ios 将字符串转换为 NSURL 在 swift 中返回 nil
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Convert String to NSURL is return nil in swift
提问by Dharmesh Kheni
I am trying to convert a Stringto NSURLand my code for that is Below:
我正在尝试将 a 转换String为NSURL，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)")
And console prints something like this:
控制台打印如下内容：
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
The remoteUrlis niland I don't know what is the problem here.
的remoteUrl是nil，我不知道这里有什么问题。
After that I try sort Stringlike this:
之后，我尝试这样排序String：
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)")
And console prints:
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)
This is working fine.
这工作正常。
So can anybody please tell me what is wrong with my first case?
那么有人可以告诉我我的第一个案例有什么问题吗？
回答by Dharmesh Kheni
As suggested by the Martin R, I see THISpost and I converted that objective-c code to swift and I got this code:
正如 Martin R 所建议的那样，我看到了这篇文章，我将目标 C 代码转换为 swift 并得到了以下代码：
var url : NSString = "https://maps.googleapis.com/maps/api/distancematrix/json?origins=\(self.latitud??e),\(self.longitude)&destinations=\(self.stringForDistance)&language=en-US"
var urlStr : NSString = url.stringByAddingPercentEscapesUsingEncoding(NSUTF8StringEncoding)!
var searchURL : NSURL = NSURL(string: urlStr)!
println(searchURL)
and this is working correctly.
这工作正常。
For swift 3.0:
对于 Swift 3.0：
let url : NSString = "https://maps.googleapis.com/maps/api/distancematrix/json?origins=\(self.latitud??e),\(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)
回答by djdance
as blwinters said, in Swift 3.0use
正如 blwinters 所说，在Swift 3.0 中使用
URL(string: urlPath.addingPercentEncoding(withAllowedCharacters: .urlQueryAllowed)!)
回答by Lizzeth Zaragoza
this work for me
这对我有用
let url : NSString = MyUrls.baseUrl + self.url_file_open as NSString
let urlStr : NSString = url.addingPercentEscapes(using: String.Encoding.utf8.rawValue)! as NSString
if let url = URL(string: urlStr as String) {
let request = URLRequest(url: url)
self.businessPlanView.loadRequest(request)
}
回答by Ilesh P
I think try this it's perfectly work for me
我认为试试这个它对我来说非常有用
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 urlStr : NSString = url.stringByAddingPercentEscapesUsingEncoding(NSUTF8StringEncoding)!
var remoteUrl : NSURL? = NSURL(string: url)
println("This is URL: \(remoteUrl!)")
回答by Arvind
You can get following error if NSURL is null and trying load http URL over web view:
如果 NSURL 为 null 并尝试通过 Web 视图加载 http URL，您可能会收到以下错误：
fatal error: unexpectedly found nil while unwrapping an Optional value
To be safe we should use:
为了安全起见，我们应该使用：
var urlStr = strLink!.stringByAddingPercentEscapesUsingEncoding(NSUTF8StringEncoding)
if var url = NSURL(string: urlStr!){
println(self.strLink!)
self.webView!.loadRequest(NSURLRequest(URL: url))
}
回答by Matthew Zourelias
SWIFT 3.0
斯威夫特 3.0
A safe way to fix a bad string being converted to NSURL is by unwrapping the urlPath string variable using "guard let"
修复转换为 NSURL 的错误字符串的安全方法是使用“guard let”解开 urlPath 字符串变量
guard let url = NSURL(string: urlPath.addingPercentEncoding(withAllowedCharacters: .urlQueryAllowed)!)
else
{
print("Couldn't parse myURL = \(urlPath)")
return
}
The variable called "urlPath" in my above example would be the url string you have already declared somewhere else in your code.
在我上面的示例中，名为“urlPath”的变量将是您已经在代码中的其他地方声明的 url 字符串。
I came across this answer because randomly I was getting the nil error with XCode breaking at the point my string was made into a NSURL. No logic as to why it was random even when I printed the url's they would look fine. As soon as I added the .addingPercentEncoding it was back working without any issues whatsoever.
我遇到了这个答案，因为在我的字符串被转换为 NSURL 时，我随机收到了 XCode 中断的 nil 错误。没有逻辑为什么它是随机的，即使我打印了 url，它们看起来也不错。一旦我添加了 .addingPercentEncoding 它就可以正常工作了，没有任何问题。
tl;dr For anyone reading this, try my above code and swap out "urlPath" for your own local string url.
tl; dr 对于阅读本文的任何人，请尝试我上面的代码并将“urlPath”替换为您自己的本地字符串 url。
