SWIFT5.0 iOS 13
试图理解UIViewRepresentable是如何工作的,并把这个简单的例子放在一起,就快到了,但也许它完全是胡说八道。是的,我知道在SwiftUI中已经有了一个SwiftUI,这只是一个测试。
不会编译,因为它说'super.init‘不是在从initialiser返回之前在所有路径上调用的,我尝试并设置了这个值,但显然不正确。
import SwiftUI
struct newView: UIViewRepresentable {
typealias UIViewType = UIView
var v = UIView()
func updateUIView(_ uiView: UIView, context: Context) {
v.backgroundColor = UIColor.yellow
}
func makeUIView(context: Context) -> UIView {
let tapGesture = UITapGestureRecognizer(target: self, action: #selector(Coordinator.handleTap(sender:)))
v.addGestureRecognizer(tapGesture)
return v
}
func makeCoordinator() -> newView.Coordinator {
Coordinator(v)
}
final class Coordinator: UIView {
private let view: UIView
init(_ view: UIView) {
self.view = view
}
required init?(coder: NSCoder) {
fatalError("init(coder:) has not been implemented")
}
@objc func handleTap(sender: UITapGestureRecognizer) {
print("tap")
}
}
}发布于 2020-06-01 11:50:22
只需使您的Coordinator是一个NSObject,它通常扮演桥/控制器/委托/参与者角色,而不是表示,所以不应该是is-a-UIView。
final class Coordinator: NSObject {
private let view: UIView
init(_ view: UIView) {
self.view = view
}再来一个..。
func makeUIView(context: Context) -> UIView {
// make target a coordinator, which is already present in context !!
let tapGesture = UITapGestureRecognizer(target: context.coordinator,
action: #selector(Coordinator.handleTap(sender:)))
v.addGestureRecognizer(tapGesture)
return v
}发布于 2020-06-01 11:43:41
这是因为您的Coordinator是UIView和您的子类。
必须调用超类的指定初始化程序“UIView”
在从init返回之前
init(_ view: UIView) {
self.view = view
super.init(frame: .zero) // Or any other frame you need
}https://stackoverflow.com/questions/62131218
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