Question

我一直在写我的第一个SwiftUI应用程序，该应用程序管理一个藏书。它的List大约有3,000个项目，可以非常有效地加载和滚动。如果使用切换控件过滤列表以仅显示我没有UI的书在更新前冻结了20到30秒，这可能是因为UI线程正忙于决定是否显示3,000个单元格中的每一个。 / p>

在SwiftUI中，有没有很好的方法来处理对大列表的更新？

var body: some View {
        NavigationView {
            List {
                Toggle(isOn: $userData.showWantsOnly) {
                    Text("Show wants")
                }

                ForEach(userData.bookList) { book in
                    if !self.userData.showWantsOnly || !book.own {
                        NavigationLink(destination: BookDetail(book: book)) {
                            BookRow(book: book)
                        }
                    }
                }
            }
        }.navigationBarTitle(Text("Books"))
    }

Answer 1

您是否尝试过将过滤后的数组传递给ForEach。像这样：

ForEach(userData.bookList.filter {  return !$0.own }) { book in
    NavigationLink(destination: BookDetail(book: book)) { BookRow(book: book) }
}

更新

事实证明，这确实是一个丑陋，丑陋的错误：

代替过滤数组，我只是在翻转开关时一起移除了ForEach，并用简单的Text("Nothing")视图替换了它。结果是一样的，需要30秒！

struct SwiftUIView: View {
    @EnvironmentObject var userData: UserData
    @State private var show = false

    var body: some View {
        NavigationView {

            List {
                Toggle(isOn: $userData.showWantsOnly) {
                    Text("Show wants")
                }

                if self.userData.showWantsOnly {
                   Text("Nothing")
                } else {
                    ForEach(userData.bookList) { book in
                        NavigationLink(destination: BookDetail(book: book)) {
                            BookRow(book: book)
                        }
                    }
                }
            }
        }.navigationBarTitle(Text("Books"))
    }
}

解决方法

我确实找到了一种可以快速运行的解决方法，但是它需要一些代码重构。 “魔术”通过封装发生。解决方法迫使SwiftUI完全丢弃列表，而不是一次删除一行。通过在两个单独的封装视图中使用两个单独的列表来做到这一点：Filtered和NotFiltered。以下是包含3000行的完整演示。

import SwiftUI

class UserData: ObservableObject {
    @Published var showWantsOnly = false
    @Published var bookList: [Book] = []

    init() {
        for _ in 0..<3001 {
            bookList.append(Book())
        }
    }
}

struct SwiftUIView: View {
    @EnvironmentObject var userData: UserData
    @State private var show = false

    var body: some View {
        NavigationView {

            VStack {
                Toggle(isOn: $userData.showWantsOnly) {
                    Text("Show wants")
                }

                if userData.showWantsOnly {
                    Filtered()
                } else {
                    NotFiltered()
                }
            }

        }.navigationBarTitle(Text("Books"))
    }
}

struct Filtered: View {
    @EnvironmentObject var userData: UserData

    var body: some View {
        List(userData.bookList.filter { $0.own }) { book in
            NavigationLink(destination: BookDetail(book: book)) {
                BookRow(book: book)
            }
        }
    }
}

struct NotFiltered: View {
    @EnvironmentObject var userData: UserData

    var body: some View {
        List(userData.bookList) { book in
            NavigationLink(destination: BookDetail(book: book)) {
                BookRow(book: book)
            }
        }
    }
}

struct Book: Identifiable {
    let id = UUID()
    let own = Bool.random()
}

struct BookRow: View {
    let book: Book

    var body: some View {
        Text("\(String(book.own)) \(book.id)")
    }
}

struct BookDetail: View {
    let book: Book

    var body: some View {
        Text("Detail for \(book.id)")
    }
}

Answer 2

我认为我们必须等到随后的beta版本中的SwiftUI List性能提高。当列表从非常大的数组（超过500个）过滤到很小的数组时，我也遇到了同样的滞后。我创建了一个简单的测试应用，为带有整数ID和带有Buttons的字符串的简单数组安排时间，以简单地更改要渲染的数组-相同的滞后时间。

