简单地说 您正在尝试使用可选变量的值为nil。 快速修复可以使用保护或如果让而不是强制打开像放!在变量的末尾

如果在CollectionView中出现此错误，请尝试创建CustomCell文件和Custom xib。

在mainVC的ViewDidLoad()中添加此代码。

    let nib = UINib(nibName: "CustomnibName", bundle: nil)
    self.collectionView.register(nib, forCellWithReuseIdentifier: "cell")

基本上，你试图在Swift只允许非nil值的地方使用nil值，通过告诉编译器信任你，那里永远不会有nil值，从而允许你的应用程序编译。

有几种情况会导致这种致命错误:

forced unwraps: let user = someVariable! If someVariable is nil, then you'll get a crash. By doing a force unwrap you moved the nil check responsibility from the compiler to you, basically by doing a forced unwrap you're guaranteeing to the compiler that you'll never have nil values there. And guess what it happens if somehow a nil value ends in in someVariable? Solution? Use optional binding (aka if-let), do the variable processing there: if user = someVariable { // do your stuff } forced (down)casts: let myRectangle = someShape as! Rectangle Here by force casting you tell the compiler to no longer worry, as you'll always have a Rectangle instance there. And as long as that holds, you don't have to worry. The problems start when you or your colleagues from the project start circulating non-rectangle values. Solution? Use optional binding (aka if-let), do the variable processing there: if let myRectangle = someShape as? Rectangle { // yay, I have a rectangle } Implicitly unwrapped optionals. Let's assume you have the following class definition: class User { var name: String! init() { name = "(unnamed)" } func nicerName() { return "Mr/Ms " + name } } Now, if no-one messes up with the name property by setting it to nil, then it works as expected, however if User is initialized from a JSON that lacks the name key, then you get the fatal error when trying to use the property. Solution? Don't use them :) Unless you're 102% sure that the property will always have a non-nil value by the time it needs to be used. In most cases converting to an optional or non-optional will work. Making it non-optional will also result in the compiler helping you by telling the code paths you missed giving a value to that property Unconnected, or not yet connected, outlets. This is a particular case of scenario #3. Basically you have some XIB-loaded class that you want to use. class SignInViewController: UIViewController { @IBOutlet var emailTextField: UITextField! } Now if you missed connecting the outlet from the XIB editor, then the app will crash as soon as you'll want to use the outlet. Solution? Make sure all outlets are connected. Or use the ? operator on them: emailTextField?.text = "my@email.com". Or declare the outlet as optional, though in this case the compiler will force you to unwrap it all over the code. Values coming from Objective-C, and that don't have nullability annotations. Let's assume we have the following Objective-C class: @interface MyUser: NSObject @property NSString *name; @end Now if no nullability annotations are specified (either explicitly or via NS_ASSUME_NONNULL_BEGIN/NS_ASSUME_NONNULL_END), then the name property will be imported in Swift as String! (an IUO - implicitly unwrapped optional). As soon as some swift code will want to use the value, it will crash if name is nil. Solution? Add nullability annotations to your Objective-C code. Beware though, the Objective-C compiler is a little bit permissive when it comes to nullability, you might end up with nil values, even if you explicitly marked them as nonnull.

Swift 5.7 +

如果让简写的影子一个现有的可选变量

以上回答清楚地解释了为什么会出现这个问题以及如何处理这个问题。但是从swift 5.7+开始，有了一种新的方式来处理这个问题。

是我的变量：整数？

以前

if let myVariable = myVariable {
    //this part get executed if the variable is not nil
}else{
    //this part get executed if the variable is nil
}

now

现在我们可以省略表达式的右边。

if let myVariable {
    //this part get executed if the variable is not nil
}else{
    //this part get executed if the variable is nil
}

以前，我们必须重复引用标识符两次，这可能导致这些可选绑定条件过于冗长，特别是在使用冗长的变量名时。

但是现在有一种简写语法，可以通过省略表达式的右边来实现可选绑定。

同样的事情也适用于guard let语句。

详情如下:

if-let简写的建议

当我试图从prepareforsegue方法中设置outlet值时，我有过这样的错误:

override func prepare(for segue: UIStoryboardSegue, sender: Any?) {
    if let destination = segue.destination as? DestinationVC{

        if let item = sender as? DataItem{
            // This line pops up the error
            destination.nameLabel.text = item.name
        }
    }
}

然后我发现我不能设置目标控制器出口的值，因为控制器还没有加载或初始化。

所以我是这样解决的:

override func prepare(for segue: UIStoryboardSegue, sender: Any?) {
    if let destination = segue.destination as? DestinationVC{

        if let item = sender as? DataItem{
            // Created this method in the destination Controller to update its outlets after it's being initialized and loaded
            destination.updateView(itemData:  item)
        }
    }
}

目的地控制器:

// This variable to hold the data received to update the Label text after the VIEW DID LOAD
var name = ""

// Outlets
@IBOutlet weak var nameLabel: UILabel!

override func viewDidLoad() {
    super.viewDidLoad()

    // Do any additional setup after loading the view.
    nameLabel.text = name
}

func updateView(itemDate: ObjectModel) {
    name = itemDate.name
}

我希望这个答案能帮助那些有同样问题的人，因为我发现标记的答案对理解可选选项及其工作方式是很好的资源，但并没有直接解决问题本身。

简单地说 您正在尝试使用可选变量的值为nil。 快速修复可以使用保护或如果让而不是强制打开像放!在变量的末尾