使用 OrientDB SQL 查找 "friends of friends"

Question

虽然这是演示图形数据库功能的最常见用例之一，但我似乎找不到使用 OrientDB "friends of friends" 获得 SQL 的良好示例或最佳实践。

让我们假设一个社交网络并尝试用 "user" 个顶点和 "is_friend_with" 个边对其建模。

定义:

Vertex class user with properties uuid (custom unique id) and name

Edge class is_friend_with with 属性 status 可以是“待定”或“已批准”

用户通过单向边相互连接。方向并不重要；只要 status="approved"，这两个用户就是朋友。

这是我想出的一个解决方案：

select from (
    select expand($all) let
        $a = (select expand(outE('is_friend_with')[status='approved'].inV('user').outE('is_friend_with')[status='approved'].inV('user')) from (select from user where uuid = '95920a96a60c4d40a8f70bde98ae1a24')),
        $b = (select expand(outE('is_friend_with')[status='approved'].inV('user').inE('is_friend_with')[status='approved'].outV('user')) from (select from user where uuid = '95920a96a60c4d40a8f70bde98ae1a24')),
        $c = (select expand(inE('is_friend_with')[status='approved'].outV('user').inE('is_friend_with')[status='approved'].outV('user')) from (select from user where uuid = '95920a96a60c4d40a8f70bde98ae1a24')),
        $d = (select expand(inE('is_friend_with')[status='approved'].outV('user').outE('is_friend_with')[status='approved'].inV('user')) from (select from user where uuid = '95920a96a60c4d40a8f70bde98ae1a24')),
        $all = unionall($a, $b, $c, $d)
) where uuid <> '95920a96a60c4d40a8f70bde98ae1a24'

(uuid='95920a96a60c4d40a8f70bde98ae1a24'的用户为起点。)

但是，我觉得它不是很优雅。我可以立即发现的一些问题是：

重复 select from user where uuid = '95920a96a60c4d40a8f70bde98ae1a24'。不幸的是，我找不到将其分配给变量然后在 "from" 子句
我被迫进行 incoming/outgoing edges/vertices 的所有组合，而不是使用 both()，因为我想检查 status="approved"
此查询还 returns 直接朋友，而不是朋友的朋友

我尝试使用遍历，但无济于事（同样，没有找到一种方法如何在遍历时检查 status="approved" 的边缘）。

请问您能为这个问题提出一些 OSQL 的解决方案吗？提前致谢。

Answer 1

你试过吗？

select 
      expand(bothE('is_friend_with')[status = 'approved'].inV().bothE('is_friend_with')[status = 'approved'].inV()) 
from user where uuid = '95920a96a60c4d40a8f70bde98ae1a24'

Answer 2

可能有更好的方法，但我认为这可以满足您的需求：

设置：

create class User extends V
create class IsFriendsWith extends E
create property User.name string
create property User.uuid string
create property IsFriendsWith.status string

create vertex User set uuid = "1", name = "Bob"
create vertex User set uuid = "2", name = "Sally"
create vertex User set uuid = "3", name = "Eric"
create vertex User set uuid = "4", name = "Jenny"
create vertex User set uuid = "5", name = "Dennis"
create vertex User set uuid = "6", name = "Mary"
create vertex User set uuid = "7", name = "John"

create edge IsFriendsWith from (select from User where uuid = "1") to (select from User where uuid = "2") set status = "approved"
create edge IsFriendsWith from (select from User where uuid = "1") to (select from User where uuid = "3") set status = "pending"
create edge IsFriendsWith from (select from User where uuid = "2") to (select from User where uuid = "4") set status = "approved"
create edge IsFriendsWith from (select from User where uuid = "5") to (select from User where uuid = "2") set status = "pending"
create edge IsFriendsWith from (select from User where uuid = "3") to (select from User where uuid = "4") set status = "approved"
create edge IsFriendsWith from (select from User where uuid = "6") to (select from User where uuid = "1") set status = "approved"
create edge IsFriendsWith from (select from User where uuid = "6") to (select from User where uuid = "7") set status = "approved"

查询：

select from (
    select 
      expand(unionall(
        outE("IsFriendsWith")[status="approved"].inV(), 
        inE("IsFriendsWith")[status="approved"].outV()
      ))
      from (
        select 
            expand(unionall(
              outE("IsFriendsWith")[status="approved"].inV(), 
              inE("IsFriendsWith")[status="approved"].outV()
            ))
        from (
          select from User where uuid = "1"
        )
      )
  )
) where uuid <> "1"

Enrico 的回答不会给您真正想要的，因为当它需要双向工作时，它只考虑一个方向的边缘。

编辑： 要排除用户碰巧已经是朋友的人，请使用以下内容（请注意，此示例假定用户的 RID 为 #26:0）

  select from (
      select 
        expand(unionall(
          outE("IsFriendsWith")[status="approved"].inV(), 
          inE("IsFriendsWith")[status="approved"].outV()
        ))
        from (
          select 
              expand(unionall(
                outE("IsFriendsWith")[status="approved"].inV(), 
                inE("IsFriendsWith")[status="approved"].outV()
              ))
          from #26:0          
        )
    )
  )
  where @rid <> #26:0 and @rid NOT IN (select both("IsFriendsWith") from #26:0)

编辑 2： 使用变量代替。

 select from (
      select 
        expand(unionall(
          outE("IsFriendsWith")[status="approved"].inV(), 
          inE("IsFriendsWith")[status="approved"].outV()
        ))
        from (
          select 
              expand(unionall(
                outE("IsFriendsWith")[status="approved"].inV(), 
                inE("IsFriendsWith")[status="approved"].outV()
              ))
          from (
             select expand($u) let $u = first((select from User where uuid = "1"))
          )
        )
    )
  )
  where @rid <> $u and @rid NOT IN $u.both("IsFriendsWith")

Answer 3

这应该更简单：

select expand(
  bothE('is_friend_with')[status = 'approved'].bothV()
  .bothE('is_friend_with')[status = 'approved'].bothV()
) from user where uuid = '95920a96a60c4d40a8f70bde98ae1a24'

使用 bothE() 和 bothV()，您可以在 edge/vertex 级别获得两个 in/out 连接。

为了排除当前用户，您可以使用：

select expand(
  bothE('is_friend_with')[status = 'approved'].bothV()
  .bothE('is_friend_with')[status = 'approved'].bothV()
  .remove( @this )
) from user where uuid = '95920a96a60c4d40a8f70bde98ae1a24'

使用 OrientDB SQL 查找 "friends of friends"

Find "friends of friends" with OrientDB SQL

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