如何使用带有xcode的方案URL启动Siri? [英] How to launch Siri using scheme url with xcode?

问题描述

我需要通过openUrl:方法启动 Siri (在越狱设备上).

I need to launch Siri (on a jailbroken device) through the openUrl: method.

例如

[[UIApplication sharedApplication] openURL:[NSURL URLWithString:@"Siri://"]];

推荐答案

Siri似乎没有任何 URL方案，因此我认为您不能以这种方式打开它. Siri也不是普通的 App ，它是SpringBoard使用的库.

There doesn't seem to be any URL Scheme for Siri, so I don't think you can open it that way. Siri is also not a normal App, it's a library that's used by SpringBoard.

无论如何，如果您想以其他方式打开它，我会尝试查看rpetrich的 libActivator 源代码.

Anyway, if you want another way to open it, I would try looking at rpetrich's libActivator source code.

如果您在这里看到，您会看到类似的内容满足您的需求. 虚拟助手是"Siri".

If you look here, you'll see something similar to what you need. The Virtual Assistant is "Siri".

- (BOOL)activateVirtualAssistant{   
   if ([%c(SBAssistantController) preferenceEnabled]) { 
      if ([%c(SBAssistantController) shouldEnterAssistant]) {
          SBAssistantController *assistant = (SBAssistantController *)[%c(SBAssistantController) sharedInstance];           
          if (assistant.assistantVisible)   
             [assistant dismissAssistant];          
          else {
             [(SpringBoard *)UIApp activateAssistantWithOptions:nil withCompletion:nil];
             return YES;
          }     
       }
   }    
   return NO;
}

此处，代码调用activateAssistantWithOptions:withCompletion:，这是 SpringBoard类本身.

Here, the code calls activateAssistantWithOptions:withCompletion:, which is a method in the SpringBoard class itself.

这种技术当然是基于MobileSubstrate hooking .

This technique, of course, is based on MobileSubstrate hooking.

免责声明:我尚未测试此代码.不过，只是看一下它，这似乎就是您所需要的.

Disclaimer: I have not tested this code. Just looking at it, though, it seems to be what you need.

这不是使用URL方案，但是我确实找到了另一种启动Siri的方式，此答案中所述.

This isn't using URL schemes, but I did figure out a different way to launch Siri, described in this answer.

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