If语句代码问题
If语句代码问题
提问于 2013-03-07 23:33:04
我的教授给了我下面的代码。
public static void main(String[] args) {
Customer customer;
Transaction transaction;
double withdrawalAmount = 0;
boolean finished = false;
while (finished == false)
{
// Menu Display and Get user input
int inputInt = 0;
while (inputInt == 0)
{
inputInt = displayMenuAndGetInput();
// if the input is out of range
if ((inputInt < 1) || (inputInt > 8))
{
System.out.println("\nThe input is out of range!");
System.out.println();
inputInt = 0;
}
} //end while
// switch to correspondence function
switch (inputInt)
{
case 1:
customer = createPersonalCustomer();
System.out.println("\nThe Personal customer has been created: \n" + newPC.toString());
customers.add(customer);
break;
case 2:
customer = createCommercialCustomer();
System.out.println("\nThe Commercial customer has been created: \n" + newCC.toString());
customers.add(customer);
break;
case 3:
transaction = recordTransaction();
if(transaction != null)
System.out.println("\nThe Transaction has been created: \n" + trans.toString());
else
System.out.println("\nThe ID could not be found.");
break;
case 4:
withdrawalAmount = makeWithdrawal();
if(withdrawalAmount > 0)
System.out.println("\nAmount withdrawn from this account: " + moneyFormat.format(acct.getMakeWithdrawal()) + "\n");
else
System.out.println("\nThe ID could not be found.");
break;
case 5:
displayCustomer();
break;
case 6:
displayCustomerSummary();
break;
case 7:
displayGrandSummary();
break;
case 8:
// exit
finished = true;
break;
default:
System.out.println("Invalid Input!");
break;
} // end switch
} // end while
}// Create a new Transaction
public static Transaction recordTransaction(){}并创建一个在以下场景中工作的循环：
输入客户id，如果客户id在数组中不匹配，则生成在情况3中读出的错误，并显示主菜单。如果客户id有效，则用户在下面输入信息。
下面是我的代码
public static Transaction recordTransaction(){
System.out.println("Enter the customer ID to create the transaction > ");
long customerID = scan.nextLong();
for (Customer c : customers) {
if (c.getCustomerID() == customerID) {
if (trans != null) {
System.out.println("\nEnter the weight of gold > ");
Transaction.goldWt = scan.nextDouble();
System.out.println("\nEnter the weight of platinum > ");
Transaction.platinumWt = scan.nextDouble();
System.out.println("\nEnter the weight of silver > ");
Transaction.silverWt = scan.nextDouble();
}
}
return null;
}无论如何，我已经运行了许多方法，我的代码要么接受无效和有效的客户ID，要么不接受无效或有效的客户id。我知道我可能忽略了一些东西，这就是为什么我拼命地请求论坛的帮助。当涉及到编程时，我有OCD倾向，这是我的第一个java入门类，所以我不是很精通这门语言。在过去的两天里，我一直被这个问题卡住了。请帮帮忙。
回答 3
Stack Overflow用户
回答已采纳
发布于 2013-03-07 23:43:44
您需要在recordTransaction()方法中实例化一个new Transaction()，并在适当的时候返回它。
public static Transaction recordTransaction(){
System.out.println("Enter the customer ID to create the transaction > "); long customerID = scan.nextLong();
for (Customer c : customers) {
if (c.getCustomerID() == customerID) {
Transaction trans = new Transaction();
System.out.println("\nEnter the weight of gold > ");
trans.goldWt = scan.nextDouble();
System.out.println("\nEnter the weight of platinum > ");
trans.platinumWt = scan.nextDouble();
System.out.println("\nEnter the weight of silver > ");
trans.silverWt = scan.nextDouble();
return trans;
}
}
return null;
}Stack Overflow用户
发布于 2013-03-07 23:44:09
关于您的代码的一些备注：
if (trans != null) {什么是“反式”？我猜它一定是引用了一个事务实例。
Transaction.goldWt = ...那不应该是"trans.goldWt“吗？(当然，植物和白银也是如此)。如果它真的是Transaction.goldWt，那么您总是在更改该单个值。
return null;你从不返回任何东西，所以调用代码总是写成"no ID“。你不应该说"return trans“吗？
Stack Overflow用户
发布于 2013-03-08 11:28:58
该问题是由正在创建的每个帐户的随机customerID的getter和setter方法生成的。
谢谢你，谢谢你！
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/15275176
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