iOS在按下“下一步”按钮时加载新行表格解析Swift

Question

我尝试创建一些应用程序，但是我遇到了问题，我有这样的结构，问题和答案。

我的课：

class QuestionManager{
 var question: String?
 var answerA: String?
 var answerB: String?
 var answerC: String?
 var answerD: String?

}

我的代码：

var questions = [QuestionManager]()
var arrayIndex = 0
let query = PFQuery(className: "test")
query.findObjectsInBackground{ (objects, error) in
if error == nil && objects != nil {
    for object in objects as? [PFObject]{
          self.questions.append(QuestionManager(
          question: object!["question"] as! String,
          answerA: object!["answer1"] as! String,
          answerB: object!["answer2"] as! String,
          answerC: object!["answer3"] as! String,
          answerD: object!["answer4"] as! String

          ))
    }
} else {
    print("Error")
}

使用此代码会出现错误：

  

无法使用类型为（（QuestionManager）''的参数列表调用“追加”

为什么我得到这个错误，也许你会看到我的错误

Answer 1

替换此

var questions = [QuestionManager]()
var arrayIndex = 0
let query = PFQuery(className: "test")
query.findObjectsInBackground{ (objects, error) in
if error == nil && objects != nil {
    for object in objects as? [PFObject] {

          let objQuestionManager = QuestionManager()

          objQuestionManager.question = object!["question"] as! String
          objQuestionManager.answerA = object!["answer1"] as! String
          objQuestionManager.answerB = object!["answer2"] as! String
          objQuestionManager.answerC = object!["answer3"] as! String
          objQuestionManager.answerD = object!["answer4"] as! String

          self.questions.append(objQuestionManager)
    }
} else {
    print("Error")
}

Answer 2

向带有4个输入值的类中添加初始化程序。

 class QuestionManager{
     var question: String?
     var answerA: String?
     var answerB: String?
     var answerC: String?
     var answerD: String?


    init(quest: String?, ansA: String? , ansB: String? , ansC: String? , ansD: String){
        question = ques
  answerA = ansA
  answerB = ansB
  answerC = ansC
  answerD = ansD


    }

    }

Answer 3

由于没有在Model类中编写init（）方法，并且正在尝试初始化模型类，因此出现此错误。您可以这样写。

class QuestionManager{
    var question = ""
    var answerA = ""
    var answerB = ""
    var answerC = ""
    var answerD = ""

    init(question: String, answerA: String, answerB: String, answerC: String, answerD: String){

        self.question = question
        self.answerA = answerA
        self.answerB = answerB
        self.answerC = answerC
        self.answerD = answerD
    }
}