异常处理不起作用

Question

我做了一个ATM程序。我有一个尝试捕获，它将要求用户输入他们的密码。PIN 码必须为 5 位数字。因此，异常将检查它是否为 5 位数字，但异常处理不起作用。无论我输入什么数字，它总是显示无效数字。

这是我的代码，try catch 位于程序顶部，异常处理 checkNumber 位于程序底部。

import java.util.ArrayList;
import java.util.Scanner;
public class BankMain
{
private double availableBal = 80;
private double totalBal = 100;
private double availableBal2 = 480;
private double totalBal2 = 500;
private double availableBal3 = 80;
private double totalBal3 = 100;
ArrayList<Integer> cardNum = new ArrayList<Integer>();
static Scanner input = new Scanner(System.in);
private String error; // String the error from the exception
{
error = "error";
}
public void cardNumbers()
{
Scanner cards = new Scanner(System.in);
Scanner input = new Scanner(System.in);
Scanner keyboard = new Scanner(System.in);
try
{
System.out.println("Please select a 5 digit card number");
cardNum.add(input.nextInt());
checkNumber();
}
catch (invalidNumber err)
{
System.out.println("Caught Error: " + err.getError());
}
System.out.println("Thank you! You're card number is " + cardNum);
System.out.println("Type 'c' to go back to main menu.");
String value = keyboard.next();
if (value.equalsIgnoreCase("c"))
{
menu();
}
else if (!keyboard.equals('c'))
{
System.out.println("Invalid Entry!");
}
}
public void menu()
{
System.out.println("ATM Menu:");
System.out.println();
System.out.println("1 = Create Account");
System.out.println("2 = Account Login");
System.out.println("3 = Exit ATM");
query();
}
public void startAtm()
{
menu();
}
public void drawMainMenu()
{
AccountMain main3 = new AccountMain();
int selection;
System.out.println("nATM main menu:");
System.out.println("1 - View account balance");
System.out.println("2 - Withdraw funds");
System.out.println("3 - Add funds");
System.out.println("4 - Back to Account Menu");
System.out.println("5 - Terminate transaction");
System.out.print("Choice: ");
selection = input.nextInt();
switch (selection)
{
case 1:
viewAccountInfo();
break;
case 2:
withdraw();
break;
case 3:
addFunds();
break;
case 4:
AccountMain.selectAccount();
break;
case 5:
System.out.println("Thank you for using this ATM!!! goodbye");
}
}
public void viewAccountInfo()
{
System.out.println("Account Information:");
System.out.println("t--Total balance: $" + totalBal);
System.out.println("t--Available balance: $" + availableBal);
drawMainMenu();
}
public void viewAccountInfo2()
{
System.out.println("Account Information:");
System.out.println("t--Total balance: $" + totalBal2);
System.out.println("t--Available balance: $" + availableBal2);
drawMainMenu();
}
public void deposit(int depAmount)
{
System.out.println("n***Please insert your money now...***");
totalBal = totalBal + depAmount;
availableBal = availableBal + depAmount;
}
public void checkNsf(int withdrawAmount)
{
if (totalBal - withdrawAmount < 0)
System.out.println("n***ERROR!!! Insufficient funds in you accout***");
else
{
totalBal = totalBal - withdrawAmount;
availableBal = availableBal - withdrawAmount;
System.out.println("n***Please take your money now...***");
}
}
public void addFunds()
{
int addSelection;
System.out.println("Deposit funds:");
System.out.println("1 - $20");
System.out.println("2 - $40");
System.out.println("3 - $60");
System.out.println("4 - $100");
System.out.println("5 - Back to main menu");
System.out.print("Choice: ");
addSelection = input.nextInt();
switch (addSelection)
{
case 1:
deposit(20);
drawMainMenu();
break;
case 2:
deposit(40);
drawMainMenu();
break;
case 3:
deposit(60);
drawMainMenu();
break;
case 4:
deposit(100);
drawMainMenu();
break;
case 5:
drawMainMenu();
break;
}
}
public void withdraw()
{
int withdrawSelection;
System.out.println("Withdraw money:");
System.out.println("1 - $20");
System.out.println("2 - $40");
System.out.println("3 - $60");
System.out.println("4 - $100");
System.out.println("5 - Back to main menu");
System.out.print("Choice: ");
withdrawSelection = input.nextInt();
switch (withdrawSelection)
{
case 1:
checkNsf(20);
drawMainMenu();
break;
case 2:
checkNsf(40);
drawMainMenu();
break;
case 3:
checkNsf(60);
drawMainMenu();
break;
case 4:
checkNsf(100);
drawMainMenu();
break;
case 5:
drawMainMenu();
break;
}
}
public void query()
{
Scanner keyboard = new Scanner(System.in);
double input = keyboard.nextInt();
if (input == 2)
{
BankMainPart2 main2 = new BankMainPart2();
System.out.println("Please enter your 5 digit card number.");
BankMainPart2.loginCard(cardNum);
}
else if (input == 1)
{
cardNumbers();
}
else if (input == 3)
{
System.out.println("Thank you, have a nice day!");
System.exit(0);
}
}
public void checkingMenu()
{
AccountMain main3 = new AccountMain();
int selection;
System.out.println("nATM main menu:");
System.out.println("1 - View account balance");
System.out.println("2 - Withdraw funds");
System.out.println("3 - Add funds");
System.out.println("4 - Back to Account Menu");
System.out.println("5 - Terminate transaction");
System.out.print("Choice: ");
selection = input.nextInt();
switch (selection)
{
case 1:
viewAccountInfo2();
break;
case 2:
withdraw();
break;
case 3:
addFunds();
break;
case 4:
AccountMain.selectAccount();
break;
case 5:
System.out.println("Thank you for using this ATM!!! goodbye");
}
}
private static void checkNumber() throws invalidNumber // run the check activation exception
{
if (String.valueOf(input).length() != 5)
{
throw new invalidNumber("invalid number");
}
else
{
System.out.println("Works!");
}
}
public static void main(String args[])
{
BankMain myAtm = new BankMain();
myAtm.startAtm();
}
}

