Question

我有两种类型的顶点（ Car 和属性）和一种类型的边（ Property ）：

CREATE CLASS Car extends V
CREATE CLASS Attribute extends V
CREATE CLASS Property extends E

属性包含可用的属性，让我们说引擎，颜色，年和的模型

Car 应该包含指向这些属性的链接：

CREATE VERTEX Attribute SET attributeName = "engine"
CREATE VERTEX Attribute SET attributeName = "color"
CREATE VERTEX Attribute SET attributeName = "year"
CREATE VERTEX Attribute SET attributeName = "model"

（有没有更快的方法来创建这些没有4个查询？）

CREATE VERTEX Car SET name = "Suzuki"
CREATE VERTEX Car SET name = "Honda"

CREATE EDGE Property from (SELECT FROM Car WHERE name = "Suzuki") to (SELECT FROM Attribute WHERE attributeName = "color") SET value = "pink"

CREATE EDGE Property from (SELECT FROM Car WHERE name = "Suzuki") to (SELECT FROM Attribute WHERE attributeName = "model") SET value = "Alto"

CREATE EDGE Property from (SELECT FROM Car WHERE name = "Suzuki") to (SELECT FROM Attribute WHERE attributeName = "year") SET value = 2005


CREATE EDGE Property from (SELECT FROM Car WHERE name = "Honda") to (SELECT FROM Attribute WHERE attributeName = "color") SET value = "black"

CREATE EDGE Property from (SELECT FROM Car WHERE name = "Honda") to (SELECT FROM Attribute WHERE attributeName = "engine") SET value = 2500

CREATE EDGE Property from (SELECT FROM Car WHERE name = "Honda") to (SELECT FROM Attribute WHERE attributeName = "model") SET value = "Civic"

CREATE EDGE Property from (SELECT FROM Car WHERE name = "Honda") to (SELECT FROM Attribute WHERE attributeName = "year") SET value = 2005

（再次，更快的方式？）

现在我想搜索任何颜色为黑色的汽车（本田应该退回）。

但是当我搜索2005年的任何一辆车和2500的引擎时，虽然铃木没有引擎属性，但它也应该被返回（所以本田和铃木）。

我尝试了以下查询：

select distinct(out.Car) as car from Property WHERE 
                    (in.attributeName="engine" AND value=2500) OR 
                    (in.attributeName="year" AND value=2005)

当然，铃木也被退回了......有什么帮助吗？

非常感谢！：）

OrientDB - 比较2个顶点，边缘属性

0 个答案: