例如,我有以下设置,
像这样的模型:
public class Post {
@Id
private String id;
private String post;
private List<Vote> votes = new ArrayList<>();
// Getters & Setters...
public double getUpVotes() {
return votes.stream().filter(vote -> vote.getDirection() == 1).mapToInt(Vote::getDirection).count();
}
}
和
public class Vote {
private short direction;
// Getters & Setters...
}
然后是像这样的存储库
@Repository
public interface PostRepository extends PagingAndSortingRepository<Post, String> {
List<Post> findAll(Pageable pageable);
}
然后说我想根据getter方法getUpVotes()
的结果对帖子进行排序
我试过下面的localhost:3005/opinion?page=0&size=20&sort=upVotes
,但它不起作用。
排序文档可以对现有字段指定升序或降序。。。
https://docs.mongodb.com/manual/reference/method/cursor.sort/#sort-asc-desc
解决方法:您可以执行MongoDB聚合,在那里您可以添加具有计算值的新字段,并按此值排序:
db.post.aggregate([
{
$addFields: {
upVotes: {
$size: {
$filter: {
input: "$votes.direction",
cond: {
$eq: [ "$$this", 1 ]
}
}
}
}
}
},
{
$sort: {
upVotes: 1
}
}
])
MongoPlayground|$project
弹簧数据
@Autowired
private MongoTemplate mongoTemplate;
...
Aggregation aggregation = Aggregation.newAggregation(addFields, sort);
List<Post> result = mongoTemplate
.aggregate(aggregation, mongoTemplate.getCollectionName(Post.class), Post.class)
.getMappedResults();