Java

Question

异常处理只接受双输入。因此，当用户输入 "k" 时，它会说 "Error! Please enter a number!"。但是，它不会允许用户重新输入，而是跳到下一个输入“平均冲量”）我怎样才能让它工作，以便它保持在同一行并允许重新输入一个值？

//主要class

        public class Main { //Master class


        public static void main( String args[] ) //Standard header for main method
        {


        kbentry input = new kbentry(); //Creates object of kbentry class

        System.out.print("\nPlease enter a number for Total Impulse: " ); //Print message to enter 1. input
        double totalImpulse = input.totalImpulse1(); //Holds the variable entered
        System.out.println("You have entered : " + totalImpulse); //Shows the variable entered


        System.out.print("\nPlease enter a number for Average Impulse: " ); //Print message to enter 2. input
        double averageImpulse = input.averageImpulse2(); //Holds the variable entered
        System.out.println("You have entered : " + averageImpulse); //Shows the variable entered
        }
        }

//kbentry class

    import java.io.BufferedReader;
    import java.io.InputStreamReader;
    import java.io.IOException;


public class kbentry{ //Class name

double totalImpulse1(){ //Method for 1. input

BufferedReader in = new BufferedReader(new InputStreamReader(System.in));     //Creates BufferedReader object for System.in

//Total Impulse entry
String strTotalImpulse = null;  // These must be initialised
double    intTotalImpulse = 0; //Setting it double

try {
  strTotalImpulse = in.readLine(); //Reads string value from the keyboard
} 
catch (IOException ioe) {  // ignore exception

   }

try {
  intTotalImpulse = Double.parseDouble(strTotalImpulse);  // convert it to double
} 
catch (NumberFormatException nfe) {
   System.out.println("Error! Please enter a number!" + nfe.toString()); //Error message if its not a double

}


return intTotalImpulse; //return value
}



double averageImpulse2(){ //Method for 2. input

BufferedReader in = new BufferedReader(new InputStreamReader(System.in));//Creates BufferedReader object for System.in

// String for AverageImpulse
String strAverageImpulse = null; // These must be initialised
double    intAverageImpulse = 0; //Setting it double

try {
strAverageImpulse = in.readLine(); //Reads string value from the keyboard
} 
catch (IOException ioe) {   // ignore exception
}
// convert it to integer
try {
  intAverageImpulse = Double.parseDouble(strAverageImpulse); // convert it to double
}
catch (NumberFormatException nfe) {
System.out.println("Error! Please enter a number!" + nfe.toString()); //Error message if its not a double

}
return intAverageImpulse; //return value
}
}

Answer 1

如果用户输入的不是 double，你将得到 NumberFormatException，因为你必须再次调用该方法

catch (NumberFormatException nfe) { 
    System.out.println("Error! Please enter a number!" + nfe.toString());  //Error message if its not a double
    //again ask for input
    System.out.print("\nPlease enter a number for Total Impulse: ");
    return totalImpulse1();
}

Java - 异常处理 - 如何重新输入无效输入

Java - Exception handling - How to re-enter invalid input

exception-handling

class

input

object