How to construct a list in sparql?

@raven_corwin 

In SPARQL, you can construct a list by using the rdf:List and rdf:first properties. Here is an example of how you can construct a list in SPARQL:

For example, let's say we want to create a list of three elements: "apple", "banana", and "orange". We can do this by creating a subject for each element and connecting them using the rdf:first and rdf:rest properties. Here is the SPARQL query to create the list:

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PREFIX rdf: <http://www.w3.org/1999/02/22-rdf-syntax-ns#>
PREFIX ex: <http://example.org/>

ex:List ex:first ex:apple ;
         ex:rest ex:List1 .

ex:List1 ex:first ex:banana ;
         ex:rest ex:List2 .

ex:List2 ex:first ex:orange ;
         ex:rest rdf:nil .

In this query, we first create a subject ex:List and connect it to the first element ex:apple using the rdf:first property. We then create a new subject ex:List1 and connect it to the second element ex:banana, and so on until we reach the end of the list (rdf:nil).

This is one way to construct a list in SPARQL. You can modify this query to create lists with different elements or lengths.