Question

我有两个数组列表。员工类和用户类的ArrayList。员工类有姓名、年龄、地址作为字段。用户类具有名称、年龄、地址作为字段。下面是两个列表

List<Employee> empList = new ArrayList<Employee>();
    empList.add(new Employee("Andi",20,"NY"));
    empList.add(new Employee("Rob",22,"london"));
    empList.add(new Employee("mark",21,"berlin"));
    
List<User> userList = new ArrayList<User>();
    userList.add(new User("Andi",20,"NY"));
    userList.add(new User("Rob",22,"london"));
    userList.add(new User("mark",21,""));

想检查用户是否与员工地址相同。如果用户没有地址，则从员工那里复制。

Answer 1

也许这可以帮到你。如果您有任何问题，请给我留言

for(User user: userList) {                        
    for(Employee employee: empList) {           
        if(user.getName().equals(employee.getName())) {
            if(user.getAddress() == null || user.getAddress().equals("")) {
                user.setAddress(employee.getAddress());
            }
        }
    }
}
userList.forEach(user -> System.out.println(user));

Answer 2

我猜你想要什么。像这样 - 如果具有此名称的用户没有地址，则获取具有匹配名称的员工。将用户地址设置为员工地址。这很有用意味着名称是员工和用户的唯一键。如果您之后需要该员工，您可以从 Optional 中将其退回。

User andi = userList.stream().filter(u -> u.getName().equals("Andi")).findFirst().orElseThrow();
if (andi.getAddress() == null){
    Optional<Employee> empForUser = empList.stream().filter(e -> andi.getName().equals(e.getName())).findFirst();
    empForUser.ifPresent(employee -> andi.setAddress(employee.getAddress()));

}

Answer 3

一次迭代使用 Hashmap 的解决方案。

import java.util.ArrayList;
import java.util.List;
import java.util.Map;
import java.util.stream.Collectors;

class User {
    String name;
    int age;
    String address;

    public User(String name, int age, String address) {
        this.name = name;
        this.age = age;
        this.address = address;
    }

    public String getName() {
        return name;
    }
    public String getAddress() {
        return address;
    }

    public void setAddress(String address) {
        this.address = address;
    }

    @Override
    public String toString() {
        return "User{" +
                "name='" + name + '\'' +
                ", age=" + age +
                ", address='" + address + '\'' +
                '}';
    }
}
class Employee{
    String name;
    int age;
    String address;

    public Employee(String name, int age, String address) {
        this.name = name;
        this.age = age;
        this.address = address;
    }


    public String getName() {
        return name;
    }

    public String getAddress() {
        return address;
    }


    @Override
    public String toString() {
        return "Employee{" +
                "name='" + name + '\'' +
                ", age=" + age +
                ", address='" + address + '\'' +
                '}';
    }
}


public class Test {

    public static void main(String args[]) throws Exception {

        List<Employee> empList = new ArrayList<Employee>();
        empList.add(new Employee("Andi",20,"NY"));
        empList.add(new Employee("Rob",22,"london"));
        empList.add(new Employee("mark",21,"berlin"));

        List<User> userList = new ArrayList<>();
        userList.add(new User("Andi",20,"NY"));
        userList.add(new User("Rob",22,"london"));
        userList.add(new User("mark",21,""));

        Map<String, Employee> map = empList.stream()
                .collect(Collectors.toMap(Employee::getName, employee -> employee));
        for(User user : userList){
            Employee employee = map.get(user.getName());
            if(employee==null){
                continue;
            }
            if(user.getAddress() == null || user.getAddress().equals("")) {
                user.setAddress(employee.getAddress());
            }

        }
        userList.forEach(System.out::println);


    }

}

检查两个不同的对象是否相等

3 个答案: