Question

我有这个while循环，当用户输入费用大于交易金额时运行。例如，如果费用为4美元而您只需支付2美元，则会说您还欠2美元并提示您输入更多付款。问题是，如果付款仍然很短，我无法弄清楚如何在付款后更新变量交易。在它要求另一笔付款之后，说你仍然缺少4美元并再支付1美元，这将给你总共3美元，该计划应该说你仍然短缺1美元。尽管如此，该计划仍然表示您的原始金额很短，即2美元。

while (transaction < feeSum)
{
    double underPay = feeSum - transaction;
    System.out.println("The transaction did not meet the fee by $" + underPay);
    System.out.println("Please enter another payment to complete the balance.");

    System.out.println("Enter a number of payments.");
    int paymentSize2 = keyboard.nextInt();
    double[] payments2 = new double[paymentSize2];

    System.out.println("Enter " + payments2.length + " payment(s).");
    double paymentSum2 = 0;
    for(int i = 0; i < payments2.length; i++)
    {
        payments2[i] = keyboard.nextDouble();
        paymentSum2 = paymentSum2 + payments[i];
        transaction +=  paymentSum2; //<<<<<<< Shouldn't this update transaction? 
    }   // The second time around it should say the trans did not meet fee by $1

    if (paymentSum2 == underPay)
    {
        System.out.println("There is now no outstanding balance.");
        break;
    }

Answer 1

我重新编写了一些代码，这可行，但它并不包含您实现的所有额外内容。我真的不明白为什么你需要这些数组。如果您支付的费用超过了您的预期，则此代码无法处理，但这可以在while循环的最后几行中轻松实现。

double underPay = feeSum - transaction;
            while (underPay != 0) {
                System.out.println("The transaction did not meet the fee by $" + underPay);
                System.out.println("Please enter another payment to complete the balance.");
                int paymentSizeNext = keyboard.nextInt();
                underPay -= paymentSizeNext;


            }
            System.out.println("There is now no outstanding balance.");

更新While循环中的变量

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