I need to launch Siri (on a jailbroken device) through the openUrl:
method.
For example
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:@"Siri://"]];
I need to launch Siri (on a jailbroken device) through the openUrl:
method.
For example
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:@"Siri://"]];
This isn't possible. Also, Xcode is an IDE - a code editor. It's not a language or a platform.
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.
Anyway, if you want another way to open it, I would try looking at rpetrich's libActivator source code.
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;
}
Here, the code calls activateAssistantWithOptions:withCompletion:
, which is a method in the SpringBoard class itself.
This technique, of course, is based on MobileSubstrate hooking.
Disclaimer: I have not tested this code. Just looking at it, though, it seems to be what you need.
This isn't using URL schemes, but I did figure out a different way to launch Siri, described in this answer.
There's no URL schema for Siri and there's no integration available at this time - there may be integration coming in iOS 7 but at this stage there is no way to perform this function.
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