将String转换为NSURL迅速返回nil

一尘不染

将String转换为NSURL迅速返回nil

swift

我正在尝试将转换为String，NSURL而我的代码如下：

var url = "https://maps.googleapis.com/maps/api/distancematrix/json?origins=\(self.latitude),\(self.longitude)&destinations=\(self.stringForDistance)&language=en-US"
    println("This is String: \(url)")
    var remoteUrl : NSURL? = NSURL(string: url)
    println("This is URL: \(remoteUrl)")

控制台会打印如下内容：

This is String: https://maps.googleapis.com/maps/api/distancematrix/json?origins=-34.4232722,150.8865837&destinations=-34.4250728,150.89314939999997|-34.4356434,150.8858692|-34.4250728,150.89314939999997|-34.4356434,150.8858692|-34.4250728,150.89314939999997|-34.4356434,150.8858692|-34.423234,150.88658899999996|-34.423234,150.88658899999996|-34.428251,150.899673|-34.4257439,150.89870229999997|-34.423234,150.88658899999996|-34.4257439,150.89870229999997|-34.425376,150.89388299999996&language=en-US

This is URL: nil

的remoteUrl是nil，我不知道这里有什么问题。

之后，我尝试String像这样进行排序：

var url : String = "https://maps.googleapis.com/maps/api/distancematrix/json?origins=-34.4232722,150.8865837&destinations=-34.4250728,150.89314939999997&language=en-US"
    println("This is String: \(url)")
    var remoteUrl : NSURL? = NSURL(string: url)
    println("This is URL: \(remoteUrl)")

和控制台打印：

This is String: https://maps.googleapis.com/maps/api/distancematrix/json?origins=-34.4232722,150.8865837&destinations=-34.4250728,150.89314939999997&language=en-US
This is URL: Optional(https://maps.googleapis.com/maps/api/distancematrix/json?origins=-34.4232722,150.8865837&destinations=-34.4250728,150.89314939999997&language=en-US)

一切正常。

那么，有人可以告诉我我的第一个案例有什么问题吗？

阅读 303

2020-07-07

共1个答案

一尘不染

正如Martin R所建议的，我看到了这篇文章，并将那个Objective-
C代码转换为swift并且得到了以下代码：

var url : NSString = "https://maps.googleapis.com/maps/api/distancematrix/json?origins=\(self.latitud‌e),\(self.longitude)&destinations=\(self.stringForDistance)&language=en-US" 
var urlStr : NSString = url.stringByAddingPercentEscapesUsingEncoding(NSUTF8StringEncoding)! 
var searchURL : NSURL = NSURL(string: urlStr)! 
println(searchURL)

并且这正常工作。

对于Swift 3.0：

let url : NSString = "https://maps.googleapis.com/maps/api/distancematrix/json?origins=\(self.latitud‌e),\(self.longitude)&destinations=\(self.stringForDistance)&language=en-US"
let urlStr : NSString = url.addingPercentEscapes(using: String.Encoding.utf8.rawValue)! as NSString
let searchURL : NSURL = NSURL(string: urlStr as String)!
print(searchURL)

2020-07-07