SiriKit将对不同场景的语音支持划分为不同的domain,目前,SiriKit支持的domain包括:VoIP电话、发消息、转账、图片搜索、网约车订车、CarPlay和餐厅预定,也就是说如果你的应用中包含有这些功能之一,就可以考虑将这些功能接入到SiriKit中啦。 实现SiriKit相关功能时,我们并不需要真正对语音进行识别,语音的识别工作会由Siri完成。Siri识别完语音后,会将语音要完成的功能抽象成Intent对象传递给我们

Voice control API - high accuracy on specific phrases [closed]

泪湿孤枕 提交于 2019-12-06 05:35:24
问题 As it currently stands, this question is not a good fit for our Q&A format. We expect answers to be supported by facts, references, or expertise, but this question will likely solicit debate, arguments, polling, or extended discussion. If you feel that this question can be improved and possibly reopened, visit the help center for guidance. Closed 7 years ago . I have several ideas for voice controlled apps. Unfortunately, based on what I've seen from Siri and Google Voice Actions, the

How can I use Siri Shortcuts to show a specific page in my app?

强颜欢笑 提交于 2019-12-06 04:46:48
问题 I want to advanced research Shortcuts technology. So here are some questions: Can Siri shortcuts be used in any type of app? Because of SiriKit only working in tourism, chatting etc. Now using shortcuts. Can I jump to my app showing specific page by Siri? 回答1: Can Siri shortcuts be used in any type of app? Because of SiriKit only working in tourism, chatting etc. Yes, any. They aren't tied to any specific domain. They are custom. Now using shortcuts. Can I jump to my app showing specific page

How do I debug my Siri (Intents) extension?

我与影子孤独终老i 提交于 2019-12-06 01:43:22
问题 I'm building a Siri aka Intents extension . I do understand that I can attach the debugger to a running process after start, however, I never see my process. I do have the authorization for my app set to active and the extension must be running at least for some time, Siri is showing clear responses. By the Apple documentation, you should run the Extension Scheme with Siri as host app. However, if I try that nothing happens on the device, it's waiting to attach for ever. Apple mentions the

Can Siri be disabled within an app?

独自空忆成欢 提交于 2019-12-05 10:07:20
问题 I'm working an on iOS game and recently tested on an iPhone 4S. Siri activates sometimes when my thumb covers the proximity sensor. This is a feature of the 4S. Instead of holding the home button, users can put the phone to their ear to activate Siri. But in my game the activation is not intended and it interrupts gameplay. Can Siri be disabled within an app? Is this an iPhone 4S bug? 回答1: Apple deprecated support for this. http://developer.apple.com/library/ios/#DOCUMENTATION/UIKit/Reference

Siri chat bubble color's in iOS

大城市里の小女人 提交于 2019-12-05 05:25:04
问题 I tried creating the Siri chat bubble in core graphics. I am at a stage where I can draw the shape. I am stuck with the color's here. Wanaa get the border color and the fill color code. here is what I did so far.. - (void)drawInContext:(CGContextRef)context { CGRect rect = gradientRectFrame; CGFloat radius = 30; CGFloat originBufferX = 0.0; CGFloat originBufferY = 0.0; CGFloat rightAngleTriangleWidth = 20.0; CGFloat rightAngleTriangleHeight = 20.0; CGFloat fullRectWidth = rect.size.width;

How can I open my SwiftUI app and perform a function using NSUseractivity from within Siri Intent Extension?

吃可爱长大的小学妹 提交于 2019-12-04 12:58:34
问题 My Problem My app has its own Shortcuts actions created using Intents Extensions. They perform background actions perfectly. For some actions, I'm trying to make the intent extension open the main (container) app when run in Shortcuts and perform a function. I'm having trouble with NSUserActivity and I'm not sure if it's the fact it's a SwiftUI project or the way I'm implementing it (or both). What I've tried I have registered my NSUserActivity name as an NSUserActivityType in my info.plist (

How can I use Siri Shortcuts to show a specific page in my app?

冷暖自知 提交于 2019-12-04 11:21:14
I want to advanced research Shortcuts technology. So here are some questions: Can Siri shortcuts be used in any type of app? Because of SiriKit only working in tourism, chatting etc. Now using shortcuts. Can I jump to my app showing specific page by Siri? Can Siri shortcuts be used in any type of app? Because of SiriKit only working in tourism, chatting etc. Yes, any. They aren't tied to any specific domain. They are custom. Now using shortcuts. Can I jump to my app showing specific page by Siri? Yes. The code below shows the easiest way on how to show a specific page by Siri, it's called

Voice control API - high accuracy on specific phrases [closed]

纵然是瞬间 提交于 2019-12-04 09:23:25
As it currently stands, this question is not a good fit for our Q&A format. We expect answers to be supported by facts, references, or expertise, but this question will likely solicit debate, arguments, polling, or extended discussion. If you feel that this question can be improved and possibly reopened, visit the help center for guidance. Closed 6 years ago . I have several ideas for voice controlled apps. Unfortunately, based on what I've seen from Siri and Google Voice Actions, the technology doesn't seem to quite be there yet. Even in a perfectly quiet environment, the accuracy is so bad,

Launch app from INExtension in SiriKit

和自甴很熟 提交于 2019-12-04 07:34:37
I want to use SiriKit to start a workout. Starting the workout requires opening the main app from the app extension. The boilerplate that Apple provides for the INStartWorkoutIntentHandling handler is func handle(startWorkout startWorkoutIntent: INStartWorkoutIntent, completion: (INStartWorkoutIntentResponse) -> Void) { let userActivity = NSUserActivity(activityType: NSStringFromClass(INStartWorkoutIntent)) let response = INStartWorkoutIntentResponse(code: .success, userActivity: userActivity) completion(response) } How can I open my own app from here? Something like myapp://workout?action