Swift:不能将类型'int'的值赋给类型'int'的值吗?的错误

I'm starting to learn swift, but ran into an error. Ive just created a very simple app from a course I'm following, that calculates a cats ages based on what the user enters. The first version of the app just times what the user enters by 7, which i managed todo no problem. I thought I would have a play with what I've written and change it todo this:

我刚开始学swift，但遇到了一个错误。我刚从我正在学习的课程中创建了一个非常简单的应用程序，它根据用户输入的内容来计算猫的年龄。第一个版本的应用程序只是乘以用户输入的数字7，我成功地做到了这一点。我想我可以把我写的剧本改编成这样:

1st human year = 15 cat years
第一人类年= 15猫年
2nd human year = +10
第二人类年= +10
3rd human year = +4
人类第三年= +4
4th human year = +4
第四人类年= +4
5th human year = +4
第五人类年= +4

so on..

等等. .

e.g. a 4 year old cat would be 33 in cat years.

一只4岁的猫在猫的年龄将是33岁。

So I want my results label to say You're cat is 33 in cat years

所以我希望我的结果标签上写着你是猫年33岁

This is the code I've written:

这是我写的代码:

    @IBAction func findAge(sender: AnyObject) {

    var enteredAge = enterAge.text.toInt()

    if enteredAge = 1 {

        var catYears = 15

        resultLabel.text = "You're cat is \(catYears) in cat years"

    } else if enteredAge = 2 {

        var catYears = 25

        resultLabel.text = "You're cat is \(catYears) in cat years"

    } else {

        var catYears = (15 + 25) + (enteredAge - 2) * 4

        resultLabel.text = "You're cat is \(catYears) in cat years"

    }

}

@IBOutlet weak var enterAge: UITextField!


@IBOutlet weak var resultLabel: UILabel!

This error I'm getting is on the line 17 "if enteredAge = 1 {" it states this "cannot assign a value of type 'int' to a value of type 'int?'"

我得到的这个错误是在第17行“如果enteredAge = 1{”它声明这个“不能为int类型的值指定类型‘int’的值”。

I don't really understand why this value cannot be a integer, any help would be great.

我不明白为什么这个值不能是整数，任何帮助都是很好的。

4 个解决方案

#1

The main error (as I said in a comment) is that you mixed up the assignment operator = and the equality operator ==. The comparison should be

主要的错误(如我在注释中所说)是您混淆了赋值运算符=和等式运算符==。比较应该

if enteredAge == 1 { ... }

The next problem is (as keithbhunter already stated in his answer), toInt() returns an optional which is nil if the string is not a valid integer, and you should use optional binding:

下一个问题是(正如keithbhunter在他的回答中已经指出的那样)，toInt()返回一个可选的，如果字符串不是有效的整数，则返回nil，您应该使用可选绑定:

if let enteredAge = enterAge.text.toInt() {
   // compute cat years ... 
} else {
   // report invalid input ...
}

Additional notes:

其他说明:

All your variables in that method can be declared as constants with let.
该方法中的所有变量都可以用let声明为常量。
There are 3 identical assignments resultLabel.text = ..., this can be simplified.
有三个相同的作业结果标签。文本=…，这可以简化。
Instead of if ... else if ... else you could use a switch statement.
而不是如果……其他的如果……否则可以使用switch语句。

Then your method would look like this:

那么你的方法是这样的:

if let enteredAge = enterAge.text.toInt() {
    let catYears : Int
    switch(enteredAge) {
    case 1:
        catYears = 15
    case 2:
        catYears = 25
    default:
        catYears = (15 + 25) + (enteredAge - 2) * 4
    }
    resultLabel.text = "You're cat is \(catYears) in cat years"
} else {
    resultLabel.text = "Please enter a valid number"
}

An alternative would be to use the conditional operator ?: (sometimes also called ternary operator):

另一种方法是使用条件运算符?:(有时也称为三元运算符):

if let enteredAge = enterAge.text.toInt() {
    let catYears = enteredAge == 1 ? 15 :
                enteredAge == 2 ? 25 : (15 + 25) + (enteredAge - 2) * 4
    resultLabel.text = "You're cat is \(catYears) in cat years"
} else {
    resultLabel.text = "Please enter a valid number"
}

#2

I am not exactly sure of the toInt() method, but I am going to guess that method returns an Int?. This means that it can return nil if it cannot convert the string to an int. You should unwrap this value to handle nil cases.

我不太确定toInt()方法，但是我猜想这个方法会返回一个Int?这意味着如果不能将字符串转换为int类型，它可以返回nil。

if var enteredAge = enterAge.text.toInt() {
    // the rest of the code
}

#3

You have already assigned the value to the enteredAge variable like follows:

您已经将值分配给enteredAge变量，如下所示:

var enteredAge = enterAge.text.toInt()

So, you cannot do assignment on the place of this if expression. and you need to have condition on the if expression as follows.

因此，如果表达式的位置不能赋值。你需要在if表达式上有条件，如下所示。

if enteredAge == 1 {

    var catYears = 15

    resultLabel.text = "You're cat is \(catYears) in cat years"

}

This may help you.

这可能会帮助你。

#4

should look like this :

应该是这样:

@IBAction func button(sender: AnyObject) {
    var inputYears = textField.text.toInt()!
    if inputYears == 1 {
    var catOld = 15
    result.text = "Your cat is \(catOld) years old"
    }
    else if inputYears == 2 {
    var catOld = 25
    result.text = "Your cat is \(catOld) years old"
    }
    else {var catOld = (15+25) + (inputYears - 2)*4
    result.text = "Your cat is \(catOld) years old"
    }

#1