database - 如何在orientdb中从一个类获取数据到另一个类

Question

我创建了两个类customer 和city。customer 类包含两个属性 name 和 location 和 city 类包含 id 和 < em>位置。我想对这两个类执行连接操作。我在 orientdb studio 中创建了一个图形关系并在下面触发了一个查询

select from customer where city.location='pune'

但是这个查询没有返回任何值，它执行但没有返回任何字段，所以，这是正确的语法，或者我在某处做错了..请给我解决方案。

Answer 1

我有这个简单的数据集可以给你一些例子:

create class Customer extends V
create class City extends V
create class livesAt extends E

create property Customer.name String
create property City.id integer
create property City.location String

create vertex Customer set name="Tom"
create vertex Customer set name="John"
create vertex City set id=1, location="London"
create vertex City set id=2, location="Pune"

create edge livesAt from (select from Customer where name="Tom") to (select from City where id=1)
create edge livesAt from (select from Customer where name="John") to (select from City where id=2)

现在您可以使用不同的查询来检索您要查找的结果。

---

查询 1a:从客户开始(如上面的查询)

select from Customer where out('livesAt').location in 'Pune'

输出:

----+-----+--------+----+-----------
#   |@RID |@CLASS  |name|out_livesAt
----+-----+--------+----+-----------
0   |#12:1|Customer|John|[size=1]
----+-----+--------+----+-----------

---

查询 1b:从客户开始

select from Customer where out('livesAt').location contains 'Pune'

输出:

----+-----+--------+----+-----------
#   |@RID |@CLASS  |name|out_livesAt
----+-----+--------+----+-----------
0   |#12:1|Customer|John|[size=1]
----+-----+--------+----+-----------

---

查询 1c:

select from Customer where out('livesAt')[location = 'Pune'].size() > 0

输出:

----+-----+--------+----+-----------
#   |@RID |@CLASS  |name|out_livesAt
----+-----+--------+----+-----------
0   |#12:1|Customer|John|[size=1]
----+-----+--------+----+-----------

---

查询 2:从城市开始(更直接)

select expand(in('livesAt')) from City where location = 'Pune'

输出:

----+-----+--------+----+-----------
#   |@RID |@CLASS  |name|out_livesAt
----+-----+--------+----+-----------
0   |#12:1|Customer|John|[size=1]
----+-----+--------+----+-----------

希望对你有帮助