Hi

[snip]

I did not see any special syntax or FILTER or Label service commands to

help with splitting apart a Label by whitespace and then applying a filter on each string.

How would I accomplish this ?

Maybe with the function STRAFTER https://www.w3.org/TR/sparql11-query/#func-strafter

Example of use:

SELECT ?item ?itemLabel ?itemLabelafter WHERE { ?item wdt:P19 wd:Q647 ; rdfs:label ?itemLabel. FILTER (lang(?itemLabel) = "fr"). BIND ( STRAFTER(?itemLabel, " ") AS ?itemLabelafter ) } LIMIT 10

Cdlt, ~nicolas

I don’t think there’s any SPARQL function that can transform a single result into multiple results (one for each string fragment, in that case). But for your request, a regex works:

SELECT ?human ?label WHERE {   ?human wdt:P31 wd:Q15632617;          rdfs:label ?label.   FILTER(LANG(?label) = "en")   FILTER(REGEX(?label, "y\s")) }

Link https://query.wikidata.org/#SELECT%20%3Fhuman%20%3Flabel%20WHERE%20%7B%0A%20%20%3Fhuman%20wdt%3AP31%20wd%3AQ15632617%3B%0A%20%20%20%20%20%20%20%20%20rdfs%3Alabel%20%3Flabel.%0A%20%20FILTER%28LANG%28%3Flabel%29%20%3D%20%22en%22%29%0A%20%20FILTER%28REGEX%28%3Flabel%2C%20%22y%5C%5Cs%22%29%29%0A%7D

On 18.09.2017 19:07, Thad Guidry wrote:

Say I have this query...

SELECT ?human ?label WHERE {   ?human wdt:P31 wd:Q15632617; rdfs:label ?label.   FILTER(LANG(?label) = "en").   FILTER(STRSTARTS(?label, "Mr. ")). }

What if I wanted to see if any one of a humans name ends with "y" such as my last name does , their first, last, doesn't matter.  I have a "d" and a "y" on the array returned from my name (if it were split by whitespace)

I did not see any special syntax or FILTER or Label service commands to help with splitting apart a Label by whitespace and then applying a filter on each string.

How would I accomplish this ?

Thad ​​ +ThadGuidry https://www.google.com/+ThadGuidry

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