'Limit rectangle to screen edge on drag gesture
I'm just getting started with SwiftUI and I was hoping for the best way to tackle the issue of keeping this rectangle in the bounds of a screen during a drag gesture. Right now it goes off the edge until it reaches the middle of the square (I think cause I'm using CGPoint).
I tried doing some math to limit the rectangle and it succeeds on the left side only but it seems like an awful way to go about this and doesn't account for varying screen sizes. Can anyone help?
struct ContentView: View {
@State private var pogPosition = CGPoint()
var body: some View {
PogSound()
.position(pogPosition)
.gesture(
DragGesture()
.onChanged { value in
self.pogPosition = value.location
// Poor solve
if(self.pogPosition.x < 36) {
self.pogPosition.x = 36
}
}
.onEnded { value in
print(value.location)
}
)
}
}
Solution 1:[1]
Here is a demo of possible approach (for any view, cause view frame is read dynamically).
Demo & tested with Xcode 12 / iOS 14
struct ViewSizeKey: PreferenceKey {
static var defaultValue = CGSize.zero
static func reduce(value: inout Value, nextValue: () -> Value) {
value = nextValue()
}
}
struct ContentView: View {
@State private var pogPosition = CGPoint()
@State private var size = CGSize.zero
var body: some View {
GeometryReader { gp in
PogSound()
.background(GeometryReader {
Color.clear
.preference(key: ViewSizeKey.self, value: $0.frame(in: .local).size)
})
.onPreferenceChange(ViewSizeKey.self) {
self.size = $0
}
.position(pogPosition)
.gesture(
DragGesture()
.onChanged { value in
let rect = gp.frame(in: .local)
.inset(by: UIEdgeInsets(top: size.height / 2.0, left: size.width / 2.0, bottom: size.height / 2.0, right: size.width / 2.0))
if rect.contains(value.location) {
self.pogPosition = value.location
}
}
.onEnded { value in
print(value.location)
}
)
.onAppear {
let rect = gp.frame(in: .local)
self.pogPosition = CGPoint(x: rect.midX, y: rect.midY)
}
}.edgesIgnoringSafeArea(.all)
}
}
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
Solution | Source |
---|---|
Solution 1 |