无法解决这个问题？您将如何解决？ “条件中的可变绑定需要初始化程序”和“如果'if'之后的预期'{'”

Question

标题概括了我的问题。我是编码的新手，不了解如何解决问题。

我已经尽力了。

import UIKit
import AVKit
import AVFoundation

class ViewController: UIViewController {

    override func viewDidAppear(_ animated: Bool) {
        super.viewDidAppear(animated)
    }

    @IBOutlet weak var PowerLVLlabel: UILabel!

    override func viewDidLoad() {
        super.viewDidLoad()
        // Do any additional setup after loading the view.

    }

    @IBAction func randomNumber(_ sender: Any) {

        let randomNumber = Int.random(in: 1000...10000)
        PowerLVLlabel.text = String(randomNumber)
        if case { randomNumber; >=9000 == (true);
            func playVideo() {
                guard let path = Bundle.main.path(forResource: "over 9000", ofType:"mp4") else {
                    debugPrint("over 9000.mp4 not found")
                    return
                }
                {let player = AVPlayer(url: URL(fileURLWithPath: path))
                let playerController = AVPlayerViewController()
                playerController.player = player
                    present(playerController, animated: true) }
                    {player.play()
}

Answer 1

尝试一下：

func playVideo(){
    guard let path = Bundle.main.path(forResource: "over 9000", ofType:"mp4") else {
        debugPrint("over 9000.mp4 not found")
        return
    }
    let player = AVPlayer(url: URL(fileURLWithPath: path))
    let playerController = AVPlayerViewController()
    playerController.player = player
    present(playerController, animated: true)
    player.play()

}

@IBAction func randomNumber(_ sender: Any) {

        let randomNumber = Int.random(in: 1000...10000)
        PowerLVLlabel.text = String(randomNumber)

        if randomNumber >= 9000{
            playVideo()
        }
    }

Answer 2

您需要从js中获取playVideo函数，并删除一些不需要的randomNumber和;

{}