Error when trying to get Json data from php script - php

Question

I get the following error when i try to get data. In the internet i read that its because the php script is invalid and don't return json data. But the php script runs fine and outputs the right data.
Error Message :
Error Domain=NSCocoaErrorDomain Code=3840 "JSON text did not start with array or object and option to allow fragments not set." UserInfo={NSDebugDescription=JSON text did not start with array or object and option to allow fragments not set.}
I tried to allow fragments but then i get just another error message.
Here is the swift code where i try to get the data :
let myUrl = NSURL(string: "http://xxxxxxxxxxx.xxx/xxxxxxxx.php")
let request = NSMutableURLRequest(URL: myUrl!)
request.HTTPMethod = "POST"
let postString = "userEmail=\(userEmail!)&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())
{
if(error != nil)
{
var alert = UIAlertController(title: "Achtung", message: error?.localizedDescription, preferredStyle: UIAlertControllerStyle.Alert)
let action = UIAlertAction(title: "Ok", style: UIAlertActionStyle.Default, handler: nil)
alert.addAction(action)
self.presentViewController(alert, animated: true, completion: nil)
}
print("1")
do {
let json = try NSJSONSerialization.JSONObjectWithData(data!, options: .MutableContainers) as? NSDictionary
if let parseJSON = json {
let userId = parseJSON["userId"] as? String
if( userId != nil)
{
print("SUCESS FUCKER")
let mainView = self.storyboard?.instantiateViewControllerWithIdentifier("main") as! FlickrPhotosViewController
let mainPageNavi = UINavigationController(rootViewController: mainView)
//open mainView
let appdele = UIApplication.sharedApplication().delegate
appdele?.window??.rootViewController = mainPageNavi
} else {
let userMassage = parseJSON["message"] as? String
let myAlert = UIAlertController(title: "Alert", message: userMassage, preferredStyle: UIAlertControllerStyle.Alert);
let okAction = UIAlertAction(title: "OK", style: UIAlertActionStyle.Default, handler: nil)
myAlert.addAction(okAction);
self.presentViewController(myAlert, animated: true, completion: nil)
}
}
} catch{
print(error)
print("FAILED CATCHED")
}
}
}).resume()
and this is the important part of the php file :
$userSecuredPassword = $userDetails["user_password"];
$userSalt = $userDetails["salt"];
if($userSecuredPassword === sha1($userPassword . $userSalt))
{
$returnValue["status"]="200";
$returnValue["userFirstName"] = $userDetails["first_name"];
$returnValue["userLastName"] = $userDetails["last_name"];
$returnValue["userEmail"] = $userDetails["email"];
$returnValue["userId"] = $userDetails["user_id"];
} else {
$returnValue["status"]="403";
$returnValue["message"]="User not found";
echo "failed";
echo json_encode($returnValue);
return;
}
echo json_encode($returnValue);
$returnValue returns this when i print it:
Array ( [status] => 200 [userFirstName] => Paul [userLastName] => Heinemeyer [userEmail] => paul_heine [userId] => 63 )

Answer 1

When you properly format your PHP code, you will see, that in the else part you have
echo "failed";
echo json_encode($returnValue);
which results in
failed{...}
As the error message already says, this "JSON text did not start with array or object ..."
Maybe there is similar output for the other if part.