我们有一个函数,根据一个CLLocationCoordinate2D数组创建路径,该数组包含4个坐标,将创建一个正方形形状的路径:
我们还有一个计算两个CGPoint之间中间点的函数:
我们想要的是每边获得四个点,然后移动到下一边。我尝试将函数组合起来,但它返回大约90个点。
这是如果坐标是硬编码的情况下的外观,每个边返回2个点:
任何想法如何每边只获得4个点?
func createPath(coordinates: [CLLocationCoordinate2D]) {
for (index, coordinate) in coordinates.enumerated() {
if index == 0 {
path.move(to: CGPoint(x: coordinate.latitude, y:coordinate.longitude))
} else {
path.addLine(to: CGPoint(x: coordinate.latitude, y: coordinate.longitude))
}
}
}
我们还有一个计算两个CGPoint之间中间点的函数:
func getMiddlePoint(firstPoint: CGPoint, secondPoint: CGPoint) -> CGPoint {
var middlePoint = CGPoint()
middlePoint.x = (firstPoint.x + secondPoint.x)/2
middlePoint.y = (firstPoint.y + secondPoint.y)/2
return middlePoint
}
我们想要的是每边获得四个点,然后移动到下一边。我尝试将函数组合起来,但它返回大约90个点。
func createPath(coordinates: [CLLocationCoordinate2D]) {
for (index, coordinate) in coordinates.enumerated() {
if index == 0 {
path.move(to: CGPoint(x: coordinate.latitude, y:coordinate.longitude))
} else {
path.addLine(to: CGPoint(x: coordinate.latitude, y: coordinate.longitude))//get the first point
path.addLine(to: getMiddlePoint(firstPoint: CGPoint(x: coordinate.latitude, y: coordinate.longitude), secondPoint: CGPoint(x: coordinate.latitude, y: coordinate.longitude))) // get the middle point
}
}
}
这是如果坐标是硬编码的情况下的外观,每个边返回2个点:
func createPath(){
path.move(to: CGPoint(x: 32.7915055, y: -96.8028408))//1
path.addLine(to: CGPoint(x: 32.79174845, y: -96.80252195))//midpoint between 1 and 2
path.addLine(to: CGPoint(x: 32.7919914, y: -96.8022031))//2
path.addLine(to: CGPoint(x: 32.791501100000005, y: -96.80235195))////midpoint between 2 and 3
path.addLine(to: CGPoint(x: 32.7910108, y: -96.8025008))//3
path.addLine(to: CGPoint(x: 32.791301700000005, y: -96.8020985))//midpoint between 3 and 4
path.addLine(to: CGPoint(x: 32.7915926, y: -96.8016962))//4
path.addLine(to: CGPoint(x: 32.79154905, y: -96.8022685))//midpoint between 4 and 1
}
任何想法如何每边只获得4个点?