测验应用与arrays没有重复的问题
这可能是一个更好的方式，而不必像我一样重复代码
我试图从一个数组中重复生成一个不同的问题。 但是，每当NextQuestion动作按下几次时，它会给我几次相同的问题，然后转到下一个，或者其他时候只按下一个，然后转到下一个。 我知道这是最好的代码，但我还在学习。
import UIKit class TestQuestionsVC: UIViewController { @IBOutlet weak var SignImage: UIImageView! @IBOutlet weak var AnswerBtn_A: MaterialButton! @IBOutlet weak var AnswerBtn_B: MaterialButton! @IBOutlet weak var AnswerBtn_C: MaterialButton! @IBOutlet weak var AnswerBtn_D: MaterialButton! var questionModel = QuestionModel() var GuessQuestionNum = "" var answer: String = "" var btnLatter = "" var previousNumber: UInt32? // used in randomNumber() override func viewDidLoad() { super.viewDidLoad() for _ in 1...5 { randomQuestion(queModel: questionModel) } } func randomQuestion(queModel queModel:QuestionModel){ let questions = queModel.questions let randomIndex = Int(arc4random_uniform(UInt32(questions.count))) answer = "\(questions[randomIndex].answers[questions[randomIndex].correctAnswerIndex])" SignImage.image = questions[randomIndex].question AnswerBtn_A.setTitle("\(questions[randomIndex].answers[0])", forState: .Normal) AnswerBtn_B.setTitle("\(questions[randomIndex].answers[1])", forState: .Normal) AnswerBtn_C.setTitle("\(questions[randomIndex].answers[2])", forState: .Normal) AnswerBtn_D.setTitle("\(questions[randomIndex].answers[3])", forState: .Normal) } override func didReceiveMemoryWarning() { super.didReceiveMemoryWarning() // Dispose of any resources that can be recreated. } @IBAction func ActBtn_A(sender: AnyObject) { GuessQuestionNum = (AnswerBtn_A.titleLabel?.text)! if GuessQuestionNum == answer { AnswerBtn_A.backgroundColor = GREEN_COLOR } else { AnswerBtn_A.backgroundColor = RED_COLOR } } @IBAction func ActBtn_B(sender: AnyObject) { GuessQuestionNum = (AnswerBtn_B.titleLabel?.text)! if GuessQuestionNum == answer { AnswerBtn_B.backgroundColor = GREEN_COLOR } else { AnswerBtn_B.backgroundColor = RED_COLOR } } @IBAction func ActBtn_C(sender: AnyObject) { GuessQuestionNum = (AnswerBtn_C.titleLabel?.text)! if GuessQuestionNum == answer { AnswerBtn_C.backgroundColor = GREEN_COLOR } else { AnswerBtn_C.backgroundColor = RED_COLOR } } @IBAction func ActBtn_D(sender: AnyObject) { GuessQuestionNum = (AnswerBtn_D.titleLabel?.text)! if GuessQuestionNum == answer { AnswerBtn_D.backgroundColor = GREEN_COLOR } else { AnswerBtn_D.backgroundColor = RED_COLOR } } @IBAction func NextQuestion(sender: AnyObject) { AnswerBtn_A.backgroundColor = BLUE_COLOR AnswerBtn_B.backgroundColor = BLUE_COLOR AnswerBtn_C.backgroundColor = BLUE_COLOR AnswerBtn_D.backgroundColor = BLUE_COLOR for _ in 1...5 { randomQuestion(queModel: questionModel) } } func randomQuestion( queModel:QuestionModel){ let questions = queModel.questions func randomNumber() -> UInt32 { var randomNumber = arc4random_uniform(UInt32(questions.count)) while previousNumber == randomNumber { randomNumber = arc4random_uniform(UInt32(questions.count)) } previousNumber = randomNumber return randomNumber } let randomIndex = Int((UInt32(randomNumber()))) answer = "\(questions[randomIndex].answers[questions[randomIndex].correctAnswerIndex])" SignImage.image = questions[randomIndex].question AnswerBtn_A.setTitle("\(questions[randomIndex].answers[0])", forState: .Normal) AnswerBtn_B.setTitle("\(questions[randomIndex].answers[1])", forState: .Normal) AnswerBtn_C.setTitle("\(questions[randomIndex].answers[2])", forState: .Normal) AnswerBtn_D.setTitle("\(questions[randomIndex].answers[3])", forState: .Normal) } } 有很多方法来处理这个问题。 我的build议是创build一个问题索引数组。 从索引数组中随机删除一个对象，并使用它来获取问题。 示例代码：
import Foundation let array = ["one", "two", "three", "four", "five", "six", "seven", "eight", "nine", "ten"] var indexes = [Int](); func randomItem() -> String { if indexes.count == 0 { print("Filling indexes array") indexes = (0...array.count-1).map{$0} } let randomIndex = Int(arc4random_uniform(UInt32(indexes.count))) let anIndex = indexes.removeAtIndex(randomIndex) return array[anIndex]; } for i in 1...20 { print("random word = \(randomItem())") }
randomItem（）函数被写入来重新填充索引数组，一旦它是空的，所以你会得到一组新的随机问题。
请注意，在重新填充数组后，可能会连续两次获取相同的问题，因为新数组将包含所有可能的索引，包括最后一个使用的索引。 你需要添加额外的逻辑来防止这种情况。