Answer 3

查看本文https://www.hackingwithswift.com/articles/210/how-to-fix-slow-list-updates-in-swiftui

简而言之，本文提出的解决方案是将 .id（UUID（））添加到列表中：

List(items, id: \.self) {
    Text("Item \($0)")
}
.id(UUID())

“”现在，使用id（）这样有一个弊端：您不会使更新动起来。记住，我们实际上是在告诉SwiftUI旧列表已经消失，现在有了新列表，这意味着它不会尝试以动画方式移动行。”

Answer 4

代替复杂的解决方法，只需清空List数组，然后设置新的filters数组。可能有必要引入一个延迟，以便随后的写入操作不会遗漏清空listArray。

List(listArray){item in
  ...
}

self.listArray = []
DispatchQueue.main.asyncAfter(deadline: .now() + .milliseconds(100)) {
  self.listArray = newList
}

Answer 5

如果您按以下方式在“ SceneDelegate”文件中初始化类，则此代码将正确运行：

class SceneDelegate: UIResponder, UIWindowSceneDelegate {

var window: UIWindow?
var userData = UserData()


func scene(_ scene: UIScene, willConnectTo session: UISceneSession, options connectionOptions: UIScene.ConnectionOptions) {
    // Use this method to optionally configure and attach the UIWindow `window` to the provided UIWindowScene `scene`.
    // If using a storyboard, the `window` property will automatically be initialized and attached to the scene.
    // This delegate does not imply the connecting scene or session are new (see `application:configurationForConnectingSceneSession` instead).

    // Create the SwiftUI view that provides the window contents.
    let contentView = ContentView()

    // Use a UIHostingController as window root view controller.
    if let windowScene = scene as? UIWindowScene {
        let window = UIWindow(windowScene: windowScene)
        window.rootViewController = UIHostingController(rootView:
            contentView
            .environmentObject(userData)
        )
        self.window = window
        window.makeKeyAndVisible()
    }
}

Answer 6

寻找如何让Seitenwerk的响应适应我的解决方案，我找到了一个Binding扩展，对我有很大帮助。这是代码：

struct ContactsView: View {
    
    @State var stext : String = ""
    @State var users : [MockUser] = []
    @State var filtered : [MockUser] = []
    
    var body: some View {
        
        Form{
            SearchBar(text: $stext.didSet(execute: { (response) in
                
                if response != "" {
                    self.filtered = []
                    self.filtered = self.users.filter{$0.name.lowercased().hasPrefix(response.lowercased()) || response == ""}
                }
                else {
                    self.filtered = self.users
                }
            }), placeholder: "Buscar Contactos")
            
            List{
             ForEach(filtered, id: \.id){ user in
                    
                    NavigationLink(destination: LazyView( DetailView(user: user) )) {
                        ContactCell(user: user)
                    }
                }
            }
        }            
        .onAppear {
                self.users = LoadUserData()
                self.filtered = self.users
        }
    }
}

这是Binding扩展名：

extension Binding {
    /// Execute block when value is changed.
    ///
    /// Example:
    ///
    ///     Slider(value: $amount.didSet { print($0) }, in: 0...10)
    func didSet(execute: @escaping (Value) ->Void) -> Binding {
        return Binding(
            get: {
                return self.wrappedValue
            },
            set: {
                execute($0)
                self.wrappedValue = $0
            }
        )
    }
}

LazyView是可选的，但我花了很多时间来展示它，因为它对列表的性能有很大帮助，并阻止swiftUI创建整个列表的NavigationLink目标内容。

struct LazyView<Content: View>: View {
    let build: () -> Content
    init(_ build: @autoclosure @escaping () -> Content) {
        self.build = build
    }
    var body: Content {
        build()
    }
}

如何在SwiftUI中有效过滤长列表？

6 个答案:

更新

解决方法