Question

我在使用此try / catch语句时遇到问题。我试图用它来抛出一条错误信息，上面写着“请输入一个整数！”如果用户输入了一封信。我使用的第一个工作，但它最终采取2行用户输入而不是一个，所以它基本上跳过应该被问的问题。在那之后，其他任何一个都没有工作，他们只是完全被跳过。我还需要为用户输入做同样的事情，如果用户输入一个字母应该是的整数，它会抛出一条错误消息“请输入一个字符串！”。我知道我非常接近！

public static void main(String[] args) {
        Scanner input = new Scanner(System.in);
        boolean validInput = false;
        int val = 0;


        System.out.print("Enter your first name: ");    
        String firstName = input.nextLine();
        System.out.print("Enter your last name: ");
        String lastName = input.nextLine();
        System.out.print("Enter your address: ");
        String address = input.nextLine();
        System.out.print("Enter your city: ");
        String city = input.nextLine();
        System.out.print("Enter your state: ");
        String state = input.nextLine();
        System.out.print("Enter your zip code + 4: ");
        String zip = input.nextLine();
        while(!validInput) {
             try {
        val = input.nextInt();
        validInput = true;
           } catch(Exception e) {
        System.out.println("Please enter an integer!");
        input.nextLine();
    }
}
        System.out.print("Enter amount owed: ");
        String amount = input.nextLine();
        while(!validInput) {
             try {
        val = input.nextInt();
        validInput = true;
           } catch(Exception e) {
        System.out.println("Please enter an integer!");
        input.next();
    }
}
        System.out.print("Enter your payment amount: ");
        String payment = input.nextLine();
        while(!validInput) {
             try {
        val = input.nextInt();
        validInput = true;
           } catch(Exception e) {
        System.out.println("Please enter an integer!");
        input.next();
    }
}
        System.out.print("Enter the date of payment: ");
        String date = input.nextLine();
        while(!validInput) {
             try {
        val = input.nextInt();
        validInput = true;
           } catch(Exception e) {
        System.out.println("Please enter an integer!");
        input.next();
    }
}
        System.out.println("\t\t\t\t\t" + "XYZ Hospital");
        System.out.println();
        System.out.println("Name Information" + "\t\t\t\t" + "Address" + "\t\t\t\t\t\t" + "Payment");
        System.out.println();
        System.out.println("Last" +"\t"+ "First" + "\t\t\t" + "Address Line 1" + "\t" + "City" + "\t" + "State" + "\t" + "Zip" + "\t" + "Amount Owed" + "\t" + "Payment Amount" + "\t" + "Payment Date");
        System.out.println();
        System.out.println(lastName + " " + firstName + "\t" + address + " " + city + ", " + state + " " + zip + "\t" + amount + "\t\t" + payment +"\t\t"+ date);
        System.out.print("Would you like to enter abother patient? Type Yes or No: ");
        String userInput = input.nextLine();
        if (userInput.equals("Yes")) {
            while (userInput.equals("Yes")) {
                System.out.print("Enter your first name: ");    
                String firstName2 = input.nextLine();
                System.out.print("Enter your last name: ");
                String lastName2 = input.nextLine();
                System.out.print("Enter your address: ");
                String address2 = input.nextLine();
                System.out.print("Enter your city: ");
                String city2 = input.nextLine();
                System.out.print("Enter your state: ");
                String state2 = input.nextLine();
                System.out.print("Enter your zip code + 4: ");
                String zip2 = input.nextLine();
                System.out.print("Enter amount owed: ");
                String amount2 = input.nextLine();
                System.out.print("Enter your payment amount: ");
                String payment2 = input.nextLine();
                System.out.print("Enter the date of payment: ");
                String date2 = input.nextLine();
                System.out.println("\t\t\t\t\t" + "XYZ Hospital");
                System.out.println();
                System.out.println("Name Information" + "\t\t\t\t" + "Address" + "\t\t\t\t\t\t" + "Payment");
                System.out.println();
                System.out.println("Last" +"\t"+ "First" + "\t\t\t" + "Address Line 1" + "\t" + "City" + "\t" + "State" + "\t" + "Zip" + "\t" + "Amount Owed" + "\t" + "Payment Amount" + "\t" + "Payment Date");
                System.out.println();
                System.out.println(lastName2 + " " + firstName2 + "\t" + address2 + " " + city2 + ", " + state2 + " " + zip2 + "\t" + amount2 + "\t\t" + payment2 +"\t\t"+ date2);
                System.out.print("Would you like to enter another patient? Type Yes or No: ");
                userInput = input.nextLine();
            }

        }
        else if (userInput.equals("No")) {
            System.out.println("Goodbye");
        }

Answer 1

在第一次while循环后，您不会将validInput重置为False。所以它不会进入下一个。

Answer 2

你应该在catch块中的catch块注释input.nextLine()中输入，然后它应该正常工作

我在代码本身中添加了解释

public static void main(String[] args) {
  Scanner input = new Scanner(System.in);
  boolean validInput = false;
  int val = 0;

  System.out.print("Enter your first name: ");
  String firstName = input.nextLine();
  System.out.print("Enter your last name: ");
  String lastName = input.nextLine();
  System.out.print("Enter your address: ");
  String address = input.nextLine();
  System.out.print("Enter your city: ");
  String city = input.nextLine();
  System.out.print("Enter your state: ");
  String state = input.nextLine();
  System.out.print("Enter your zip code + 4: ");
  String zip = input.nextLine();
  while (!validInput) {
    try {
      val = input.nextInt();
      validInput = true; //if exception occurs this line wont be executed
    } catch (Exception e) {
      System.out.println("Please enter an integer!");
      // input.nextLine(); --remove this line
    }
  }
  System.out.print("Enter amount owed: ");
  String amount = input.nextLine();
  //reset the value of validInput
  //validInput will true after the execution of above while loop
  validInput = false;
  while (!validInput) {
    try {
      val = input.nextInt();
      validInput = true; //if exception occurs this line wont be executed
    } catch (Exception e) {
      System.out.println("Please enter an integer!");
      // input.next(); -- remove this line
    }
  }
}

Answer 3

快速解决方案可能是，我在使用Scanner.hasNextInt之前检查了下一个令牌是什么：

import java.util.Scanner;

public class java_so {
    private static int privateGetInt(String request, Scanner input) {
        // cant be false, but we return when done.
        while (true) {
            // print the question
            System.out.println(request);
            // check what the next token is, if an int we can then retrieve
            if (input.hasNextInt() == true) {
                int out = input.nextInt();
                input.nextLine(); // clear line, as nextInt is token orientated, not line,
                return out;
            } else {
                // else for when not an int, clear line and try again
                input.nextLine();
            }
        }
    }

    public static void main(String[] args) {
            Scanner input = new Scanner(System.in);

            System.out.print("Enter your first name: ");    
            System.out.println(input.nextLine());
            // call function above 
            System.out.println(privateGetInt("enter an Int", input));
            System.out.print("Enter your last name: ");
            System.out.println(input.nextLine());
            System.out.print("Enter your address: ");
            System.out.println(input.nextLine());
    }
}

看起来有点像调用input.nextInt之后你可能没有摆脱剩下的行（可能只是"\n"）所以下次调用input.getLine()只能得到它，跳上一个。关注input.getInt并input.getLine清除此内容。

try / catch java的问题

3 个答案: