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)")
控制台会输出如下内容:
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?
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,然后得到了以下代码:
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)
正如blwinters所说,在Swift 3.0中使用
URL(string: urlPath.addingPercentEncoding(withAllowedCharacters: .urlQueryAllowed)!)
我认为尝试一下对我来说是完美的选择
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!)")
如果NSURL为null并尝试通过Web视图加载http URL,则会出现以下错误:
fatal error: unexpectedly found nil while unwrapping an Optional value
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))
}
这项工作对我来说
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)
}
SWIFT 3.0
修复将不良字符串转换为NSURL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
}
在我上面的示例中,名为“ urlPath”的变量将是您已经在代码中其他位置声明的url字符串。
我遇到了这个答案,因为在将我的字符串制成NSURL的那一刻,我随机遇到了XCode中断的错误,并且报错。为什么没有随机性,即使我打印了网址也没有问题。一旦我添加了.addingPercentEncoding,它就可以正常工作而没有任何问题。
tl; dr对于阅读本文的任何人,请尝试我上面的代码,并将“ urlPath”换成您自己的本地字符串url。
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