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ios - Swift,无法读取url中的数据

转载 作者:搜寻专家 更新时间:2023-11-01 07:30:02 35 4
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我运行了这个,但当我尝试发送数据时没有任何反应。所以我调试了这个,它无法从 url 中读取任何内容。它说“无法读取数据”。我检查了 url 是否正确,我检查了这个服务器是否正常以及我的 php 代码。只有当我升级到 swift 2 或 Xcode7 时,我才会遇到这个问题。感谢您的帮助!

let myUrl = NSURL(string: "http://localhost/SwiftAppAndMySQL/scripts/registerUser.php");
let request = NSMutableURLRequest(URL:myUrl!);
request.HTTPMethod = "POST";

let postString = "userEmail=\(userEmail!)&userFirstName=\(userFirstName!)&userLastName=\(userLastName!)&userPassword=\(userPassword!)";

request.HTTPBody = postString.dataUsingEncoding(NSUTF8StringEncoding);

NSURLSession.sharedSession().dataTaskWithRequest(request, completionHandler: { (data:NSData?, response:NSURLResponse?, error:NSError?) -> Void in

dispatch_async(dispatch_get_main_queue())
{

//spinningActivity.hide(true)

if error != nil {
self.displayAlertMessage(error!.localizedDescription)
return
}

do {
let json = try NSJSONSerialization.JSONObjectWithData(data!, options: .MutableContainers) as? NSDictionary

if let parseJSON = json {

let userId = parseJSON["userId"] as? String

if( userId != nil)
{
let myAlert = UIAlertController(title: "Alert", message: "Registration successful", preferredStyle: UIAlertControllerStyle.Alert);

let okAction = UIAlertAction(title: "OK", style: UIAlertActionStyle.Default){(action) in

self.dismissViewControllerAnimated(true, completion: nil)
}

myAlert.addAction(okAction);
self.presentViewController(myAlert, animated: true, completion: nil)
} else {
let errorMessage = parseJSON["message"] as? String
if(errorMessage != nil)
{
self.displayAlertMessage(errorMessage!)
}

}

}
} catch{
print(error)
}

}

}).resume()

最佳答案

I experience this problem only when I upgraded to swift 2 or Xcode7.

Xcode 7 包含带有 iOS 9 的 iOS 模拟器。iOS 9 有一个新功能——App Transport Security (ATS),它可以防止不安全的连接 (http)。更多信息:

关于ios - Swift,无法读取url中的数据,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/33132835/

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