英文:
How to assign multiple data types to a single variable in java?
问题
现在我正在创建一个银行应用程序，我不知道如何将例如 string name; double balance; 等分配给正确的 int PIN;。将会有许多带有不同PIN和不同值的帐户。我尝试创建多个对象：
perInfo card1 = new perInfo();
card1.PIN = 1994;
card1.balance = 24.68;
card1.isValid = true;
perInfo karta2 = new perInfo();
card2.PIN = 2002;
card2.balance = 522.2;
card2.isValid = false;
但我认为这样做太麻烦了，而且会影响应用程序的性能。我还尝试过创建一个列表：
public bApp(int pin, double balance) {
this.pin = pin;
this.balance = balance;
}
List<bApp> pass = new ArrayList<>();
pass.add(new bApp(1994, 568.45));
pass.add(new bApp(2002, 13866.24));
但这没有起作用，因为我无法调用PIN来检查用户是否提供了正确的PIN。数组也不适用于这种情况。
英文:
Right now I'm creating a bankApp and I don't have an idea how can I assign e.g. string name; double balance; etc. to a right one int PIN;. There is gonna be many accounts with different PINs and different values assigned to it . I tried making many objects :
perInfo card1 = new perInfo();
card1.PIN = 1994;
card1.balance = 24.68;
card1.isValid = true;
perInfo karta2 = new perInfo();
card2.PIN = 2002;
card2.balance = 522.2;
card2.isValid = false;
but I think it's too much work to do and it'll worsen performance of the app. I also tried making a list
public bApp(int pin, double balance){
this.pin = pin;
this.balance = balance;}
List<bApp> pass = new ArrayList<>();
pass.add(new bApp(1994, 568.45));
pass.add(new bApp(2002, 13866.24));
but it didn't work ,because I couldn't call the PIN to check that if user has provided the right one PIN . Arrays are also not suited for this.
答案1
得分: 0
你正在寻找的数据结构是一个映射（Map）。看一下 HashMap，你会想要以你用来查找的任何值作为键。
例如，如果你想通过 PIN 码查找用户：
Map<Integer, bApp> passes = new HashMap<>();
passes.put(1994, new bApp(1994, 568.45));
passes.put(2002, new bApp(2002, 13866.24));
英文:
The data structure you are looking for is a Map. Take a look at HashMap and you'll want to key off of whatever value you are using to lookup.
For example if you wanted to lookup a user by pin:
Map<Integer, bApp> passes = new HashMap<>();
passes.put(1994, new bApp(1994, 568.45));
passes.put(2002, new bApp(2002, 13866.24));
答案2
得分: 0
你可以使用HashMap来实现这个，将pin作为键存储在HashMap中，并将对象存储在其中。这将允许您仅使用pin来访问每张卡。然而，这将不允许重复的pin。我建议您通过唯一的ID来引用每个账户，并在对象内部检查pin。
HashMap<Integer, Account> accounts = new HashMap<Integer, Account>();
accounts.put(12345678, new Account(1994, 568.4));
然后，您可以使用唯一的ID获取账户，并检查pin是否正确。
Account acc = accounts.get(uniqueID);
if(acc.pin == enteredPin){
//无论您需要做什么
}
英文:
You could use a HashMap for this and use the pin as a key and store the object in the HashMap. This would allow you to access each card using only the pin. However, this would not allow for duplicate pins. I would recommend that you reference each account by a unique ID and check the pin within the object itself.
HashMap<Integer, Account> accounts = new HashMap<Integer, Account>();
accounts.put(12345678, new Account(1994, 568.4));
You can then get the account using the unique ID and check if the pin is correct.
Account acc = accounts.get(uniqueID);
if(acc.pin == enteredPin){
//Whatever you need to do
}
答案3
得分: 0
我认为只需要一个包含不同卡片的数组，每个 PIN 设置为数组地址会更简单：
bApp[] cards = new bApp[NUMBER_OF_CARDS];
cards[0] = new bApp(0, 568.45);
// ...
cards[2002] = new bApp(2002, 13866.24);
英文:
I think it would be simpler to just have an array of the different cards, with each pin being set to the array address:
bApp[] cards = new bApp[NUMBER_OF_CARDS];
cards[0] = new bApp(0, 568.45);
// ...
cards[2002] = new bApp(2002, 13866.24);
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