Converting PFObject (Parse) into JSON in Swift? - ios

Question

Is there a way to convert a PFObject from Parse into JSON? I saved as JSON, but when I try to load I'm getting [AnyObject] back. Casting to JSON won't work:
class func loadPeople() -> [String : Person] {
var peopleDictionary: [String : Person] = [:]
let query = PFQuery(className: "userPeeps")
query.findObjectsInBackgroundWithBlock { (objects, error) -> Void in
if error == nil {
//this only returns the first entry, how do I get them all?
if let peopleFromParse = objects?.first?.objectForKey("userPeeps") as? JSON {
for name in peopleFromParse.keys {
if let personJSON = peopleFromParse[name] as? JSON,
let person = Person(json: personJSON) {
peopleDictionary[name] = person
}
}
}
below is my save function, which works and saves the JSON into Parse like I want:
class DataManager {
typealias JSON = [String: AnyObject]
class func savePeople(people: [String : Person]) {
var peopleDictionary = people
var peopleJSON: JSON = [:]
for name in peopleDictionary.keys {
peopleJSON[name] = peopleDictionary[name]!.toJSON()
}
let userPeeps = PFObject(className: "userPeeps")
userPeeps.setObject(peopleJSON, forKey: "userPeeps")
userPeeps.saveInBackgroundWithBlock { (succeeded, error) -> Void in
if succeeded {
println("Object Uploaded")
} else {
println("Error: \(error) \(error!.userInfo!)")
}
}
}

Answer 1

So the answer (as Paulw11 points out above) is that "objects" is sort of a wrapper for the real data, so it was necessary to iterate through the array and store each value as JSON:
var peopleDictionary: [String : Person] = [:]
//1 load the dictionary of JSON for key people from Parse
let query = PFQuery(className: "userPeeps")
query.findObjectsInBackgroundWithBlock { (objects, error) -> Void in
if error == nil {
if let unwrappedObjects = objects {
for object in unwrappedObjects {
if let peopleFromParse = object as? JSON {
for name in peopleFromParse.keys {
if let personJSON = peopleFromParse[name] as? JSON,
let person = Person(json: personJSON) {
peopleDictionary[name] = person
}
}
}
}
}