Question

我有这张桌子：

// QandA
+----+--------+----------------------------------------+------+---------+
| id |  title |                  content               | type | related |
+----+--------+----------------------------------------+------+---------+
| 1  | title1 | content of question1                   | 0    | NULL    |
| 2  |        | content of first answer for question1  | 1    | 1       |
| 3  | title2 | content of question2                   | 0    | NULL    |
| 4  | title3 | content of question3                   | 0    | NULL    |
| 5  |        | content of second answer for question1 | 1    | 1       |
| 6  |        | content of first answer for question3  | 1    | 4       |
| 7  | title4 | content of question4                   | 0    | NULL    |
| 8  |        | content of first answer for question2  | 1    | 3       |
+----+--------+----------------------------------------+------+---------+
-- type colum: it is 0 for questions and 1 for answers.
-- related column: it is NULL for questions and {the id of its own question} for answers.

我还有另外两张表：

// interface_tags
+---------+--------+
| post_id | tag_id |
+---------+--------+
| 1       | 1      |
| 1       | 5      |
| 3       | 4      |
| 4       | 1      |
| 4       | 2      |
| 4       | 5      |
| 7       | 2      |
+---------+--------+

// tags
+----+----------+
| id | tag_name |
+----+----------+
| 1  | PHP      |
| 2  | SQL      |
| 3  | MySQL    |
| 4  | CSS      |
| 5  | Java     |
| 6  | HTML     |
| 7  | JQuery   |
+----+----------+

这是我的疑问：

SELECT id,
       title,
       content
       (SELECT i_t.tag_name
        FROM tags t
        JOIN interface_tags i_t
        ON t.id = i_t.tag_id
        WHERE i_t.post_id = :id1) tag
FROM QandA
WHERE id = :id2 OR related = :id3

-- Note: :id1, :id2 and :id3 are identical

如您所见，我的查询同时选择了问题（id = :id2）及其所有答案（related = :id3）。有一个子查询可以获取问题的所有标签。但它将按行执行。因此，子查询将针对问题和答案执行。有很多浪费过程，因为答案没有标签。

我的问题是什么？如何避免执行该子查询以获得答案？我的意思是我只想针对这个问题执行该子查询。

编辑：这是预期的输出：

-- :id = 1

// QandA
+----+--------+----------------------------------------+------------+
| id |  title |                  content               |   tag      |
+----+--------+----------------------------------------+------------+
| 1  | title1 | content of question1                   | PHP, JAVA  |
| 2  |        | content of first answer for question1  |            |
| 5  |        | content of second answer for question1 |            |
+----+--------+----------------------------------------+------------+

Answer 1

你需要CASE：

SELECT id,
       title,
       content,
       CASE
            WHEN type = 0 THEN      -- get tags only when post type is QUESTION
                (SELECT i_t.tag_name
                FROM tags t
                JOIN interface_tags i_t
                ON t.id = i_t.tag_id
                WHERE i_t.post_id = :id1)
            ELSE        -- else consider it an ANSWER and don't get tags
                NULL
        END tag
FROM QandA
WHERE id = :id2 OR related = :id3

Answer 2

你试过这个吗？

SELECT q.id,
       q.title,
       q.content,
       GROUP_CONCAT(i_t.tag_name)       
FROM QandA q
LEFT JOIN interface_tags i_t
    ON q.id = i_t.post_id
LEFT JOIN tags t
    ON t.id = i_t.tag_id    
WHERE 
    q.id = :id2 OR q.related = :id3
GROUP BY
    q.id

这是您想要的可能解决方案。

Answer 3

为什么不用UNION执行两个单独的查询？

SELECT id, title, content,
       (SELECT GROUP_CONCAT(t.tag_name)
        FROM tags t
        JOIN interface_tags i_t
        ON t.id = i_t.tag_id
        WHERE i_t.post_id = :id1) tag
FROM QandA WHERE id = :id2
UNION ALL
SELECT id, title, content, NULL
FROM QandA WHERE related = :id3

sqlfiddle

Answer 4

您可以使用CASE表达式，并添加1个参数，因此我们有：id1 ...：id4（全部相同）

SELECT 
    id,
    title,
    content
    CASE WHEN id = :id1 
        THEN (SELECT i_t.tag_name
            FROM tags t
            JOIN interface_tags i_t
            ON t.id = i_t.tag_id
            WHERE i_t.post_id = :id2) 
    END tag
FROM 
    QandA
WHERE 
    id = :id3 OR related = :id4;

如何避免某些行的子查询执行？

4 个答案: