我正在尝试快速执行以下操作 - 尝试反向解码数组中的地址列表并打印它们的纬度/经度坐标。我的代码如下。
let addressArray = ["Address 1", "Address 2"]
var coordinatesArray = [CLLocationCoordinate2D]()
override func viewDidLoad() {
super.viewDidLoad()
createAddressList()
printAddressList()
}
func printAddressList() {
for i in 0 ..< addressArray.count {
print("Address = \(addressArray[i]) Coordinates = \(coordinatesArray[i].latitude),\(coordinatesArray[i].latitude)")
}
func createAddressList() {
for i in 0 ..< addressArray.count {
let address = addressArray[i]
geocoder.geocodeAddressString(address, completionHandler: {(placemarks, error) -> Void in
print("Address = \(address)");
if let placemark = placemarks?.first {
let coordinate = placemark.location?.coordinate
self.coordinatesArray.append(coordinate!)
}
})
}
}
}
代码只打印第一个被解码的地址,然后什么也没有发生。
我确实有一个像下面这样的修复方法,就是像这样从 viewDidLoad 方法中移动 printAddressList 调用
func createAddressList() {
if count < self.addressArray.count {
let address = addressArray[count]
geocoder.geocodeAddressString(address, completionHandler: {(placemarks, error) -> Void in
print("Address = \(address)");
if let placemark = placemarks?.first {
let coordinate = placemark.location?.coordinate
self.coordinatesArray.append(coordinate!)
}
print("Count = \(self.count)")
self.count += 1
self.createAddressList()
})
} else {
printAddressList()
}
}
尽管后一种解决方案有效,但我发现它并不干净,想知道在使代码可读和干净的同时执行此操作的正确方法。
最佳答案
使用这个结构怎么样?
let workGroup = dispatch_group_create()
for i in 0..<addressArray.count {
dispatch_group_enter(workGroup)
performGeoCoding({ successCallback :
dispatch_group_leave(workGroup)
})
}
dispatch_group_notify(workGroup, dispatch_get_main_queue()){
successCallback()
printAddressList()
}
关于 dispatch_group 有很好的教程 here .
关于ios - Swift:for循环中的异步调用,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/39633807/