假设我有一个实体类Article
@Data
@Document(collection = "articles")
public final class Article {
@Id
private String id;
private String name;
private String url;
private String author;
private List<Comment> comments;
}
实体类Comment
@Data
public final class Comment {
private String userId;
private String content;
}
已经通过测试用例添加进去几条数据,假设其中一个Article的_id为5b5694f38ec9b636c4e52d69
,
可以通过MongoRepository
获取到这个id的Article,也可以通过MongoTemplate
获取
final String id = "5b5694f38ec9b636c4e52d69";
Query query = new Query()
.addCriteria(Criteria.where("id").is(id));
Article article = mongoTemplate.findOne(query, Article.class);
这样可以获取到,现在我想只获取这个Article的List<Comment>
Document queryObject = new Document();
queryObject.put("id", id);
Document fieldsObject = new Document();
fieldsObject.put("id", false);
fieldsObject.put("comments", true);
Query query = new BasicQuery(queryObject, fieldsObject);
如果我find时返回List
List comments = mongoTemplate.findOne(query, List.class);
这样毫无疑问会出现异常,返回Article
Article article = mongoTemplate.findOne(query, Article.class);
List<Comment> comments = Optional.of(article).orElseThrow(ArticleNotFoundException::new)
.getComments();
这样可以获取List<Comment>
,但是这样的获取意义不是很大,也无法获取指定数量的Commit
(subList()也没有什么意义)。
有没有什么方式直接返回List<Commit>
呢,这样便可以使用limit()
和skip()
;
如果没有,那有没有一种方式可以实现limit()
skip()
同样的效果?
通过查找文档找到了答案
MongoCollection<Document> collection =
mongoClient.getDatabase(YOUR_DATABASE_NAME).getCollection(YOUR_COLLECTION_NAME);
FindIterable<Document> findIterable = collection.find()
.filter(eq("_id", new ObjectId(id)))
.projection(fields(include("comments"), excludeId(), slice("comments", 2)));
Document document = findIterable.first();
if (document != null) {
Object o = document.get("comments");
if (o instanceof List) {
List comments = (List) o;
assert comments.size() == 2;
}
}
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