Question

我有一个有效的查询，它会返回Tom Hanks的所有合作演员（基本上是朋友问题的朋友）。

SELECT COUNT(name) as name FROM ( SELECT expand(out('ACTS_IN').in('ACTS_IN')) FROM Actor WHERE name = 'Tom Hanks' ) WHERE name <> 'Tom Hanks'

现在我正在尝试构建一个查询，让我成为合作演员的共同演员（朋友的朋友的朋友）

我的查询，还没有完成

SELECT count(DISTINCT(name)) as coactor FROM ( SELECT expand(out('ACTS_IN').in('ACTS_IN').out('ACTS_IN').in('ACTS_IN').out('ACTS_IN').in('ACTS_IN')) FROM Person WHERE name = 'Tom Hanks' ) WHERE name <> 'Tom Hanks'

问题是，你可以像这样堆叠 .in 和 .out ：

expand(out('ACTS_IN').in('ACTS_IN').out('ACTS_IN').in('ACTS_IN').out('ACTS_IN').in('ACTS_IN'))

Answer 1

在您的情况下，您可以使用TRAVERSE()函数检索由边ACTS_IN链接的所有记录。

<强>结构：

create class Person extends V
create class Movie extends V
create class acts_In extends E
create class directed extends E
create class friend extends E
create class rated extends E

create property Person.name String
create property Person.surname String
create property Movie.title String

create vertex Person set name="Tom", surname="Hanks"
create vertex Person set name="Robin", surname="Wright"
create vertex Person set name="Helen", surname="Hunt"
create vertex Person set name="Robert", surname="Zemeckis"
create vertex Person set name="Russell", surname="Crowe"
create vertex Person set name="Ben", surname="Affleck"
create vertex Person set name="Kevin", surname="Macdonald"
create vertex Person set name="John"
create vertex Person set name="Mark"
create vertex Person set name="Paul"
create vertex Person set name="Mel", surname="Gibson"
create vertex Person set name="Nancy", surname="Meyers"
create vertex Movie set title="Forrest Gump"
create vertex Movie set title="Cast Away"
create vertex Movie set title="State of Play"
create vertex Movie set title="What Women Want"

create edge acts_In from (select from Person where name="Tom" and surname="Hanks") to (select from Movie where title="Forrest Gump")
create edge acts_In from (select from Person where name="Tom" and surname="Hanks") to (select from Movie where title="Cast Away")
create edge acts_In from (select from Person where name="Robin" and surname="Wright") to (select from Movie where title="Forrest Gump")
create edge acts_In from (select from Person where name="Robin" and surname="Wright") to (select from Movie where title="State of Play")
create edge acts_In from (select from Person where name="Helen" and surname="Hunt") to (select from Movie where title="Cast Away")
create edge acts_In from (select from Person where name="Helen" and surname="Hunt") to (select from Movie where title="What Women Want")
create edge acts_In from (select from Person where name="Mel" and surname="Gibson") to (select from Movie where title="What Women Want")
create edge acts_In from (select from Person where name="Russell" and surname="Crowe") to (select from Movie where title="State of Play")
create edge acts_In from (select from Person where name="Ben" and surname="Affleck") to (select from Movie where title="State of Play")
create edge friend from (select from Person where name="Mel" and surname="Gibson") to (select from Person where name="Helen" and surname="Hunt")
create edge friend from (select from Person where name="Ben" and surname="Affleck") to (select from Person where name="Russell" and surname="Crowe")
create edge directed from (select from Movie where title="What Women Want") to (select from Person where name="Nancy" and surname="Meyers")
create edge directed from (select from Movie where title="Cast Away") to (select from Person where name="Robert" and surname="Zemeckis")
create edge directed from (select from Movie where title="Forrest Gump") to (select from Person where name="Robert" and surname="Zemeckis")
create edge directed from (select from Movie where title="State of Play") to (select from Person where name="Kevin" and surname="Macdonald")
create edge rated from (select from Movie where title="What Women Want") to (select from Person where name="Paul")
create edge rated from (select from Movie where title="Cast Away") to (select from Person where name="John")
create edge rated from (select from Movie where title="Forrest Gump") to (select from Person where name="Mark")
create edge rated from (select from Movie where title="State of Play") to (select from Person where name="John")

查询1 ：检索“汤姆汉克斯”中的所有演员/合作演员/ ecc ... （除了他自己）

SELECT expand(DISTINCT) FROM (SELECT DISTINCT(@rid) FROM (TRAVERSE BOTH('ACTS_IN') 
FROM (SELECT FROM Person WHERE name = 'Tom' AND surname = 'Hanks') MAXDEPTH 4) 
WHERE name <> 'Tom' and surname <> 'Hanks')

查询2 ：检索汤姆汉克斯＆＃39;的演员/合作者/ ecc ...的数量。（除了他自己）

SELECT COUNT(DISTINCT(@rid)) AS co_actors FROM (TRAVERSE BOTH('ACTS_IN') 
FROM (SELECT FROM Person WHERE name = 'Tom' AND surname = 'Hanks') MAXDEPTH 4) 
WHERE name <> 'Tom' and surname <> 'Hanks'

OrientDB SQL recursivness

1 个答案: