По-просто решение, но дава резултати само когато няма пропуски в item.position
s:
SELECT DISTINCT sentence.sentenceid
FROM sentence
JOIN item ON sentence.sentenceid = item.sentenceid
JOIN word ON item.wordid = word.wordid
JOIN item AS next_item ON sentence.sentenceid = next_item.sentenceid
AND next_item.position = item.position + 1
JOIN word AS next_word ON next_item.wordid = next_word.wordid
WHERE word.spelling = 'word1'
AND next_word.spelling = 'word2'
По-общо решение, използващо прозоречни функции :
SELECT DISTINCT sentenceid
FROM (SELECT sentence.sentenceid,
word.spelling,
lead(word.spelling) OVER (PARTITION BY sentence.sentenceid
ORDER BY item.position)
FROM sentence
JOIN item ON sentence.sentenceid = item.sentenceid
JOIN word ON item.wordid = word.wordid) AS pairs
WHERE spelling = 'word1'
AND lead = 'word2'
Редактиране :Също общо решение (разрешени пропуски), но само с обединения:
SELECT DISTINCT sentence.sentenceid
FROM sentence
JOIN item ON sentence.sentenceid = item.sentenceid
JOIN word ON item.wordid = word.wordid
JOIN item AS next_item ON sentence.sentenceid = next_item.sentenceid
AND next_item.position > item.position
JOIN word AS next_word ON next_item.wordid = next_word.wordid
LEFT JOIN item AS mediate_word ON sentence.sentenceid = mediate_word.sentenceid
AND mediate_word.position > item.position
AND mediate_word.position < next_item.position
WHERE mediate_word.wordid IS NULL
AND word.spelling = 'word1'
AND next_word.spelling = 'word2'