如何检查答案的作者是否与qustion的作者相同？

Question

这是我的表结构：

// QandA
+----+------+-----------+-------------------+-------------------------------+-------------+
| id | type | author_id |       title       |            content            | question_id |
+----+------+-----------+-------------------+-------------------------------+-------------+
| 1  | 0    | 10        | title of question | content of question           | NULL        |
| 2  | 1    | 33        | NULL              | content of answer             | 1           |
| 3  | 0    | 51        | title of question | content of question           | NULL        |
| 4  | 1    | 10        | NULL              | content of answer             | 1           |
| 5  | 1    | 23        | NULL              | content of answer             | 3           |
+----+------+-----------+-------------------+-------------------------------+-------------+

-- type column: 0 is question, 1 is answer
-- title column: it is always NULL for answers
-- question_id column: it is NULL for questions and the id of its own question for answers

如果author_id是问题的作者，我想设置蓝色背景色。像这样：

所以我试图做的就是检测答案的作者是否和问题的作者一样？我怎么能这样做？

这是我的问题：

SELECT id, (author_id = {Id don't know}) AS set_highlight
FROM QandQ
WHERE id = :id OR question_id = :id
ORDER BY type ASC, id ASC

这是$id = 1的预期结果：

+----+---------------+
| id | set_highlight |
+----+---------------+
| 1  | 1             |
| 2  | 0             |
| 4  | 1             |
+----+---------------+

Answer 1

试试这个，这不是完整的预期结果。这只是满足您需求的方式。

   select answer.*,question.* from(select * from yourtable where `type`=1)answer
    inner join 
    (select * from yourtable where `type`=0)question on question.id=answer.question_id and answer.author_id=question.author_id
WHERE question.id = :id OR answer.question_id = :id

Answer 2

你可以用这个

SELECT id, 
    (CASE WHEN author_id = (
            SELECT author_id 
            FROM QandA
            WHERE id = :id
                AND type = 0) 
        THEN 1
        ELSE 0 END) AS set_highlight
FROM QandA
WHERE id = :id OR question_id = :id
ORDER BY type ASC, id ASC;