SPARQL查詢以選擇RECIPE

Question

好的，假設我需要提取所有包含2/3成分的食譜。 配方表示為鏈接數據，這是http://linkedrecipes.org/schema使用的本體。 我知道如何用咖喱找到食譜：

PREFIX rdfs: <htp://ww.w3.org/2000/01/rdf-schema#>
PREFIX recipe: <htp://linkedrecipes.org/schema/>

SELECT ?label ?recipe 
WHERE{ 
?food rdfs:label ?label2 . 
?food recipe:ingredient_of ?recipe .
?recipe a recipe:Recipe . 
?recipe rdfs:label ?label.  

FILTER (REGEX(STR(?label2), 'curry', 'i'))
}

但是如何找到例如咖喱和雞肉的食譜？

Answer 1

這應該找到咖喱和雞肉：

PREFIX rdfs: <htp://ww.w3.org/2000/01/rdf-schema#>
PREFIX recipe: <htp://linkedrecipes.org/schema/>

SELECT ?label ?recipe { 
    ?recipe a recipe:Recipe . 
    ?recipe rdfs:label ?label.  

    ?curry recipe:ingredient_of ?recipe .
    ?curry rdfs:label ?curry_label . 
    FILTER (REGEX(STR(?curry_label), 'curry', 'i'))

    ?chicken recipe:ingredient_of ?recipe .
    ?chicken rdfs:label ?chicken_label . 
    FILTER (REGEX(STR(?chicken_label), 'chicken', 'i'))
}