我有一个对象的List,我需要在Morphia中执行查询才能从给定对象中的List的另一个String嵌入中获得唯一的结果。

对象类别:

@Entity
public class Fruits {
  @NotNull private Long id;
  @NotNull private List<String> categories;
}


Json格式的数据:

水果:

 {[
  "id": 1234566,
  "categories": ["A", "B", "C", "D"]
 ],
 [
  "id": 32434,
  "categories": ["A", "C", "E", "F"]
 ],
 [
  "id": 32434,
  "categories": ["A", "L", "M", "N"]
 ]
}


聚合的结果应该是:

[A,C,B,D,E,F,L,M,N]


输出为排序形式。我该如何在吗啡上做到这一点?我试图搜索官方文档,但找不到提示。

任何帮助或暗示将是可观的。谢谢

编辑1

List<Fruits> fruitList = fruitControllerDao.search(fruitList);

List<Category> categories = new ArrayList<>();
datastore.createAggregation(Fruits.class)
         .unwind("categories")
         .group(Group.id(Group.grouping("categories")))
         .sort(Sort.ascending("_id.categories"))
         .project(Projection.projection("_id").suppress(),
          Projection.projection("categories", "_id.categories"))

         .aggregate(Category.class).forEachRemaining(categories::add);


而我创造了

class Category {
   private String category;
}


fruitList中的数据不可用,我需要在aggregation(签出JSON格式)本身上应用fruitList

最佳答案

您可以使用以下代码通过聚合管道来获取唯一和排序类别。

准备查询过滤器:

Query<Fruits> query = datastore.createQuery(Fruits.class);
Iterator<Fruits> fruitIterator = fruitList .iterator();
CriteriaContainer orQuery = query.or();
while(fruitIterator.hasNext()) {
      Fruits fruit = fruitIterator.next();
      orQuery.add(query.and(query.criteria("id").equal(fruit.getId()), query.criteria("categories").equal(fruit.getCategories())));
 }


以下查询过滤器上的查询$matches$unwinds categories,后跟$group删除重复项和排序。最后一步是从分组阶段和$project到类别字段中读取值。

 List<Category> categories = new ArrayList<>();
  datastore.createAggregation(Fruits.class)
        .match(query)
        .unwind("categories")
        .group(Group.id(Group.grouping("categories")))
        .sort(Sort.ascending("_id.categories"))
        .project(Projection.projection("_id").suppress(), Projection.projection("category", "_id.categories"))
        .aggregate(Category.class).forEachRemaining(categories::add);


class Category {
    private String category;
}


更新:

Mongo Shell查询

   [{ "$match" : { "$or" : [ { "$and" : [ { "_id" : 1234566} , { "categories" : [ "A" , "B" , "C" , "D"]}]} , { "$and" : [ { "_id" : 32434} , { "categories" : [ "A" , "C" , "E" , "F"]}]} , { "$and" : [ { "_id" : 32434} , { "categories" : [ "A" , "L" , "M" , "N"]}]}]}}, { "$unwind" : "$categories"}, { "$group" : { "_id" : { "categories" : "$categories"}}}, { "$sort" : { "_id.categories" : 1}}, { "$project" : { "_id" : 0 , "category" : "$_id.categories"}}]

关于java - 如何在吗啡中进行聚集?,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/43617897/

10-09 03:32