Question: When I am getting a value back from an API that I am hitting, when I print the data it appears as 

```
Optional((
  jknjknjkn
))

```

for example. I use a selector to run a method when the data is returned

```
func result(data: AnyObject){
   println(data["info"])
}

```

What is printed is the Optional thing above. How can I get the value without that Optional() thing wrapping it?

Here is the raw data if I just print data

```
<AppUser: 0x78ec5650, objectId: I6nxFSZx9h, localId: (null)> {
AppID =     (
    jknkjnjknkn
);
info =     (
    jknjknjkn
);
}

```

Answer: 
In Swift, an optional is a variable that can either hold a value or `nil` (no value at all). To get the value from it, you have to *unwrap* it with an exclamation point:

```
func result(data: AnyObject){
   println(data["info"]!)
}

```

Note that if `data["info"]` was `nil`, your app would crash with this error message:

```
fatal error: unexpectedly found nil while unwrapping Optional value

```

If you are concerned that the expression might produce `nil`, you can use *optional binding*:

```
func result(data: AnyObject){
   if let info = data["info"] {    //info is now the unwrapped version of `data["info"]
        println(info)              //will only be executed if data["info"] is not nil
    }
}

```
Disregard this, I am trying to use Parse in a swift project and their documentation has multiple ways to show the code. I just copied a different code block and now I am able to do what I need to do just fine.