Question

public class CHECK {
public CHECK(){
    String []wrkrs = {"Денис", "Саша", "Наталья", "Анатолий", "Юра", "Коля", "Катя", "Дима", "Антон","Тамара"};
    int [] wrkrsPhone = {22626,22627,22628,22629,22630,22631,22632,22633,22634,22635};
    String a = JOptionPane.showInputDialog(null, "Hello,friend!Do you wanna know, is that guy at work?Enter name:");


    if(Arrays.asList(wrkrs).contains(a)){
        JOptionPane.showMessageDialog(null, "That guy is at work!");
        JOptionPane.showMessageDialog(null, "calling " + wrkrsPhone[wrkrsPhone.toString().indexOf(a)]);
    }else{
        JOptionPane.showMessageDialog(null, "Такого сотрудника нет!");
    }
}

我有两个数组，包含整数和字符串。如您所见，我想将字符串数组中的元素数（例如，wrkrs number 3）添加到int数组中，称为wrkrs phone。我该怎么做？我尝试过IndexOf，但它没有用。

输出，我想要的是：

Enter name:
Юра
That guy is at work!
Calling Юра + wrkrsPhone(Юра).

Answer 1

更好的解决方案是拥有一个包含工作人员姓名和电话号码的工人类。

然后，您可以使用HashMap<String,Worker>代替数组来存储数据。

这使搜索更有效：

Map<String,Worker> workersMap = new HashMap<>();
workersMap.put ("Денис", new Worker ("Денис", 22626));
...
Worker worker = workersMap.get(a);
if (worker != null) {
    call (worker.getPhone()); // or do whatever you want to do with the phone number
}

这比Arrays.asList(wrkrs).contains(a)更有效，后者在列表上执行线性搜索。

Answer 2

...
List list = Arrays.asList(wrkrs);
if(list.contains(a)){
    JOptionPane.showMessageDialog(null, "That guy is at work!");
    JOptionPane.showMessageDialog(null, "calling " + wrkrsPhone[list.indexOf(a)]);
}
...

您最好使用Map或提取具有Contact和name属性的类phone，但我认为这就是您所寻求的：）

Java中的Int和String数组

2 个答案: