来自主题的 sparql 完整树

Question

例如，当我有一个人图时，例如约翰和约翰有一个工作地址家庭地址电话号码关系等。

是否有可能在不知道它是什么的情况下检索与 john 和 john 的子类相关的所有内容？

这样我就可以检索例如以下内容

John < address < house_number
     < mobile_number
     < company < address
               < function < office number < etc...

And retrieve this via:
     "John" rdfs:everything ?everything ... as deep as the tree goes.

还是我需要知道图表？

Answer 1

你的术语有点不对劲，因为地址、手机号码等不是 John 的子类。RDF 是关于三元组（标记为有向边），而约翰的地址是具有形式 的三元组的对象John hasAddress AddressOfJohn。用基于图的术语来说，听起来您是在问是否可以检索从 John 开始的某个有向路径可到达的所有事物。使用具体数据更容易，所以让我们假设您有这些数据：

@prefix : <http://stackoverflow.com/q/20878795/1281433/> .

:john :hasAddress :addressOfJohn ;
      :hasMobileNumber :mobileNumber ;
      :hasCompany :company .

:addressOfJohn :hasHouseNumber :houseNumber .

:company :hasAddress :addressOfCompany ;
         :hasFunction :function .

SPARQL 1.1 added support for Property Paths, which a sort of like a regular expression language for sequences of properties, including typical operators like * (any number of repetition) and + (one or more). Unfortunately, you can't use variables in property paths, so you can't just do

:john ?p+ ?object

However, as noted in a similar question on answers.semanticweb.com SPARQL: property paths without specified predicates, you can construct a disjunction that will match every property. For instance, since : is a defined prefix and is thus an IRI, the pattern :|!: will match any property (since everything is either : or it isn't). That mans that (:|!:)+ is a genuine wildcard path. Thus, we can write a query like the following and get the corresponding results:

prefix : <http://stackoverflow.com/q/20878795/1281433/>

select ?object where { 
  :john (:|!:)+ ?object 
}

---------------------
| object            |
=====================
| :company          |
| :function         |
| :addressOfCompany |
| :mobileNumber     |
| :addressOfJohn    |
| :houseNumber      |
---------------------