如何比较两个在java中至少有一个属性不同的类似对象的数组列表？

Question

我有两个数组列表.每个都有User类型的对象列表.

User类如下所示

    public class User {

    private long id;

    private String empCode;

    private String firstname;

    private String lastname;

    private String email;

    public User( String firstname, String lastname, String empCode, String email) {
        super();
        this.empCode = empCode;
        this.firstname = firstname;
        this.lastname = lastname;
        this.email = email;
    }

    // getters and setters

}


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

public class FindSimilarUsersWithAtLeastOneDifferentProperty {

    /**
     * @param args
     */
    public static void main(String[] args) {
        // TODO Auto-generated method stub

        List<User> list1 = new ArrayList<User>();

        list1.add(new User("F11", "L1", "EMP01", "u1@test.com"));
        list1.add(new User("F2", "L2", "EMP02", "u222@test.com"));
        list1.add(new User("F3", "L3", "EMP03", "u3@test.com"));
        list1.add(new User("F4", "L4", "EMP04", "u4@test.com"));
        list1.add(new User("F5", "L5", "EMP05", "u5@test.com"));
        list1.add(new User("F9", "L9", "EMP09", "u9@test.com"));
        list1.add(new User("F10", "L10", "EMP10", "u10@test.com"));

        List<User> list2 = new ArrayList<User>();

        list2.add(new User("F1", "L1", "EMP01", "u1@test.com"));
        list2.add(new User("F2", "L2", "EMP02", "u2@test.com"));
        list2.add(new User("F6", "L6", "EMP06", "u6@test.com"));
        list2.add(new User("F7", "L7", "EMP07", "u7@test.com"));
        list2.add(new User("F8", "L8", "EMP08", "u8@test.com"));
        list2.add(new User("F9", "L9", "EMP09", "u9@test.com"));
        list2.add(new User("F100", "L100", "EMP10", "u100@test.com"));

        List<User> resultList = new ArrayList<User>();
        // this list should contain following users
        // EMP01 (common in both list but differs in firstname)
        // EMP02 (common in both list but differs in email)
        // EMP10 (common in both list but differs in firstname, lastname and email)


    }

}

如果您看到示例代码,则这两个列表有四个用户,其代码为EMP01,EMP02,EMP09和EMP10.

因此,我们只需要比较这四个用户的属性.

如果任何用户至少有一个不同的属性,则应将其添加到结果列表中.

请告知我该如何解决这个问题？

Answer 1

实现equals,hashcode inUser

    @Override
    public boolean equals(Object obj) {
        if (obj == null)
            return false;
        if (!(obj instanceof User))
            return false;
        User u = (User) obj;
        return this.empCode == null ? false : this.empCode
                .equals(u.empCode);
    }

    @Override
    public int hashCode() {
        return this.empCode == null ? 0 : this.empCode.hashCode();
    }

    @Override
    public String toString() {
        return "Emp Code: " + this.empCode;
    }

然后使用retainAll

list2.retainAll(list1);-->EMP01, EMP02, EMP09, EMP10

Answer 2

我认为这是你应该做的 -

for(User user1 : list1) {
    for(User user2 : list2) {
        if(user1.getEmpCode().equals(user2.getEmpCode())) {
            if(!user1.getFirstName().equals(user2.getFirstName()) ||
               !user1.getLastName().equals(user2.getLastName()) ||
               !user1.getEmail().equals(user2.getEmail())) {
                resultList.add(user1);
            }
        }
    }
}

User覆盖equal并hashCode仅用于此目的可能没有意义。它们应该以在域方面更有意义的方式被覆盖。