ArrayList - 使用String获取“Contain”和“lastIndexOf”

Question

我正在尝试使用String获取方法“contains”和“lastIndexOf”的正确输出，但它不会为我提供正确的输出，在这种情况下，false或true表示“包含”和位置“lastIndexOf”的元素。 我该怎么办？ 非常感谢。

public class Employee {
    public static final int size = 0;
    String firstName;
    String surname;
    int yearOfBirth;
    String PPSNumber;
    String email;
    String phoneNumber;

    public Employee(String firstName, String surname, int yearOfBirth, String PPSNumber, String email, String phoneNumber) {
        this.firstName = firstName;
        this.surname = surname;
        this.yearOfBirth = yearOfBirth;
        this.PPSNumber = PPSNumber;
        this.email = email;
        this.phoneNumber = phoneNumber;
    }
}

// // ____________________________________________________

import java.util.ArrayList;

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

        employeeList.add(new Employee("Charlie", "Charles", 1991, "234567b", "charlie@b.ie", "7654321"));
        employeeList.add(new Employee("David", "Davies", 1992, "5213452d", "david@d.ie", "352135613"));
        employeeList.add(new Employee("Levi", "Silva", 1990, "1234", "Levi@b.ie", "333333"));
        employeeList.add(new Employee("Gus", "Silva", 1993, "4321", "Gus@b.ie", "444444"));

        for (Employee getName : employeeList) {

            System.out.print(getName.firstName + ",         ");

        }

        System.out.println(" ");

        // contains(Employee)
        Employee Emp = new Employee("Gus", "Silva", 1993, "4321", "Gus@b.ie", "444444");
        System.out.println("Contains String: " + employeeList.contains(Emp));    // can't make give me the right contain answer


        // lastIndexOf()    
        Employee Charlie1 = new Employee("Charlie", "Charles", 1991, "234567b", "charlie@b.ie", "7654321");
        System.out.println("lastIndexOf:  " + employeeList.lastIndexOf(Charlie1));    // can't find the lastIndexOf


    }
}

Answer 1

您需要在Employee （或任何其他类）中实现equals和hashCode ，以便contains ， sort或其他方法有效。

例如，此构思向导默认使用所有字段：

class Employee {
    public static final int size = 0;
    String firstName;
    String surname;
    int yearOfBirth;
    String PPSNumber;
    String email;
    String phoneNumber;

    public Employee(String firstName, String surname, int yearOfBirth, String PPSNumber, String email, String phoneNumber) {
        this.firstName = firstName;
        this.surname = surname;
        this.yearOfBirth = yearOfBirth;
        this.PPSNumber = PPSNumber;
        this.email = email;
        this.phoneNumber = phoneNumber;
    }

    @Override
    public boolean equals(Object o) {
        if (this == o) return true;
        if (o == null || getClass() != o.getClass()) return false;

        Employee employee = (Employee) o;

        if (yearOfBirth != employee.yearOfBirth) return false;
        if (firstName != null ? !firstName.equals(employee.firstName) : employee.firstName != null) return false;
        if (surname != null ? !surname.equals(employee.surname) : employee.surname != null) return false;
        if (PPSNumber != null ? !PPSNumber.equals(employee.PPSNumber) : employee.PPSNumber != null) return false;
        if (email != null ? !email.equals(employee.email) : employee.email != null) return false;
        return phoneNumber != null ? phoneNumber.equals(employee.phoneNumber) : employee.phoneNumber == null;
    }

    @Override
    public int hashCode() {
        int result = firstName != null ? firstName.hashCode() : 0;
        result = 31 * result + (surname != null ? surname.hashCode() : 0);
        result = 31 * result + yearOfBirth;
        result = 31 * result + (PPSNumber != null ? PPSNumber.hashCode() : 0);
        result = 31 * result + (email != null ? email.hashCode() : 0);
        result = 31 * result + (phoneNumber != null ? phoneNumber.hashCode() : 0);
        return result;
    }
}

现在，只需添加所需或必填字段，您将看到contains工作方式。

Answer 2

作为lastIndexOf的javadoc状态，如果找不到该对象，则得到-1 。 但是找不到它的原因是，你正在处理一个Employee列表而不是一个String列表。 您当然可以在Employee -class上实现equals和hashCode ，但这样你就不够灵活了。 相反，你应该只是过滤你需要的元素列表，例如

// Get (or show) all employees, whose name is "some string" using Stream API
employeeList.stream()
            .filter(employee -> employee.firstName.contains("some string"))
            .collect(Collectors.toList()); // or: .forEach(System.out::println);

关于获取条目的最后一个索引....你真的需要这样的功能吗？ 如果是这样，并且您没有要查询的实际对象（“Gus”的Employee对象）并且您不想实现hashCode和equals ，我建议您只是遍历列表并在第一次出现时break ：

int lastIndex = -1;
for (int i = employeeList.size() - 1; i >= 0 ; i--) {
  Employee anEmployee = employeeList.get(i);
  if (anEmployee.firstName.equals("Gus")) {
    lastIndex = i;
    break;
 }
}
System.out.printf("lastIndexOf: %s%n", lastIndex);

Answer 3

您的employeeList被声明为ArrayList<Employee> ，因此您的employeeList.lastIndexOf("Gus")将无法工作，因为该列表包含Employee对象，而"Gus"是一个String因此它在列表中没有任何位置。

关于“包含”，我无法在您的代码中找到对ArrayList.contains(*)调用（前提是您正在讨论的方法）。 你能指出来吗？