要指出太多缺陷。

Question

我编写了一部分代码来获取用户输入，将其与字符串值匹配，然后使用相关的double值进行计算：

double [] currency = new double[] {0.05,0.10,0.20,0.50,1.00,2.00,5.00,10.00,20.00,50.00,100.00};
String [] currencytext = {"$0.05","$0.10","$0.20","$0.50","$1.00","$2.00","$5.00","$10.00","$20.00","$50.00","$100.00"};
Scanner keyboard = new Scanner(System.in);

for (int i = 0; i < currencytext.length; i++) {
    boolean valid = false;
    while(!valid){
        System.out.format("$%.2f remains to be paid. Enter coin or note: ",sum);
        String payment = keyboard.next();
        if(payment.equals(currencytext[i])){
            sum = sum - currency[i];
            if(sum == 0) {
                System.out.print("You gave " + payment);
                System.out.print("Perfect! No change given.");
                System.out.print("");
                System.out.print("Thank you" + name + ".");
                System.out.print("See you next time.");
                }
                }
        if(!(payment.equals(currencytext[i]))) {
            System.out.print("Invalid coin or note. Try again. \n");
                }
        if(payment.equals(currencytext[i]) && currency[i] > sum){
            System.out.print("You gave " + payment);
            System.out.print("Your change:");
        }
    }
}

问题在于，当它获得用户输入时，除了$ 0.05之外，它不匹配任何字符串值。在我看来，它不是正确地迭代数组，但我无法弄清楚原因。有人能在这里看到问题吗？

Answer 1

要指出太多缺陷。

然而，

当currencytext[i]与payment不匹配时，它会执行以下代码：

System.out.print("Invalid coin or note. Try again. \n");
System.out.format("$%.2f remains to be paid. Enter coin or note: ",sum);
payment = keyboard.next();

因此，它会在所有执行此操作时输入与currencytext[i]不匹配的次数。而且，在这个区块中，你有

payment = keyboard.next();

因此，它要求在此块本身中输入新输入。因此，您将获得除$0.05之外的所有输入的所述输出。

就$0.05而言，您的第一个if块成功执行，并且不打印任何输出。因此，它移动到while循环的下一次迭代，同样payment保持不变（$0.05），但currencytext[i]变为$0.10。所以他们不匹配，你得到了所说的输出。

如何解决此问题：

使用此代码，您需要执行批次更正。

我建议你再从头开始。

Answer 2

如果它不合适，则将有效设置为true，因此代码有机会检查@Override protected void onCreate(Bundle savedInstanceState) { super.onCreate(savedInstanceState); setContentView(R.layout.activity_main); // Find Views by IDs : ourButton = (ImageButton) findViewById(R.id.gobutton); operand1 = (EditText) findViewById(R.id.edit_Name); ourButton.setOnClickListener(new View.OnClickListener() { @Override public void onClick(View v) { if (operand1.getText().toString().trim().length() > 0) { Intent myIntent = new Intent(MainActivity.this, ScreenName.class); MainActivity.this.startActivity(myIntent); } else { Toast.makeText(MainActivity.this, "You did not enter a username", Toast.LENGTH_LONG).show(); } } }); }处的第一项currencytext[0]。然后$0.05也是如此，您的代码在那里打印行。您的!payment.equals(currencytext[i])也未正确嵌套。

Answer 3

这是您的问题的可能解决方案

    Scanner keyboard = new Scanner(System.in);

    double [] currency = new double[] {0.05,0.10,0.20,0.50,1.00,2.00,5.00,10.00,20.00,50.00,100.00};
    String [] currencytext = {"$0.05","$0.10","$0.20","$0.50","$1.00","$2.00","$5.00","$10.00","$20.00","$50.00","$100.00"};

    String payment = keyboard.next();

    double sum = 100; // <- Working example - Read sum from keyboard entry

    while (sum > 0) {

        boolean paymentFound = false;

        for (int i = 0; i < currencytext.length; i++) {

            if (payment.equals(currencytext[i])) {

                sum = sum - currency[i];
                paymentFound = true;

                if (sum == 0) {
                    System.out.println("You gave " + payment);
                    System.out.println("Perfect! No change given.");

                    // System.out.print("Thank you" + name + ".");

                    System.out.println("See you next time.");
                    break;
                } else if (sum < 0) {
                    System.out.println("You gave " + payment);
                    System.out.println("Your change:" + (-1 * sum));
                    break;
                }
            }
        }

        if (!paymentFound) {
           System.out.println("Invalid coin or note. Try again. \n");

        }

        if (sum > 0) {
            System.out.format("$%.2f remains to be paid. Enter coin or note: ", sum);
            payment = keyboard.next();
        }
    }

while循环将继续，直到付款已满。

for循环遍历数组，直到找到合适的付款

如果找到合适的付款，我们会从sum中扣除。在这两种情况下，我们使用break退出for循环。没有必要继续搜索。

如果找不到合适的付款[!paymentFound]，我们会继续询问。

    if (!paymentFound) {
       System.out.println("Invalid coin or note. Try again. \n");

    }

    if (sum > 0) {
        System.out.format("$%.2f remains to be paid. Enter coin or note: ", sum);
        payment = keyboard.next();
    }

程序将在（sum＆lt; 0）结束时结束，在这种情况下，while循环退出。我使用println代替print来改善邮件易读性。

Answer 4

我不知道你是如何阅读输入的。您可以做的一项改进是在for循环中写入读取输入代码。

Scanner scanner = new Scanner(System.in);
for (... ) {
....
String payment = scanner.nextLine();
....
}

正确的用户输入与数组值不匹配

4 个答案:

要指出太多缺陷。

然而，