So, still cannot see how to unparse, but if I already know the structure of
the query a bit, I can manipulate as follows
import rdflib
from rdflib.plugins.sparql.parser import parseQuery
q = parseQuery(something)
if q[1].where.part[2].triples[0][2] == rdflib.Variable('parent'):
q[1].where.part[2].triples[0][2] =
rdflib.URIRef('http://example.org/fubar#4')
This doesn't quite help with the SPARQLWrapper, because there is no great
way to "unparse" the query, but it almost does.
On Thursday, April 4, 2019 at 2:31:32 PM UTC-4, Dan Davis wrote:
>
> I will answer my own question, then. In
> rdflib/plugins/sparql/algebra.py, the function translateQuery uses the
> traverse function to translate the second part of the query (after the
> prologue), and then further translates to create the query algebra. The
> answer to my question then would be:
>
> * The traverse function acts on the Query syntax producing by parsing
> the query. There is no real included capability to do a transform on the
> algebra.
> * To use traverse yourself, first use
> rdflib.plugins.sparql.parser.parseQuery to parse the query, and then use
> traverse to modify that query.
> * To translate it into algebra, pass it to
> rdflib.plugins.sparql.algebra.translateQuery directly.
>
>>
>>
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