在Java中合并两个对象列表

Question

如何合并两个用户列表，如果用户是普通用户则合并其属性。

我有两个用户列表，从数据库获取第一个列表，然后从Web服务获取第二个列表，如果它们的名字和姓氏相同，则用户合并属性并返回合并的列表。

import java.util.List;
import java.util.Objects;

 class User {
     String firstName;
     String lastName;
     Integer count;
     Integer marks;
     String status;

     @Override
     public boolean equals(Object o) {
         if (this == o) return true;
         if (!(o instanceof User)) return false;
         User user = (User) o;
         return Objects.equals(firstName, user.firstName) &&
                 Objects.equals(lastName, user.lastName);
     }

     @Override
     public int hashCode() {
         return Objects.hash(firstName, lastName);
     }

    //constructor getter setter....

 }




 public class MainTest {

     public static void main(String[] args) {

         List<User> source1 = List.of(new User("A", "A", 1, 0, null),
                 new User("B", "B", 2, 0, null),
                 new User("C", "C", 3, 0, null));

         List<User> source2 = List.of(new User("A", "A", 0, 10, "FAIL"),
                 new User("B", "B", 0, 20, "FAIL"),
                 new User("D", "D", 0, 30, "PASS"));


         List<User> combinedExpectedList = List.of(new User("A", "A", 1, 10, "FAIL"),
                 new User("B", "B", 2, 20, "FAIL"),
                 new User("C", "C", 3, 0, null),
                 new User("D", "D", 0, 30, "PASS"));

         //List<User> userList3= Stream.concat(source1.stream(), source2.stream()).distinct().collect(Collectors.toList());//Adding 4 records but marks and status not copied
         //BeanUtils.copyProperties(model2, model1); - can't user don't have library

         for (User t : source2) {
             for (User s : source1) {
                 if (t.getFirstName().equals(s.getFirstName()) && t.getLastName().equals(s.getLastName())) {
                     t.setCount(s.getCount());
                 }
             }
         }// Not adding record with firstName=C and LastName=C

         System.out.println(combinedExpectedList.equals(source2));
     }
 }

Answer 1

假设您要保留非零标记和非null状态元素，则不能直接使用distinct。您可以如下使用Collectors.toMap，

List<User> userList3 = Stream.concat(source1.stream(), source2.stream())
                .collect(Collectors.toMap(e -> e.firstName + e.lastName, e -> e,
                        (User u1, User u2) -> {
                            if (u1.marks == 0)
                                u1.marks = u2.marks;
                            if (u1.status == null)
                                u1.status = u2.status;
                            return u1;
                        })).values().stream().collect(Collectors.toList());

注意，您可能需要重新排序元素。

Answer 2

类似于this answer，但此处的源列表中的“用户”保持不变：

List<User> values = new ArrayList<>(Stream.concat(source1.stream(), source2.stream())
        .collect(Collectors.toMap(e -> e.firstName + e.lastName, Function.identity(),
                (User u1, User u2) ->
                        new User(u1.firstName, u1.lastName, u1.count,
                                (u1.marks == 0) ? u2.marks : u1.marks,
                                (u1.status == null) ? u2.status : u1.status
                        )
        )).values());