Question

所以我所拥有的代码链接到一个带有按钮的故事板。该按钮假设发出一种名为Laughing的声音效果。但按下按钮时。应用程序崩溃并发送此错误＆＃34;线程1：信号SIGABRT

    //
     //  ViewController.swift
     //  LaLaLaughing
     //
     //  Created by MightyShin on 5/16/16.
     //  Copyright © 2016 Reverze. All rights reserved.
     //

     import UIKit
     import AVFoundation

    class ViewController: UIViewController {

    var audioPlayer = AVAudioPlayer()

    override func viewDidLoad() {
        super.viewDidLoad()
        // Do any additional setup after loading the view, typically from a nib.
    }

    override func didReceiveMemoryWarning() {
        super.didReceiveMemoryWarning()
        // Dispose of any resources that can be recreated.
    }


    @IBAction func LaughSound(sender: AnyObject) {

        var alertSound = NSURL(fileURLWithPath: NSBundle.mainBundle().pathForResource("Laughing", ofType: "mp3")!)

        do {
            // Preperation
            try AVAudioSession.sharedInstance().setCategory(AVAudioSessionCategoryPlayback)
        } catch _ {
        }
        do {
            try AVAudioSession.sharedInstance().setActive(true)
        } catch _ {
        }

        // Play the sound
        do {
            audioPlayer = try AVAudioPlayer(contentsOfURL: alertSound)
        } catch _{
        }

        audioPlayer.prepareToPlay()
        audioPlayer.play()
    }
}

Answer 1

Sigabrt通常关注连接。检查所有插座和操作。如果您创建了一个插座然后将其类型更改为textfield，或者将其名称（例如nameDoctorLabel）更改为doctorNameLabel，则可能会出现问题。

您可能还手动编写了IBoutlets和IbActions，忘记将它们与故事板中各自的对象链接。

同时检查IBAction左侧是否有填充的灰点。

检查连接检查器中的连接。

Xcode Swift - 按下按钮时应用程序崩溃

1 个答案: