openURL in APP Extension

时间:2022-02-06 06:16:37
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。