番石榴的排序方式根据另一个列表？ [英] Guava way of sorting List according to another list?

问题描述

我有列表<整数> 包含我的用户ID。在数据库查询之后，我正在检索 List< User> 。我想根据第一个Id列表订购此列表。 列表<用户> 可能不包含部分ID。什么是Guava排序这个列表的方法？

I have List<Integer> consisting Ids of my Users. And after a database query, I am retrieving List<User>. I would like to order this list according to first Id list. List<User> may not include some of the Ids. What is the Guava way for sorting this list?

推荐答案

使用Guava的完全功能方式将结合 Ordering＃explicit（） with 订购＃onResultOf（）

The fully "functional" way, using Guava, would combine Ordering#explicit() with Ordering#onResultOf()

public class UserService {

    @Inject private UserDao userDao;

    public List<User> getUsersWithIds(List<Integer> userIds) {
        List<User> users = userDao.loadUsersWithIds(userIds);
        Ordering<User> orderById = Ordering.explicit(userIds).onResultOf(UserFunctions.getId());
        return orderById.immutableSortedCopy(users);
    }

}

你可以声明一个内联的匿名函数，但我喜欢在单独的类中将我的函数声明为静态工厂方法，以获得更清晰的代码（Java函数声明的详细程度隐藏在实用程序类中）：

You could declare an anonymous function inline, but I like to declare my functions as static factory methods in a separate class, for a cleaner code (the verbosity of Java's function declarations is hidden in the utility class):

/**
 * Static factory methods to create {@link Function}s for {@link User}s.
 */
public final class UserFunctions {
    private UserFunctions() { /* prevents instantiation */ }

    /**
     * @return a {@link Function} that returns an {@link User}'s id.
     */
    public static Function<User, Integer> getId() {
        return GetIdFunction.INSTANCE;
    }

    // enum singleton pattern
    private enum GetIdFunction implements Function<User, Integer> {
        INSTANCE;

        public Integer apply(User user) {
            return user.getId();
        }
    }

}

---

正如@Arne在评论中所提到的，这可以在Java 8中简化，使用方法引用而不是 UserFunctions 类：

public class UserService {

    @Inject private UserDao userDao;

    public List<User> getUsersWithIds(List<Integer> userIds) {
        List<User> users = userDao.loadUsersWithIds(userIds);
        Ordering<User> orderById = Ordering.explicit(userIds).onResultOf(User::getId);
        return orderById.immutableSortedCopy(immutableSortedCopy(users));
    }

}

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