问题描述

好吧，在Spring Data中，这种查询不是trivial。

Spring Data Repository没有针对的解决方案MongoDB Aggregation。因此，您无法在MongoRepository中实现任何方法，例如aggregateBy...

Spring Data提供了MongoTemplate允许您执行复杂查询的类，就像在标准MongoDB shell中所做的那样。

因此，由于您只想对exclude不符合某些条件的子文档进行分类，因此我们需要定义集合pipelines。

zip codes are Numeric (In your example is string)
And, to exclude subdocument, we filter by `zip`
There is no any other filter

MongoDB聚合为：

db.person.aggregate([
    {$unwind: "$address"},
    {$match: {"address.zip": 12345}},
    {$group: { _id: { "firstName":"$firstName", "lastName":"$lastName", _id:"$_id" }, address: { $push: "$address" } } },
    {$project: {_id:0, "firstName":"$_id.firstName", "lastName":"$_id.lastName", "address": "$address"}}
])

如果所有筛选器均成功，则我们得到：

[ 
    {
        "address" : [ 
            {
                "zip" : 12345
            }, 
            {
                "zip" : 12345
            }
        ],
        "firstName" : "George",
        "lastName" : "Washington"
    }
]

现在，以Spring Data的方式，您需要在项目中添加一些更改：

首先，找到您mongo-config.xml需要添加的位置：

<!-- Define the mongoDbFactory with your database Name  -->
<mongo:db-factory uri="mongodb://user:pass@localhost:27017/db"/>

<!-- Define the MongoTemplate  -->
<bean id="mongoTemplate" class="org.springframework.data.mongodb.core.MongoTemplate">
    <constructor-arg name="mongoDbFactory" ref="mongoDbFactory" />
</bean>

MongoTemplate是Spring的MongoDB支持的核心类，提供与数据库交互的功能集。该模板...提供了 和 之间的映射。更多信息

其次，在您的@Service课程中，添加以下代码以供加载@PostConstruct

@Autowired
private MongoOperations mongoOperations;

...

public List<Person> findByAddressZipCode(int zip) {

    List<AggregationOperation> list = new ArrayList<AggregationOperation>();
    list.add(Aggregation.unwind("address"));
    list.add(Aggregation.match(Criteria.where("address.zip").is(zip)));
    list.add(Aggregation.group("firstName", "lastName").push("address").as("address"));
    list.add(Aggregation.project("firstName", "lastName", "address"));
    TypedAggregation<Person> agg = Aggregation.newAggregation(Person.class, list);
    return mongoOperations.aggregate(agg, Person.class, Person.class).getMappedResults();
}

这两个，Person并且Address应该有默认的空构造！

解决方法

@Repository
public interface PersonRepository extends MongoRepository <Person,String> {
    List<Person> findByAddressZipCode(@Param("zip") int zip);
}

@Document
public class Person {
    @Id
    private String id;

    private String firstName;
    private String lastName;
    private List<Address> address;
}

public class Address {
    private int zip;
}

{
 "firstName":"George","lastName":"Washington","address":[{
     "zip":"12345"
  },{
     "zip":"98765"
  },{
     "zip":"12345"
  }]
}

{
 "firstName":"George",{
     "zip":"12345"
  }]
}

{
 "firstName":"George",{
     "zip":"12345"
  }]
}