Question

我对编程很新（大约一年），而且我遇到了逻辑上的差距。假设我正在构建员工数据库：如何在不对员工1，员工2等进行硬编码的情况下创建员工实例。

我被建议使用数组，但在Java中，数组必须有一个设置长度，所以我不确定如何实现这一点。任何帮助都会很棒！ -Matt

编辑：感谢大家的帮助。我不知道ArrayList类型！

Answer 1

ArrayLists可以改变大小。这是一个将20个对象添加到数组List的简单循环：

ArrayList<Object> array = new ArrayList<>();
for(int i =0; i < 20; i ++) {
    array.add(new Object());
}

这是一种简单的方法，可以一次创建大量对象，而且您不必具有指定的大小，因此您可以轻松更改ArrayList

中包含的对象数量

Answer 2

您可以使用Employee

类型的ArrayList

https://docs.oracle.com/javase/8/docs/api/java/util/ArrayList.html

Answer 3

使用ArrayList（在这种情况下无需指定列表的长度）

List<Employee> employees = new ArrayList<>();
while (resultSet.next()) {
    Employee employee = new Employee();
    employee.setId(resultSet.getInt("id"));
    employee.setFirstname(resultSet.getString("firstname"));
    .....
    employees.add(employee);
}

Answer 4

创建一个类来保存员工信息。

Employee.java

public class Employee {
    String firstName;
    String lastName;
    String deptName;

    public Employee() {
    }

    public Employee(String firstName, String lastName, String deptName) {
        this.firstName = firstName;
        this.lastName = lastName;
        this.deptName = deptName;
    }

    public String getFirstName() {
        return firstName;
    }

    public void setFirstName(String firstName) {
        this.firstName = firstName;
    }

    public String getLastName() {
        return lastName;
    }

    public void setLastName(String lastName) {
        this.lastName = lastName;
    }

    public String getDeptName() {
        return deptName;
    }

    public void setDeptName(String deptName) {
        this.deptName = deptName;
    }

    @Override
    public String toString() {
        return "Employee{" +
               "firstName='" + firstName + '\'' +
               ", lastName='" + lastName + '\'' +
               ", deptName='" + deptName + '\'' +
               '}';
    }
}

创建多名员工并打印出来的示例代码

Employees.java

import java.util.ArrayList;
import java.util.List;

public class Employees {

    public static void main(String[] args) {
        List<Employee> employeeList = new ArrayList<>();

        employeeList.add(new Employee("FN-1","LN-1","Dept-1"));
        employeeList.add(new Employee("FN-2","LN-2","Dept-2"));
        employeeList.add(new Employee("FN-3","LN-3","Dept-3"));

        //loop
        for (int i = 4; i < 8; i++) {
            employeeList.add(new Employee("FN-" + i, "LN-" + i, "Dept-" + i));
        }

        //print
        employeeList.forEach(System.out::println);
    }
}

<强>输出：

Employee{firstName='FN-1', lastName='LN-1', deptName='Dept-1'}
Employee{firstName='FN-2', lastName='LN-2', deptName='Dept-2'}
Employee{firstName='FN-3', lastName='LN-3', deptName='Dept-3'}
Employee{firstName='FN-4', lastName='LN-4', deptName='Dept-4'}
Employee{firstName='FN-5', lastName='LN-5', deptName='Dept-5'}
Employee{firstName='FN-6', lastName='LN-6', deptName='Dept-6'}
Employee{firstName='FN-7', lastName='LN-7', deptName='Dept-7'}

实现类

4 个答案: