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我已经看到了其他答案，但是在使用多个 try catch 块之后，我的程序在第一个异常处终止。我希望它排除所有四个变量值，然后使用我在每个 catch 块中保留的标志来处理任何异常。

这是我的代码：

import java.text.DecimalFormat;
import java.util.Scanner;

public class Brokerage2 {
   public static void main(String[] args) {
      float bro = 0, buy = 0, sell = 0;
      int quant = 0, flag = 0;
      double subtot, pro, bb, bst, sb, ssst, seltot, buytot, surcharge;
      Scanner sc = new Scanner(System.in);
      try {
         bro = sc.nextFloat();
         if (bro > 0.03 || bro < 0.0) {
            flag = 1;
         }
      } catch (Exception e) {
         flag = 1;
      }
      try {
         buy = sc.nextFloat();
      } catch (Exception e) {
         flag = 1;
      }
      try {
         sell = sc.nextFloat();
      } catch (Exception e) {
         flag = 1;
      }
      try {
         quant = sc.nextInt();
      } catch (Exception e) {
         flag = 1;
      }
      bb = (buy * quant) * bro * 0.01;
      bst = 0.1036 * bb;
      buytot = bst + bb;
      sb = sell * quant * bro * 0.01;
      ssst = sell * quant * 0.00025;
      seltot = sb + ssst + bst;
      surcharge = ((buy * quant) + (sell * quant)) * 0.00006;
      subtot = surcharge + buytot + seltot;
      pro = (sell - buy) * quant;
      if (flag == 1) {
         System.out.println("Invalid Input");
      } else if (pro > subtot) {
         System.out.println("profit");
         System.out.println(Double.parseDouble(new DecimalFormat("##.##").format(pro - subtot)));
      } else if (subtot > pro) {
         System.out.println("loss");
         System.out.println(Double.parseDouble(new DecimalFormat("##.##").format(subtot - pro)));
      }
   }
}
4

1 回答 1

1

如果扫描仪接收到无效输入，则不再等待输入。对多个输入使用同一个扫描仪并不是一个好主意。相反，每次都创建一个新的扫描仪，它会等待每个扫描仪的输入。

import java.text.DecimalFormat;

import java.util.Scanner;

public class brokerage2 {

    public static void main(String[] args) {
        float bro = 0, buy = 0, sell = 0;
        int quant = 0, flag = 0;
        double subtot, pro, bb, bst, sb, ssst, seltot, buytot, surcharge;

        Scanner sc = new Scanner(System.in);
        try {
            System.out.println("Enter float: ");
            bro = sc.nextFloat();
            if (bro > 0.03 || bro < 0.0) {
                flag = 1;

            }

        } catch (Exception e) {
            flag = 1;

        }

        try {
            sc = new Scanner(System.in);
            System.out.println("Enter float: ");
            buy = sc.nextFloat();

        }

        catch (Exception e) {
            flag = 1;
        }
        try {
            sc = new Scanner(System.in);
            System.out.println("Enter float:");
            sell = sc.nextFloat();
        } catch (Exception e) {
            flag = 1;
        }
        try {
            sc = new Scanner(System.in);
            System.out.println("Enter int: ");
            quant = sc.nextInt();

        } catch (Exception e) {
            flag = 1;
        }

        bb = (buy * quant) * bro * 0.01;
        bst = 0.1036 * bb;
        buytot = bst + bb;

        sb = sell * quant * bro * 0.01;
        ssst = sell * quant * 0.00025;
        seltot = sb + ssst + bst;

        surcharge = ((buy * quant) + (sell * quant)) * 0.00006;

        subtot = surcharge + buytot + seltot;

        pro = (sell - buy) * quant;
        if (flag == 1) {
            System.out.println("Invalid Input");
        }

        else if (pro > subtot) {
            System.out.println("profit");
            System.out.println(Double.parseDouble(new DecimalFormat("##.##")
                    .format(pro - subtot)));
        }

        else if (subtot > pro) {
            System.out.println("loss");
            System.out.println(Double.parseDouble(new DecimalFormat("##.##")
                    .format(subtot - pro)));

        }

    }
}
于 2013-08-13T14:34:38.767 回答