想知道是否有人可以告诉我为什么我的try-catch不起作用。当我输入一个整数时,我输入一个字符,程序只是以异常退出。它与我在另一个程序中编写的try-catch相同。不知道为什么它不起作用。 :(
import java.io.*;
import java.util.*;
public class Lab8FileIO {
public static void main(String[] args) throws IOException {
int menuChoice = 0;
String filename, userInput, choice;
boolean playAgain = true;
char response;
Scanner keyboard = new Scanner(System.in);
while (playAgain)
{
// Display menu and get user selection
displayMenu();
do
{
try
{
menuChoice = keyboard.nextInt();
if ((menuChoice < 1) || (menuChoice > 4)) // Make sure user enters a number within the specified range
System.out.print("You must enter a number from 1 to 4!");
}
catch (InputMismatchException e)
{
System.out.print("You must enter a number from 1 to 4!");
keyboard.nextInt();
}
} while (menuChoice < 1 && menuChoice > 4);
// Get input/output filename from user
System.out.print("\nEnter filename in the format filename.txt: ");
filename = keyboard.next();
switch (menuChoice)
{
// Write to file
case 1:
PrintWriter outputFile = new PrintWriter(filename);
userInput = getUserInput();
System.out.println("I am writing your input to filename: " + filename + "\n");
outputFile.println(userInput);
outputFile.close();
break;
// Read from file
case 2:
File file = new File(filename);
Scanner inputFile = new Scanner(file);
System.out.println("I am reading from filename: " + filename + "\n");
while (inputFile.hasNextLine())
{
String line = inputFile.nextLine();
System.out.println(line);
}
inputFile.close();
break;
// Append to file
case 3:
FileWriter appendFile = new FileWriter(filename, true);
userInput = getUserInput();
System.out.println("I am appending your input to filename: " + filename + "\n");
appendFile.write(userInput);
appendFile.close();
break;
// Exit
case 4:
System.out.println("Goodbye!");
System.exit(0);
break;
}
// Ask user to play again
do
{
System.out.print("\nWould you like to continue (Y/N)? ");
choice = keyboard.next();
response = Character.toUpperCase(choice.charAt(0));
} while (response != 'Y' && response != 'N'); // Keep asking until user enters correct response
if (response == 'Y') // Play again
{
playAgain = true;
System.out.println();
}
else if (response == 'N') // Quit
{
playAgain = false;
System.out.println("\nThanks for playing!");
}
}
}
public static void displayMenu()
{
System.out.println("File IO Options:");
System.out.println("1) Write");
System.out.println("2) Read");
System.out.println("3) Append");
System.out.println("4) Quit");
System.out.print("What would you like to do: ");
}
public static String getUserInput()
{
String str;
Scanner keyboard = new Scanner(System.in);
System.out.println("Start typing. Hit enter when you're done:");
str = keyboard.nextLine();
return str;
}
}
这是我得到的错误:
文件IO选项:
1)写上
2)阅读
3)附加
4)退出
你想做什么:a
您必须输入1到4之间的数字!线程“main”中的例外情况
java.util.InputMismatchException
在java.util.Scanner.throwFor(Scanner.java:840)
在java.util.Scanner.next(Scanner.java:1461)
在java.util.Scanner.nextInt(Scanner.java:2091)
在java.util.Scanner.nextInt(Scanner.java:2050)
在Lab8FileIO.main(Lab8FileIO.java:41)
答案 0 :(得分:5)
您的计划失败,因为在您的catch
屏蔽中,您有keyboard.nextInt();
,这会引发新的未处理的InputMismatchException
catch (InputMismatchException e)
{
System.out.print("You must enter a number from 1 to 4!");
// You can remove this line, since it will be called anyways in the next loop iteration
//keyboard.nextInt();
// You can artificially set menuChoice = 1; for example for the loop to continue
}