2

斯威夫特 5.0 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,几乎就在那里,但也许它完全是胡说八道。是的,我知道 SwiftUI 中已经有一个 tapGesture,这只是一个测试。

不会编译,因为它说在从初始化程序返回之前没有在所有路径上调用“super.init”,我尝试设置但显然不正确。

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")
  }
 }

}
4

2 回答 2

3

只是让你Coordinator是一个NSObject,它通常扮演桥梁/控制器/代表/演员角色,但不是演示,所以不应该is-a-UIView

final class Coordinator: NSObject {
  private let view: UIView

init(_ view: UIView) {
    self.view = view
}

还有一个……

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
}
于 2020-06-01T11:50:22.150 回答
1

那是因为 yourCoordinatorUIViewand you的子类

必须调用超类“UIView”的指定初始化程序

从以下地点返回之前init

init(_ view: UIView) {
    self.view = view
    super.init(frame: .zero) // Or any other frame you need
}
于 2020-06-01T11:43:41.190 回答