这是我的错误消息:
无法调用' getObjectInBackgroundWithId'使用类型'的参数列表(String,(PFObject!,NSError!) - > Void)'
这就是我的代码:
func CallData(){
var query : PFQuery = PFQuery(className: "QuestionsandAnswers")
query.getObjectInBackgroundWithId("iXCrgCLKk2"){
(Object : PFObject!, error : NSError!) -> Void in
if (error == nil){
self.Question = Object["Question"] as String!
self.Answers = Object["Answers"] as Array!
self.Answer = Object["Answer"] as String!
if (self.Answers.count > 0){
self.QuestionLabel.text = self.Question
}
} else {
NSLog("There is something wrong!")
}
}
}
我尝试更改感叹号/问号/ PFObject
和NSError
后面没有任何内容,但没有任何效果。但是,如果我在PFObject
和NSError
后面加上一个问号,我会收到一条新的错误消息:
' AnyObject'不能转换为' String!&#39 ;;你的意思是使用'作为!'迫使低垂?
如果我将代码更改为as!
,那么我会收到此错误:
无法指定类型'字符串的不可变值!'
有关如何解决此问题的任何想法?任何帮助表示赞赏!
答案 0 :(得分:0)
看看: http://blog.parse.com/announcements/building-apps-with-parse-and-swift/
我认为这会有用
var query = PFQuery(className: "QuestionsandAnswers")
query.getObjectInBackgroundWithId("iXCrgCLKk2", block: { (object:PFObject?, error:NSError?) -> Void in
if let error = error {
//error
} else if let object = object{
//do something with your object as PFObject
// such as
var Question:String = object.objectForKey("Question") as! String
}
})