如何在SwiftUI中为按钮创建触觉反馈？
如何在SwiftUI中为按钮创建触觉反馈？
提问于 2019-06-25 15:06:15
在SwiftUI中，我正在尝试在点击按钮时实现触觉反馈。因此，我正在尝试使用simultaneousGesture，但我仍然在努力。我想不出水龙头什么时候开始响。
也没有为Swift UI实现触觉反馈，所以我猜我会从UIKit中混合它？
我试图实现TapGesture的更新方法，但它似乎什么也做不了。这就是我到目前为止所得到的。谢谢你的任何提示。
struct HapticButton : View {
@GestureState var isDetectingTap = false
var body: some View {
let tap = TapGesture()
.updating($isDetectingTap) { (body, stateType, transaction) in
// nothing happens below
print(body)
print(stateType)
print(transaction)
}.onEnded { _ in
// this one works but it is to late
// I need to figure out the beginning of the tap
print("Button was tapped, will invoke haptic feedback, maybe with UIKit")
}
return Button(action: {
print("Action executed")
}) {
HStack {
Image("icon")
Text("Login")
}
}.simultaneousGesture(tap)
}
}回答 4
Stack Overflow用户
发布于 2019-09-27 00:06:26
下面是纯SwiftUI解决方案：
HStack {
Image("icon")
Text("Login")
}
.onTouchGesture(
touchBegan: { self.generateHapticFeedback = true },
touchEnd: { _ in self.generateHapticFeedback = false }
)只需将此代码片段添加到您的项目中的某个位置：
struct TouchGestureViewModifier: ViewModifier {
let touchBegan: () -> Void
let touchEnd: (Bool) -> Void
@State private var hasBegun = false
@State private var hasEnded = false
private func isTooFar(_ translation: CGSize) -> Bool {
let distance = sqrt(pow(translation.width, 2) + pow(translation.height, 2))
return distance >= 20.0
}
func body(content: Content) -> some View {
content.gesture(DragGesture(minimumDistance: 0)
.onChanged { event in
guard !self.hasEnded else { return }
if self.hasBegun == false {
self.hasBegun = true
self.touchBegan()
} else if self.isTooFar(event.translation) {
self.hasEnded = true
self.touchEnd(false)
}
}
.onEnded { event in
if !self.hasEnded {
let success = !self.isTooFar(event.translation)
self.touchEnd(success)
}
self.hasBegun = false
self.hasEnded = false
})
}
}
extension View {
func onTouchGesture(touchBegan: @escaping () -> Void = {},
touchEnd: @escaping (Bool) -> Void = { _ in }) -> some View {
modifier(TouchGestureViewModifier(touchBegan: touchBegan, touchEnd: touchEnd))
}
}Stack Overflow用户
发布于 2019-06-27 07:31:59
您可以像这样使用UIFeedbackGenerator：
let generator = UINotificationFeedbackGenerator()
generator.notificationOccurred(.error)或者，当您使用SwiftUI时，您可以像这样使用CoreHaptics：
let engine = try CHHapticEngine()
try engine.start()
let hapticEvent = CHHapticEvent(eventType: .hapticTransient, parameters: [
CHHapticEventParameter(parameterID: .hapticSharpness, value: sharpness), CHHapticEventParameter(parameterID: .hapticIntensity, value: intensity),
], relativeTime: 0)
let audioEvent = CHHapticEvent(eventType: .audioContinuous, parameters: [
CHHapticEventParameter(parameterID: .audioVolume, value: volume),
CHHapticEventParameter(parameterID: .decayTime, value: decay),
CHHapticEventParameter(parameterID: .sustained, value: 0),
], relativeTime: 0)
let pattern = try CHHapticPattern(events: [hapticEvent, audioEvent], parameters: [])
let hapticPlayer = try engine.makePlayer(with: pattern)
try hapticPlayer?.start(atTime: CHHapticTimeImmediate)Stack Overflow用户
发布于 2021-06-23 02:00:18
您还可以将Haptic生成器封装在单例中：
import UIKit
class Haptics {
static let shared = Haptics()
private init() { }
func play(_ feedbackStyle: UIImpactFeedbackGenerator.FeedbackStyle) {
UIImpactFeedbackGenerator(style: feedbackStyle).impactOccurred()
}
func notify(_ feedbackType: UINotificationFeedbackGenerator.FeedbackType) {
UINotificationFeedbackGenerator().notificationOccurred(feedbackType)
}
}调用站点：
// As a button action
Button(action: {
// some action
Haptics.shared.play(.heavy)
}, label: {
Text("Button")
})
// As a tap gesture on a View
SomeView()
.onTapGesture(perform: {
Haptics.shared.play(.light)
})
// In a function
func someFunc() {
Haptics.shared.notify(.success)
}选项：
Haptics.shared.play(.heavy)
Haptics.shared.play(.light)
Haptics.shared.play(.medium)
Haptics.shared.play(.rigid)
Haptics.shared.play(.soft)
Haptics.shared.notify(.error)
Haptics.shared.notify(.success)
Haptics.shared.notify(.warning)页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/56748539
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