代码主外部距离我的3分

Question

因此我有一个学校项目，需要我编写代​​码并实现方法来证明用户输入的3个坐标集合将是某个三角形。

我试图写一行代码来获得我的3点P1，P2，P3之间的距离。我不知道我是如何参考我建立的坐标，如x1，x2，x3，y1，y2，y3。 这里是我的代码：

import javax.swing.JOptionPane; 

public class EpsteinProj3 
{ 
    public static void main (String args[]) 
    { 
    String input1 = JOptionPane.showInputDialog("Enter coordinate A"); 
    String input2 = JOptionPane.showInputDialog("Enter coordinate B"); 
    String input3= JOptionPane.showInputDialog("Enter coordinate C"); 
    System.out.println("Your Triangle is: " + input1 + " " + input2 + " " + input3); 

    //double temp = Double.parseDouble(input1); 
    //System.out.println("" + temp); 

    //double temp1 = Double.parseDouble(input2); 
    //System.out.println("" + temp1); 

    //double temp2 = Double.parseDouble(input3); 
    // System.out.println("" + temp2); 

     Double x1, x2, x3; 
     Double y1, y2, y3; 

     x1 = Double.parseDouble(input1.substring(1, input1.indexOf(","))); 
     y1 = Double.parseDouble(input1.substring(input1.indexOf(",") +1, input1.length()-1)); 
     Point P1 = new Point(x1, y1); 

     x2 = Double.parseDouble(input2.substring(1, input2.indexOf(","))); 
     y2 = Double.parseDouble(input2.substring(input2.indexOf(",") +1, input2.length()-1)); 
     Point P2 = new Point(x2, y2); 

     x3 = Double.parseDouble(input3.substring(1, input3.indexOf(","))); 
     y3 = Double.parseDouble(input3.substring(input3.indexOf(",") +1, input3.length()-1)); 
     Point P3 = new Point(x3, y3); 


     System.out.println("Point A is " + P1.toString()); 
     System.out.println("Point B is " + P2.toString()); 
     System.out.println("Point C is " + P3.toString()); 

     //scalene, isosceles, right 


} 

public static double distanceBetween() 
{ 
    double Dis1 = Math.sqrt((x1-x2)^2 + (y1- y2)^2); 
    Dis1 = Math.sqrt((x1-x2)^2 + (y1- y2)^2); 
} 

} 

Answer 1

那是因为你开始从字符串的第一个元素读书。 如果您确信该格式将永远是（POSX，波西），然后

x = Double.parseDouble(input1.substring(1, input1.indexOf(","))); 
    y = Double.parseDouble(input1.substring(input1.indexOf(",") +1, input1.length()-1)); 

Answer 2

我使用解析点following正则表达式。这允许可选(或)和,后的可选空间。

请注意，此示例使用java.awt.Point，因此您应该将Integer.parseInt替换为Double.parseDouble以实现摆动。

public class StringToPoint { 
private static final Pattern POINT_PATTERN = Pattern.compile("(?:\\()?([0-9]+)?,(?:)?([0-9]+)?(?:\\))?"); 

private static Point parsePoint(String point){ 
    Matcher matcher = POINT_PATTERN.matcher(point); 
    if (!matcher.matches() || matcher.groupCount()!=2){ 
     throw new IllegalArgumentException(String.format("%s cannot be parsed as point",point)); 
    } 
    return new Point(
     Integer.parseInt(matcher.group(1)), 
     Integer.parseInt(matcher.group(2)) 
    ); 
} 

@Test 
public void test(){ 
    TestCase.assertEquals(new Point(123,456), parsePoint("(123,456)")); 
    TestCase.assertEquals(new Point(123,456), parsePoint("(123, 456)")); 
    TestCase.assertEquals(new Point(123,456), parsePoint("123, 456")); 
    TestCase.assertEquals(new Point(123,456), parsePoint("123,456")); 
} 
}