我跑这个,当我尝试发送数据时没有任何反应。所以我调试了这个,它不能从URL中读取任何东西。它说“无法读取数据”。我检查了它的URL是否正确,我检查了它的好处以及我的php代码。我仅在升级到swift 2或Xcode7时遇到此问题。谢谢您的帮助!Swift,无法读取url中的数据
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()
查找ATS,并期待ifau的答复。我认为他的回答很大程度上帮助你 – Lorenzo