Question

您正在为学校制作一个程序，以此形式阅读用户的输入

“String / double / int / int / int”或“String / double / int / int”，具体取决于饮料的形状。

到目前为止，我已经为Parser类提供了代码。

public class DrinkParser {

    public static Drink parseStringToDrink(String lineToParse){

        String regex = "[/]";
        String [] split = lineToParse.split(regex);

        if ("Box".equals(split[0]) || "box".equals(split[0])){

           DrinkInBox dIB = new DrinkInBox(split[1], split[2], split[3], split[4], split[5]);


        }

        if("Cylinder".equals(split[0]) || "cylinder".equals(split[0])){

            DrinkInCylinder dIC = new DrinkInCylinder(split[1], split[2], split[3], split[4]);

        }

    }



}

并且继承了DrinkInBox和DrinkInCylinder类的代码......我觉得它的相关性。

public class DrinkInBox extends Drink {

    private int height;
    private int width;
    private int depth;

    public DrinkInBox(String drinkId, double unitPrice, int height, int width, int depth){

        super(drinkId, unitPrice);
        this.height = height;
        this.width = width;
        this.depth = depth;

    }

    public void computeTotalPrice(){

        volume = height * width * depth;
        totalPrice = volume * unitPrice;

    }

    public String toString(){

        return "\nThe Drink in a Box\nThe Height:\t\t" +height+ "\nThe Width:\t\t" +width+ "\nThe Depth:\t\t" +depth+ "\nThe DrinkId:\t\t" +drinkId+ "\n The Volume:\t\t" +volume+ "\nThe Unit Price:\t\t" +unitPrice+ "\n The Total Price:\t" +totalPrice+ "\n\n";

    }

}

继承人在DrinkIn Cylinder Class

public class DrinkInCylinder extends Drink {

    private int radius;
    private int height;

    public DrinkInCylinder(String drinkId, double unitPrice, int radius, int height){

        super(drinkId, unitPrice);
        this.radius = radius;
        this.height = height;

    }

    public void computeTotalPrice(){

        volume = (int) (Math.PI * (radius*radius) * height);
        totalPrice = volume * unitPrice;

    }

    public String toString(){

        return "\nThe Drink in a Cylinder\nThe Radius:\t\t" +radius+ "\nThe Height:\t\t" +height+ "\nThe DrinkId:\t\t" +drinkId+ "\n The Volume:\t\t" +volume+ "\nThe Unit Price:\t\t" +unitPrice+ "\n The Total Price:\t" +totalPrice+ "\n\n";

    }

}

我知道很多，但我感谢你们能提供的任何帮助。

Answer 1

 DrinkInBox dIB = new DrinkInBox(split[1], split[2], split[3], split[4], split[5]);

split是String[]，因此split[n]，0 <= n <= split.length-1将返回String。现在让我们来看看你的DrinkInBox构造函数：

public DrinkInBox(String drinkId, double unitPrice, int height, int width, int depth)

您传递5 String作为参数，但您没有DrinkInBox构造函数，其中包含5个String参数。您必须相应地解析每个拆分，使其与DrinkInBox构造函数的参数类型匹配。

 DrinkInBox dIB = new DrinkInBox(split[1], Double.parseDouble(split[2]), Integer.parseInt(split[3]), Integer.parseInt(split[4]), Integer.parseInt(split[5]));

您的代码中还会出现另一种情况。你需要做同样的事情。

另外，只是一个改进的提示，你应该熟悉继承和多态。 DrinkInBox和DrinkInCylinder都有一个共同的超类。你可以改写为：

Drink drink = split[0].equalsIgnoreCase("box") ? new new DrinkInBox(split[1], Double.parseDouble(split[2]), Integer.parseInt(split[3]), Integer.parseInt(split[4]), Integer.parseInt(split[5])) : split[0].equalsIgnoreCase("cylinder") ? new DrinkInCylinder(split[1], Double.parseDouble(split[2]), Integer.parseInt(split[3]), Integer.parseInt(split[4]), Integer.parseInt(split[5])) : null;

如果它不是方框或圆柱，drink将是null。您可以通过执行DrinkInBox检查来查看是DrinkInCylinder还是instanceof：

if(drink instanceof DrinkInBox) 
if(drink instanceof DrinkInCylinder)

在String拆分类的中间，无法将String转换为double？

1 个答案: