嗨,我在使用Jongo使用$unwind
运算符从“人”集合中获取地址列表时遇到问题。
如您所见,我将Person类定义如下:
public class Person {
@Id
private long personId;
private String name;
private int age;
private List<Address> addresses;
//getters and setters
并且Address类的定义如下:
public class Address {
private String houseNumber;
private String road;
private String town;
private String postalCode;
//gettes and setters
查询非常简单:
public List<Address> getAddressByPersonId(long id) {
List<Address> list = persons.aggregate("{$project:{addresses:1}}")
.and("{$match:{_id:#}}",id)
.and("{$unwind: '$addresses'}")
.as(Address.class);
return list;
}
我的收藏:
> db.persons.find()
{ "name" : "Bob", "age" : 34, "addresses" : [ { "houseNumber" : "12",
"road" : "High Street", "town" : "Small Town", "postalCode" : "BC2 3DE"
}, { "houseNumber" : "12", "road" : "High Street", "town" :
"Small Town", "postalCode" : "BC2 3DE" } ], "_id" : NumberLong(1) }
>
我也为此进行了JUnit测试:
@Before
public void setUp(){
service = new PersonServiceImpl();
//store a person into the database (manually) before the tests
MongoCollection persons = Database.getInstance().getCollection(CollectionNames.PERSONS);
//create a new person
Person p = new Person();
p.setPersonId(1L);
p.setName("Bob");
p.setAge(34);
//create two addresses for this person
Address a = new Address();
a.setHouseNumber("33");
a.setRoad("Fake Road");
a.setPostalCode("AB1 2CD");
a.setTown("Big Town");
p.addAddress(a);
a.setHouseNumber("12");
a.setRoad("High Street");
a.setPostalCode("BC2 3DE");
a.setTown("Small Town");
p.addAddress(a);
persons.save(p);
}
@After
public void tearDown(){
service = null;
}
@Test
public void getAddressesByPersonIdTest(){
List<Address> list = service.getAddressByPersonId(1L);
for (Address item : list){
item.print();
}
Assert.assertTrue(list.size() > 0);
}
}
哪个输出
### Address: null null, null, null
### Address: null null, null, null
我不太清楚问题是什么。显然,测试不会失败(因此
list.size()
大于零...但显示为null)。 print()
方法已经过测试并且可以工作。我想获取两个Bob的地址,但是查询返回空对象。我想念什么吗?我应该以其他方式使用$ unwind吗?请建议
最佳答案
Aggregation Framework返回与输入格式完全相同的格式-展开后唯一的区别是addresses
字段下嵌入了多少个地址。在您的情况下,您将获得两个文档:
{ "_id": 1,
"name" : "Bob",
"age" : 34,
"addresses" : { "houseNumber" : "12",
"road" : "High Street",
"town" : "Small Town",
"postalCode" : "BC2 3DE"
}
}
这不能非常有效地转换为Address类,因为如您所见,Address对象是
addresses
字段的值,而不是顶级对象。尚不清楚的是为什么您需要进行汇总-您正在获得一个人的地址列表。当您通过ID对人员执行简单的
find()
时,字段“地址”已经是List
对象的Address
。在MongoDB find() has the option to return just selected fields中,或排除命名字段(类似于SQL的SELECT *效率不如SELECT col1,col2高效),因此,如果执行the Java equivalent of
db.persons.find({filter-condition}, {"_id":0, "addresses":1})
,则将返回仅包含地址列表的文档。