0

背景:

我正在使用 Spring-boot 和 MongoDB。

public class User {
   @DBRef
   private List<Contact> contacts;
   ...


public class Contact {

   @DBRef
   private List<Booking> bookings;
   ...

   public Contact(){}

   public Contact(Booking booking){
       this.bookings = new ArrayList<Booking>();
       this.bookings.add(booking);
       ...
   }

@Override
public List<Contact> findAllContactsForUser(String id) {
    Query query = new Query().addCriteria(Criteria
            .where("_id").is(id));
    query.fields().include("contacts");
    User user = mongoTemplate.findOne(query, User.class);
    return user.getContacts();
}

findAllContactsForUser 的输出:

[
   {
     "field":"value",
     "bookings": [
                   "field":"value"
                 ]
   }
]

问题:

我想要的结果:

[
   {
     "field":"value",
     "bookings": null
   }
]

如何从 findAllContactsForUser 的查询中排除预订关系?

4

1 回答 1

1

您只需要contacts.bookings在查询中排除该字段:

@Override
public List<Contact> findAllContactsForUser(String id) {
    Query query = new Query().addCriteria(Criteria.where("_id").is(id));
    query.fields().exclude("contacts.bookings");
    User user = mongoTemplate.findOne(query, User.class);
    return user.getContacts();
}

顺便说一句,默认情况下不需要.include(contacts), 字段。

于 2017-01-01T14:48:09.040 回答