mysql - 选择MariaDB/MySql中的任何一个未回答的多项选择题

SQL Fiddle（小提琴对我不起作用）
我想获取id为99的用户尚未回答的任何一个多项选择题。他只回答了一个问题-A question见存储user_answer而不是问题id的choice id表。
我的预期结果应该是这样的：

Question    Question_id Choice_id  Choice
B question       2        3        B choice 1
B question       2        4        B choice 2

或者

C question       3        5        C choice 1
C question       3        6        C choice 2
C question       3        7        C choice 3

使用下面的Sql，我在获取B或C问题及其选择时遇到困难

SELECT q.question_id,q.question,qc.choice,qc.choice_id
FROM question_choice qc
INNER JOIN question q ON q.question_id = qc.question_id
WHERE qc.question_id NOT IN
(SELECT question_id
   FROM user_answer ua
   INNER JOIN question_choice qc ON qc.choice_id = ua.choice_id
   WHERE ua.user_id = 99)
GROUP BY qc.choice_id
ORDER BY q.question_id;

我不知道该在limit clause中放些什么，以便只获取一个带有可变选择数的未回答问题。
表格：

CREATE TABLE question
(question_id INT, question VARCHAR(100));

CREATE TABLE question_choice
(question_id INT, choice_id INT,choice VARCHAR(100));

CREATE TABLE user_answer
(user_id INT,choice_id INT);

INSERT INTO question (question_id,question)
VALUES
(1,"A question"),
(2,"B question"),
(3,"C question");

INSERT INTO question_choice (choice_id,choice,question_id)
VALUES
(1,"A choice 1",1),
(2,"A choice 2",1),
(3,"B choice 1",2),
(4,"B choice 2",2),
(5,"C choice 1",3),
(6,"C choice 2",3),
(7,"C choice 3",3);

INSERT INTO user_answer(user_id,choice_id)
VALUES
(99,2);

最佳答案

你在正确的轨道上。但这可能有助于你考虑达成解决方案。
首先，可以使用以下查询获取未回答的问题列表：

SELECT q.*
FROM question q
WHERE NOT EXISTS (SELECT question_id
                  FROM user_answer ua INNER JOIN
                       question_choice qc
                       ON qc.choice_id = ua.choice_id
                  WHERE ua.user_id = 99 AND qc.question_id = q.question_id
                 );

注意，我用NOT IN替换了NOT EXISTS。因为它们对NULL值的处理方式不同，我强烈建议使用NOT EXISTS——对NULL值的处理更直观。
然后，您可以在查询中使用它来获取有关未回答问题的其他信息：

SELECT q.question_id, q.question, qc.choice, qc.choice_id
FROM (SELECT q.*
      FROM question q
      WHERE NOT EXISTS (SELECT question_id
                        FROM user_answer ua INNER JOIN
                             question_choice qc
                             ON qc.choice_id = ua.choice_id
                        WHERE ua.user_id = 99 AND qc.question_id = q.question_id
                       )
     ) q JOIN
     question_choice qc
     ON qc.question_id = q.question_id;

如果愿意，可以通过删除子查询来研究如何“简化”此查询。