在 Action Extension 中打开 URL

最近个人app一个小需求，在Safari中的Action Extension 需要跳转 Containing App，试过一些接口，要么被废弃，要么被限制，要么无效。花了一俩天寻找与试错，明明很小的需求，却浪费了那么多时间精力，必须记录下：

xcode8.3 iOS8.0～iOS12可行
最后结果：
1、swift版本：

let url = NSURL(string: "webcapture://")
let selectorOpenURL = sel_registerName("openURL:")
let context = NSExtensionContext()
context.open(url! as URL, completionHandler: nil)

var responder = self as UIResponder?

while (responder != nil){
    if responder?.responds(to: selectorOpenURL) == true{
        responder?.perform(selectorOpenURL, with: url)
    }
    responder = responder!.next
}

2、OC版本：

    NSString *scheml = @"webcapture://";
    NSURL *url = [NSURL URLWithString:scheml];
    SEL selectorOpenURL = sel_registerName("openURL:");
    NSExtensionContext *context = [[NSExtensionContext alloc] init];
    [context openURL:url completionHandler:nil];
    
    UIResponder *responder = (UIResponder *)self;
    while (responder != nil) {
        if ([responder respondsToSelector:selectorOpenURL]) {
            [responder performSelector:selectorOpenURL withObject:url];
        }
        responder = responder.nextResponder;
    }