无法解析为VS Code中的类型[Java]

Question

我在VS Code中运行我的Dog.class和DogDemo.java文件时遇到了问题。我输入的完全是Java书中写的。我想出了错误：

Dog无法解析为类型

在DogDemo。main（DogDemo.java:7）

我也尝试过清理java工作区，但不起作用。

狗。类

public class Dog
{
    public String name;
    public String breed;
    public age = 0;
    

    public void writeOutput()
    {
        System.out.println("Name " + name);
        System.out.println("Breed: " + breed);
        System.out.println("Age in calendar years" + age);
        System.out.println("Age in human years" + getAgeInHumanYears());
        System.out.println();
    }

    public int getAgeInHumanYears()
    {
        int humanAge = 0;

        if (age <= 2) 
        {
            humanAge = age * 11;
        }

        else
        {
            humanAge = 22 + ((age-2) * 5);
        }

        return humanAge;
    }
}

DogDemo.java

public class DogDemo 
{
    public static void main(String[] args) 
    {
        Dog kumo = new Dog();

        kumo.name = "kumo";
        kumo.age = 42;
        kumo.breed = "Corgi";
        kumo.writeOutput();

        Dog scooby = new Dog();
        scooby.name = "Scooby";
        scooby.age = 9;
        scooby.breed = "Great Dane";
        System.out.println(scooby.name + " is a " + scooby.breed +
        ".");`
        System.out.print("He is " + scooby.age + " years old, or ");

        int humanYears = scooby.getAgeInHumanYears();
        System.out.println(humanYears + " in human years.");
    }
}

Answer 1

狗

class Dog {
    String name;
    String breed;
    int age;

    void writeOutput()
    {
        System.out.println("Name " + name);
        System.out.println("Breed: " + breed);
        System.out.println("Age in calendar years" + age);
        System.out.println("Age in human years" + getAgeInHumanYears());
        System.out.println();
    }

    int getAgeInHumanYears()
    {
        int humanAge = 0;

        if (age <= 2)
        {
            humanAge = age * 11;
        }

        else
        {
            humanAge = 22 + ((age-2) * 5);
        }

        return humanAge;
    }
}

DogDemo

public class DogDemo {

    public static void main(String[] args) {
        Dog kumo = new Dog();

        kumo.name = "kumo";
        kumo.age = 42;
        kumo.breed = "Corgi";
        kumo.writeOutput();

        Dog scooby = new Dog();
        scooby.name = "Scooby";
        scooby.age = 9;
        scooby.breed = "Great Dane";
        System.out.println(scooby.name + " is a " + scooby.breed + ".");
        System.out.print("He is " + scooby.age + " years old, or ");

        int humanYears = scooby.getAgeInHumanYears();
        System.out.println(humanYears + " in human years.");
    }
}

我所做的改变：

1. 从数据类中删除了public关键字Dog
2. 您错过了打印名称品种的主要功能
3. 将int类型添加到Dog属性的age中

请确保您的类在相同的包中，如果不是，请将其导入到您的类中。

编辑：

我想其他人也有和你一样的问题，你可以在这里找到

你也可以按照这里的步骤，这个留档是由VS代码提供的。你可以看到缺少了哪些步骤。

Answer 2

您忘记说明变量的类型age

    public age = 0;

应该成为

    public int age = 0;

另外在DogDemo.java17-18行你有

        System.out.println(scooby.name + " is a " + scooby.breed +
        ".");`

您有一个回调，这是无效的，应该删除成为

        System.out.println(scooby.name + " is a " + scooby.breed +
        ".");

Answer 3

在Dog类中，您忘记了年龄变量的类型（可能是一个整数），因此正确答案是：

public class Dog
{
    public String name;
    public String breed;
    public int age = 0;

在你的DogDemo课程中，你有一行输入错误：

System.out.println(scooby.name + " is a " + scooby.breed +
    ".");``

分号后的 " ' " 应删除为：

System.out.println(scooby.name + " is a " + scooby.breed + ".");

然后它对我来说很好：）继续编码！