Answer 1

这个代码片段看起来不错：

if (String.valueOf(input).length() != 5)
{
throw new invalidNumber("invalid number");
}
else
{
System.out.println("Works!");
}

只要您没有意识到input不是在某处本地声明的double变量：

double input = keyboard.nextInt()

相反，它是java.util.Scanner(！

static Scanner input = new Scanner(System.in)

Scanner.toString()肯定不是您想要的 PIN。

Answer 2

为什么你有三个不同的 Scanner 类实例。这就是使程序混乱的原因。

在compareNumber()方法中，您实际上是在检查input的值，这是 Scanner 的一个实例。最好这样使用它：-

checkNumber(input.nextInt())

并将数字添加到checkNumber(int num)方法的列表中。

当然，我并不是说这是一种很好的编码方式。但它会暂时解决你的问题。

否则，您的代码存在很多问题。

这是您的尝试捕获块：-

try {
System.out.println("Please select a 5 digit card number");
cardNum.add(input.nextInt());
checkNumber();
} catch (invalidNumber err) {
System.out.println("Caught Error: " + err.getError());
}

这是您的checkNumber()方法：-

private static void checkNumber() throws invalidNumber 
{
if (String.valueOf(input).length() != 5) {
throw new invalidNumber("invalid number");
}
else {
System.out.println("Works!");
}
}

现在您必须看到您正在使用input作为参数来String.valueOf(input)。

但是您在 try-catch 块之前已将"input"声明为扫描仪的实例。

Scanner cards = new Scanner(System.in);
Scanner input = new Scanner(System.in);
Scanner keyboard = new Scanner(System.in);

此代码位于您的codeNumbers()方法中。

因此，显然，您的input实际上永远不能包含用户输入，而是表示对象new Scanner(System.in)的hashcode。

因此，最好将整数输入从用户传递到checkNumber()方法。

代码中需要修改。

因此，您的 checkNumber()' 将被修改为： -

private static void checkNumber(int number) throws invalidNumber 
{
if (String.valueOf(number).length() != 5) {
throw new invalidNumber("invalid number");
}
else {
System.out.println("Works!");
}
}

您在try-catch块中对此方法的调用将更改为：-

try {
System.out.println("Please select a 5 digit card number");
int number = input.nextInt();
cardNum.add(number);
checkNumber(number);
} catch (invalidNumber err) {
System.out.println("Caught Error: " + err.getError());
}

Answer 3

您的代码检查String.valueOf(input)的长度是否为 5 个字符。但input不是用户输入的数字。它是类型Scanner的对象，用于分析用户输入的内容。所以String.valueOf(input)的结果可能是java.util.Scanner@B09876.