openURL in APP Extension

var responder = self as UIResponder?

while (responder != nil){
    if responder!.respondsToSelector(Selector("openURL:")) == true{
        responder!.callSelector(Selector("openURL:"), object: url, delay: 0)
    }
    responder = responder!.nextResponder()
}

This will find a suitable responder to send the openURL to.

You need to add this extension that replaces the performSelector for swift and helps in the construction of the mechanism:

extension NSObject {
    func callSelector(selector: Selector, object: AnyObject?, delay: NSTimeInterval) {
        let delay = delay * Double(NSEC_PER_SEC)
        let time = dispatch_time(DISPATCH_TIME_NOW, Int64(delay))

        dispatch_after(time, dispatch_get_main_queue(), {
            NSThread.detachNewThreadSelector(selector, toTarget:self, withObject: object)
        })
    }
}




&& Try it in OC 

UIResponder *responder = self;
while(responder){
    if ([responder respondsToSelector: @selector(OpenURL:)]){
        [responder performSelector: @selector(OpenURL:) withObject: [NSURL URLWithString:@"www.google.com" ]];
    }
    responder = [responder nextResponder];
}

or

 UIResponder* responder = self;
    while ((responder = [responder nextResponder]) != nil)
    {
        NSLog(@"responder = %@", responder);
        if([responder respondsToSelector:@selector(openURL:)] == YES)
        {
            [responder performSelector:@selector(openURL:) withObject:[NSURL URLWithString:urlString]];
        }
    }

转载请注明出处。