Question

我使用了来自Error creating bean with name 'personRepository': Invocation of init method failed; nested exception is com.mongodb.util.JSONParseException:的代码，现在尝试调用

我得到的响应为空。

此外，下面的查询不会提取任何内容。

Person p = personRepository.findByAddresses_City("London NW1");
System.out.println("PERSON FOR ADDRESS = "+p);

这里：

db.getCollection（'地址'）.find（{}）

Query query = new Query(Criteria.where("address.$id").is(2L));
List<Person> l = mongoTemplate.find(query, Person.class);
System.out.println(l);

和/* 1 */ { "_id" : NumberLong(1), "address" : "221b Baker Street", "city" : "London NW1", "state" : "London", "zipcode" : NumberLong(12345), "_class" : "com.example.demo.model.Address" }

db.getCollection('person').find({})

如何解决此错误？

/* 1 */
{
    "_id" : NumberLong(1),
    "name" : "Achilles",
    "age" : 0,
    "addresses" : [],
    "_class" : "com.example.demo.model.Person"
}

/* 2 */
{
    "_id" : NumberLong(2),
    "name" : "Hektor",
    "age" : 0,
    "addresses" : [ 
        {
            "$ref" : "address",
            "$id" : NumberLong(1),
            "$db" : "address"
        }
    ],
    "_class" : "com.example.demo.model.Person"
}

和

@Document(collection = "address")
public class Address {

    @Id
    private long addressId;
    private String address;
    private String city;
    private String state;
    private long zipcode;

    public Address() {
        System.out.println("CAlling default cons");
    }

    @PersistenceConstructor
    public Address(long addressId, String address, String city, String state, long zipcode) {
        this.addressId = addressId;
        this.address = address;
        this.city = city;
        this.state = state;
        this.zipcode = zipcode;
    }
    // setter and getter... toString()..
}

Answer 1

这按设计工作。 MongoDB不允许通过查询进行应用程序级联接，您需要对更复杂的查询使用聚合框架。因此，存储库查询仅允许按完整值（即Address对象）或标识符查找DBRef。

如果将where子句固定为address.addressId，第二个示例应该可以工作。

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使用Spring Data Mongo时@DBRef不会提取数据

1 个答案: