Прочетете XML данни в йерархични таблици

Можем да (з)използваме ROW_NUMBER() за генериране на идентификатори извън XQuery. Преамбюл:

WITH questions AS (
    SELECT
        ROW_NUMBER() OVER (ORDER BY (SELECT NULL)) AS ID,
        n.q.value('(./questionText)[1]', 'nvarchar(max)') AS questionText,
        n.q.value('(./result)[1]', 'nvarchar(50)') AS result,
        n.q.query('answer') AS answers
    FROM
        @xml.nodes('/quizresult/question') AS n (q)
), questions_and_answers AS (
    SELECT ID, questionText, result, answer.query('.') AS answer
    FROM questions
    CROSS APPLY answers.nodes('answer') AS a(answer)
)

А сега извлечете въпросите с

SELECT ID, questionText, result 
FROM questions

И отговорите с

SELECT ID AS questionID,
    q.answer.value('answer[1]', 'nvarchar(max)') AS answer,
    q.answer.value('answer[1]/@number', 'int') AS number,
    q.answer.value('answer[1]/@value', 'int') AS val,
    q.answer.value('answer[1]/@chosen', 'bit') AS chosen
FROM questions_and_answers AS q