Hi, What exactly do you mean by "it does nothing"?
I assume you are referring to the video linked to by this introductory blog - http://www.topquadrant.com/2011/04/21/spinmap-sparql-based-ontology-mapping-with-a-graphical-notation/. Correct? The steps described in the blog are still correct, except that instead of creating a SPIN file and then importing SPINMap vocabulary, TBC now supports the direct creation of SPINMap files. When this options is chosen, all the right vocabularies are imported automatically. Before you start mapping properties, you need to map classes. Select class A, switch to the diagram view and drag and drop class B into the diagram. Draw a connection between class A and class B. This means that instances of class A get transformed into instances of class B – for each resource of rdf:type :A, TopBraid will create a resource of rdf:type B. The URIs of the newly created resources are determined by the mapping function you select. The simplest one, spinmapl:self, simply re-uses existing resources and adds ?this rdf:type :B triple. spinmapl:changeNamespace will create new resource by using the URIs of members of class A and changing the namespace to the one you specify. For example, if there is a triple {<www.example.com/A/R123> rdf:type :A}, it will create a triple {<www.example.com/B/R123> rdf:type :B}, if you specified <www.example.com/B/ as the namespace to use. And so on. Now you can start mapping properties. Simply draw a connection between a:name and b:name. Again, you will be asked to create a mapping functions. At the property level, mapping functions take the source property value as the input and produce some output that becomes the value of the target property. Since you are saying that a:name is the same as b:name, you should use spin map:equal. This does not extra transformations, simply gives the newly created resource that is a member of class B, the same property value for b:name as the source resource had for a:name. If this doesn’t work, you need to be more specific about what you are doing and what is happening as a result. Regards, Irene Polikoff On Wed, Sep 9, 2015 at 5:25 PM, User12 <[email protected]> wrote: > Hi, > > I'm newbie with TopBraid i have two classes (Class A and ClassB) the first > class contain a data property a:name and the second one contain also the > same property b:name > I would like to do a mapping to say that a:name is the same that b: name > > How can do it i try to do the same thing than Topbraid video but it do > nothing > > Any example or explication can help me.. Thank you ^^ > > -- > You received this message because you are subscribed to the Google Group > "TopBraid Suite Users", the topics of which include Enterprise Vocabulary > Network (EVN), Reference Data Manager (RDM), TopBraid Composer, TopBraid > Live, TopBraid Insight, SPARQLMotion, SPARQL Web Pages and SPIN. > To post to this group, send email to [email protected] > --- > You received this message because you are subscribed to the Google Groups > "TopBraid Suite Users" group. > To unsubscribe from this group and stop receiving emails from it, send an > email to [email protected]. > For more options, visit https://groups.google.com/d/optout. > -- You received this message because you are subscribed to the Google Group "TopBraid Suite Users", the topics of which include Enterprise Vocabulary Network (EVN), Reference Data Manager (RDM), TopBraid Composer, TopBraid Live, TopBraid Insight, SPARQLMotion, SPARQL Web Pages and SPIN. To post to this group, send email to [email protected] --- You received this message because you are subscribed to the Google Groups "TopBraid Suite Users" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. For more options, visit https://groups.google.com/d/optout.
