Question

所以，我尝试制作一个简单的毕达哥拉斯应用程序，可以在输入其他2个边的长度后测量三角形的边长。为了测量那一侧，用户应该在三角形的三个边之一中输入“x”或“X”。这是我的代码：

    import java.util.Scanner;
    public class pythagoras_1 {
    public static void main (String Args[]){
    //string data type to check user input 'x' or not
    String  shypo,scat1,scat2;
    double  hypo, cat1, cat2;
    boolean hypoSelected, cat1Selected, cat2Selected;

    Scanner console = new Scanner (System.in);

    System.out.print( "Hypotenuse = ");
    shypo = console.next();

    System.out.print("Cathetus 1 = ");
    scat1 = console.next();

    System.out.print("Cathetus 2 = ");
    scat2 = console.next();

    hypoSelected = shypo.equals("x")||shypo.equals("X");
    cat1Selected = scat1.equals("x")||scat1.equals("X");
    cat2Selected = scat2.equals("x")||scat2.equals("X");

    if(hypoSelected){
        cat1 = Double.parseDouble(scat1);
        cat2 = Double.parseDouble(scat2);
        System.out.println("Hypotenuse = "+ hypoC(cat1,cat2));
    }
    else if(cat1Selected){
        hypo = Double.parseDouble(shypo);
        cat2 = Double.parseDouble(scat2);
        System.out.println("cathetus 1 = "+ Cat1C(hypo,cat2));
    }
    else if(cat2Selected){  
        hypo = Double.parseDouble(shypo);
        cat1 = Double.parseDouble(scat1);
        System.out.println("cathetus 2 = "+ Cat2C(hypo,cat1));
    }
    else{
        System.err.println("ERROR");
    }
}

//HERE THE METHOD FUNCTION
public static double hypoC(double cat1,double cat2){
    double ohypo;
    ohypo = Math.sqrt((cat1*cat1)+(cat2*cat2));
    return ohypo;
}
public static double Cat1C(double hypo,double cat2){
    double ocat1;
    ocat1 = Math.sqrt((hypo*hypo)-(cat2*cat2));
    return ocat1;
}
public static double Cat2C(double hypo,double cat1){
    double ocat2;
    ocat2 = Math.sqrt((hypo*hypo)-(cat1*cat1));
    return ocat2;
}
}

但它仅适用于斜边（第一个if语句），如果我在cathetus 1或2上输入'x'，应用程序结束而不执行代码。我该如何解决这个问题？

这是scat1或scat 2值为'x'

时的输出

 run:
 Hypotenuse = 5
 Cathetus 1 = 4
 Cathetus 2 = x
 ERROR
 BUILD SUCCESSFUL (total time: 5 seconds)

这是shypo为'x'时的输出

 run:
 Hypotenuse = x
 Cathetus 1 = 4
 Cathetus 2 = 3  
 Hypotenuse = 5.0
 BUILD SUCCESSFUL (total time: 6 seconds)

Answer 1

错误我在你的代码中输入了错字和解析异常，因为它试图将x解析为double，这就是它没有执行代码的原因。检查你的程序日志。

更正了程序的主要功能

public static void main(String Args[]) {
        String shypo, scat1, scat2;
        Scanner console = null;
        try {
            console = new Scanner(System.in);
            // input of each side
            System.out.print("hypotenuse = ");
            shypo = console.next();

            System.out.print("cathetus 1 = ");
            scat1 = console.next();

            System.out.print("cathetus 2 = ");
            scat2 = console.next();

            if ("x".equalsIgnoreCase(shypo)) {
                System.out.println("Hypotenuse = " + hypoC(Double.parseDouble(scat1), Double.parseDouble(scat2)));
            } else if ("x".equalsIgnoreCase(scat1)) {
                System.out.println("cathetus 1 = " + Cat1C(Double.parseDouble(shypo), Double.parseDouble(scat2)));
            } else if ("x".equalsIgnoreCase(scat2)) {
                System.out.println("cathetus 2 = " + Cat2C(Double.parseDouble(shypo), Double.parseDouble(scat1)));
            } else {
                System.err.println("ERROR");
            }
        } finally {
            if (console != null)
                console.close();
        }
    }

Answer 2

你有很多错别字：

cat2Selected = scat1.equals("x")||scat1.equals("X");

将scat1更改为scat2

这是关键问题：

  hypo = Double.parseDouble(shipo);

shipo应为shypo，否则在hypo或Cat1C的调用中Cat2C未定义。而且，'x'不会作为Double进行解析。您可以将hypo，cat1和cat2的定义放在条件中。

最后：

 ohipo = Math.sqrt((cat1*cat1)+(cat2*cat2));

ohipo应为ohypo。

Java：只有第一个if语句正常工作

2 个答案: