我在文档内部有一个数组。
{
"id" : "id_1",
"name" : "name_1";
"additionalData" : [
{
"additionalDataId" : "id_1_1",
"additionalDataName" : "name_1_1",
"longText" : "A long story about..."
},
{
"additionalDataId" : "id_1_2",
"additionalDataName" : "name_1_2",
"longText" : "A longer story about danger..."
},
{
"additionalDataId" : "id_1_3",
"additionalDataName" : "name_1_3",
"longText" : "A longer story about danger and courage"
},
]
}
要检索名称为"name_1_2"
的数组元素,请使用mongo查询。
db.collection.find( { name: "name_1"},
{ _id: 0, additionalData: { $elemMatch: { "additionalDataName": "name_1_2" } }
})
如何使用mongoTemplate
执行相同操作?
我尝试过
Aggregation aggregation = Aggregation.newAggregation(
Aggregation.match(
Criteria.where("name").is("name_1").and("additionalData.additionalDataName").is("name_1_2")
),
Aggregation.project("additionalData"),
); mongoTemplate.aggregate(aggregation,“ CustmObjects”,Object.class);
我也尝试使用ArrayOperators.Filter.filter
,我用了这个答案。
Aggregation aggregation = Aggregation.newAggregation(
Aggregation.match(Criteria.where("name").is("name_1")),
Aggregation.project("additionalData").and(
ArrayOperators.Filter.filter("additionalData").as("item")
.by(ComparisonOperators.valueOf("item.additionalDataName").equalTo("name_1_2"))
)
);
mongoTemplate.aggregate(aggregation, "CustmObjects", Object.class);
https://stackoverflow.com/a/46769125/4587961
无论如何,我得到的结果是数组的所有元素。 请帮忙!
答案 0 :(得分:1)
如果我正确理解了这个问题,这将提供所需的结果。
Query query = new Query(new Criteria().andOperator(
Criteria.where("name").is("name_1"),
Criteria.where("additionalData.additionalDataName").is("name_1_2")
));
query.fields().include("additionalData").exclude("_id");
List<Document> results = template.find(query, collectionName, org.bson.Document.class);