Question

我的任务：

通过确保用户输入英尺的正整数，英寸的非负十进制数和重量的正十进制数来执行输入验证。有关详细信息，请参阅示例会话特别要注意回波打印高度和重量值的格式以及生成的体质量指数值。

我的程序有效，但我无法弄清楚如何让它看起来像下面的示例程序。我必须让用户以英尺为单位输入高度，然后输入英寸。我不知道如何在空间分割用户输入。

示例会话：

Enter height using feet space inches (e.g., 5 6.25): hi there
Invalid feet value. Must be an integer.
Invalid inches value. Must be a decimal number.
Re-enter height using feet space inches (e.g., 5 6.25): 0 9
Invalid feet value. Must be positive.
Re-enter height using feet space inches (e.g., 5 6.25): 5.25 0
Invalid feet value. Must be an integer.
Re-enter height using feet space inches (e.g., 5 6.25): 5 9.25
Enter weight in pounds: 0
Invalid pounds value. Must be positive.
Re-enter weight in pounds: 150.5
height = 5'-9.25"
weight = 150.5 pounds
body mass index = 22.1

到目前为止我的代码：

package ch14;
import java.util.Scanner;

public class BMI {


public static void main(String[] args) {

    String inputString;
    double bmi;
    double pounds;
    double inches;
    String feet;

    Scanner stdIn = new Scanner(System.in);

    System.out.print("Please enter height using feet space inches (e.g., 5 6.25): ");
    inputString = stdIn.nextLine();


    try
    {
        Double.parseDouble(inputString);
    }
    catch(NumberFormatException nfe)
    {
        System.out.println ("Invalid inch value. Number must be a decimal.");
        System.out.print ("Re-enter height in inches: ");
        inputString = stdIn.nextLine ();
    }

    inches=Double.parseDouble(inputString);

    if(inches<=0)
    {
        System.out.println("Invalid inch value. Must be a positive number.");
        System.out.print("Re-enter height in inches:");
        inputString = stdIn.nextLine();
        inches=Double.parseDouble(inputString);
    }

    System.out.print("enter weight(in pounds) ");
    inputString = stdIn.nextLine();

    try
    {
        Double.parseDouble(inputString);
    }
    catch(NumberFormatException nfe)
    {
        System.out.println("Invalid pound value. Number must be a decimal.");
        System.out.print("Re-enter the weight in pounds: ");
        inputString = stdIn.nextLine();
    }

    pounds=Double.parseDouble(inputString);

    if(pounds <= 0)
    {
        System.out.println("Invalid pound value. Must be a positive number.");
        System.out.print("Re-enter the weight in pounds:");
        inputString = stdIn.nextLine();
        pounds=Double.parseDouble(inputString);
    }

    System.out.println("Height = " + inches + "\"");
    System.out.println("Weight = " + pounds + " pounds");
    System.out.printf("Body Mass Index = %.1f\n",(pounds * 703.)/(inches * inches));

    }//end main
}

Answer 1

您使用.split分割，因此，对于单个空格......

mystring.split("\\s");

任意数量的空格......

mystring.split("\\s+");

这两个操作都返回一个字符串数组，然后可以像任何其他数组一样访问它们，例如......

String mystring = "Hello World";
String[] words = mystring.split("\\s");
System.out.println(words[0]); //<--- Would print "Hello"

对于您的特定问题，您可能希望将这些String对象转换为Double或者其他东西。您可以使用Double.parseDouble()执行此操作。例如，

Double number = Double.parseDouble("1.0");

以上代码行将"1.0" String对象转换为值1.0的双精度。

java输入验证：如何在空间中拆分用户输入？

1 个答案: