问题
I guess I'm struggling with generics. I want to create simple UIView extension to find recursively a superview of class passed in the function param. I want the function to return optional containing obviously either nil, or object visible as instance of provided class.
extension UIView {
func superviewOfClass<T>(ofClass: T.Type) -> T? {
var currentView: UIView? = self
while currentView != nil {
if currentView is T {
break
} else {
currentView = currentView?.superview
}
}
return currentView as? T
}
}
Any help much appreciated.
回答1:
Swift 3/4
This is a more concise way:
extension UIView {
func superview<T>(of type: T.Type) -> T? {
return superview as? T ?? superview.compactMap { $0.superview(of: type) }
}
func subview<T>(of type: T.Type) -> T? {
return subviews.compactMap { $0 as? T ?? $0.subview(of: type) }.first
}
}
Usage:
let tableView = someView.superview(of: UITableView.self)
let tableView = someView.subview(of: UITableView.self)
回答2:
No need to pass in the type of the class you want (at least in Swift 4.1)…
extension UIView {
func firstSubview<T: UIView>() -> T? {
return subviews.compactMap { $0 as? T ?? $0.firstSubview() as? T }.first
}
}
来源:https://stackoverflow.com/questions/37705819/swift-find-superview-of-given-class-with-generics