Sparql - 如果未绑定任何一个变量,则连接失败

Question

您好我正在使用AllegroGraph和Sparql查询来检索结果.这是一个重现我的问题的示例数据.考虑以下数据,其中一个人有姓,中名和姓.

<http://mydomain.com/person1> <http://mydomain.com/firstName> "John"^^<http://www.w3.org/1999/02/22-rdf-syntax-ns#XMLLiteral>
<http://mydomain.com/person1> <http://mydomain.com/middleName> "Paul"^^<http://www.w3.org/1999/02/22-rdf-syntax-ns#XMLLiteral>
<http://mydomain.com/person1> <http://mydomain.com/lastName> "Jai"^^<http://www.w3.org/1999/02/22-rdf-syntax-ns#XMLLiteral>
<http://mydomain.com/person1> <http://www.w3.org/1999/02/22-rdf-syntax-ns#type> <http://mydomain.com/person>

<http://mydomain.com/person6> <http://mydomain.com/middleName> "Mannan"^^<http://www.w3.org/1999/02/22-rdf-syntax-ns#XMLLiteral>
<http://mydomain.com/person6> <http://mydomain.com/lastName> "Sathish"^^<http://www.w3.org/1999/02/22-rdf-syntax-ns#XMLLiteral>
<http://mydomain.com/person6> <http://www.w3.org/1999/02/22-rdf-syntax-ns#type> <http://mydomain.com/person>

现在我需要通过组合所有3个名称来计算人名.名称是可选的,一个人可能没有任何名字,中间名和姓.

查询我试过了

select ?person ?name ?firstName ?middleName ?lastName where 
{
  ?person rdf:type  <http://mydomain.com/person>.
    optional {?person <http://mydomain.com/firstName> ?firstName}.
    optional {?person <http://mydomain.com/middleName> ?middleName}.
    optional {?person <http://mydomain.com/lastName> ?lastName}.    
    bind (concat(str(?firstName),str(?middleName),str(?lastName)) as ?name).
} 

但是结果集不包含person6(Mannan Sathish)的名称,因为第一个名称不存在.如果我没有绑定,请告诉我是否可以忽略firstName.

Answer 1

如果未绑定变量,则str(...)将导致评估错误,并且整个BIND失败.

COALESCE 可用于为表达式提供默认值.

bind ( COALESCE(?firstName, "") As ?firstName1)
bind ( COALESCE(?middleName, "") As ?middleName1)
bind ( COALESCE(?lastName, "") As ?lastName1)
bind (concat(str(?firstName1),str(?middleName1),str(?lastName1)) as ?name