Question

我需要获得10条最新消息，并且最后有3条评论。我做2个查询： 1）为用户获得10条最新消息。

SELECT *
FROM news
WHERE o_uid = 17
AND o_cid = 0
ORDER BY time DESC
LIMIT 10

2）在此之后，我希望在一个查询中获得对所选新闻的所有评论。我做了＆＃34; IN＆＃34;由新闻ID设置以检索评论。我写的：

SELECT c.*
FROM `comments` AS c
WHERE c.`id` >= Ifnull(
                         (SELECT Min(s.id)
                          FROM
                            (SELECT id
                             FROM comments
                             WHERE nid = c.nid
                             ORDER BY id DESC LIMIT 3) AS s), ~ 0)
  AND c.`nid` IN ('17',
                  '16')

但这不符合错误＆＃34;未知栏＆＃39; c.nid＆＃39;在＆＃39; where子句＆＃39;＆＃34;。通过这个子查询，我希望在每个新闻的3个最后评论ID中检索最小ID，并在我得到评论ID为＆gt; =所选id（最后3个）的所有评论之后。请帮助我检索每条新闻的最后三条评论。感谢。

Answer 1

考虑以下内容......

DROP TABLE IF EXISTS news;

CREATE TABLE news(news_id INT NOT NULL AUTO_INCREMENT PRIMARY KEY);

INSERT INTO news VALUES (1),(2),(5),(6),(7),(9);

DROP TABLE IF EXISTS comments;

CREATE TABLE comments(comment_id INT NOT NULL AUTO_INCREMENT PRIMARY KEY, news_id INT NOT NULL);

INSERT INTO comments (news_id) VALUES
(2),(2),(5),(6),(3),(3),(7),(9),(9),(9),(7),(2),(3),(9);

SELECT * FROM news;
+---------+
| news_id |
+---------+
|       1 |
|       2 |
|       5 |
|       6 |
|       7 |
|       9 |
+---------+

SELECT * FROM comments;
+------------+---------+
| comment_id | news_id |
+------------+---------+
|          1 |       2 |
|          2 |       2 |
|          3 |       5 |
|          4 |       6 |
|          5 |       3 |
|          6 |       3 |
|          7 |       7 |
|          8 |       9 |
|          9 |       9 |
|         10 |       9 |
|         11 |       7 |
|         12 |       2 |
|         13 |       3 |
|         14 |       9 |
+------------+---------+

SELECT n.*
     , c.*
  FROM news n
  LEFT
  JOIN 
     ( SELECT x.* 
         FROM comments x 
         JOIN comments y 
           ON y.news_id = x.news_id 
          AND y.comment_id >= x.comment_id 
        GROUP 
           BY x.news_id 
            , x.comment_id HAVING COUNT(*) <= 3
     ) c
    ON c.news_id = n.news_id;
 +---------+------------+---------+
 | news_id | comment_id | news_id |
 +---------+------------+---------+
 |       1 |       NULL |    NULL |
 |       2 |          1 |       2 |
 |       2 |          2 |       2 |
 |       2 |         12 |       2 |
 |       5 |          3 |       5 |
 |       6 |          4 |       6 |
 |       7 |          7 |       7 |
 |       7 |         11 |       7 |
 |       9 |          9 |       9 |
 |       9 |         10 |       9 |
 |       9 |         14 |       9 |
 +---------+------------+---------+

每个mysql最后三个

1 个答案: