在Swift 3中将可选字符串转换为double [英] Convert optional string to double in Swift 3

问题描述

我有一个选项字符串，想将其转换为双精度.
这适用于Swift 2，但是自从转换为Swift 3以来，我的值为0.

I have a option string and want to convert that to double.
this worked in Swift 2 , but since converted to Swift 3, I am getting value of 0.

var dLati = 0.0
dLati = (latitude as NSString).doubleValue

我已检查，并且纬度具有可选的字符串值，例如-80.234543218675654，但dLati值为0

***************好的，为清楚起见进行了新更新*****************
我有一个其中有一个按钮的viewcontroller，当触摸该按钮时，它将调用另一个viewcontroller并将一些值传递给它 这是第一个viewcontroller

I have check and latitude has a optional string value of something like -80.234543218675654 , but dLati value is 0

*************** ok, new update for clarity *****************
I have a viewcontroller which i have a button in it, and when the button is touched, it will call another viewcontroller and pass a few values to it here is the code for the first viewcontroller

var currentLatitude: String? = ""
var currentLongitude: String? = ""
var deviceName = ""
var address = ""
// somewhere in the code, currentLatitude and currentLongitude are get set  
override func prepare(for segue: UIStoryboardSegue, sender: Any?) {
    if segue.identifier == "map" {
       let destViewController : MapViewController = segue.destination as! MapViewController
       print(currentLongitude!)  // Print display: Optional(-80.192279355363768)
       print(currentLatitude!) // Print display: Optional(25.55692663937162) 
       destViewController.longitude = currentLongitude!
       destViewController.latitude = currentLatitude!
       destViewController.deviceName = deviceName
       destViewController.address = address
    }
}

这是第二个名为MapViewController的视图控制器的代码

Here is the code for the second view controller called MapViewController

   var longitude: String? = " "
   var latitude: String? = ""
   .
   .
   override func viewDidLoad() {
       if let lat = latitude {
          print(lat) // Print display: optiona(25.55692663937162)
          dLati = (lat as NSString).doubleValue
          print(dLati)  // Print display: 0.0
       }
       .
       .
   }

谢谢 博尔纳

推荐答案

一种无需使用Foundation类型即可实现此目的的安全方法是使用Double的初始化程序:

A safe way to achieve this without needing to use Foundation types is using Double's initializer:

if let lat = latitude, let doubleLat = Double(lat) {
  print(doubleLat)  // doubleLat is of type Double now
}

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