Question

在查看帖子的页面上，我认为为下一篇和之前的帖子提供链接是个不错的主意。我认为我应该在1个查询中获得这3条记录，因为我很聪明，无论如何我都有很多时间浪费。所以你可能猜到我做不到，但我有兴趣找到解决方案。这就是我所拥有的

SELECT a.id,
       a.title,
       a.body,
       p.id    AS prev_id,
       p.title AS prev_title,
       n.id    AS next_id,
       n.title AS next_title
FROM   posts a
       LEFT JOIN posts p
              ON p.id < a.id
       LEFT JOIN posts n
              ON n.id > a.id
WHERE  a.id = ?
LIMIT  1

问题在于prev_id和prev_title我总是在表格中获得第一个记录。我尝试添加ORDER BY，但它似乎并没有真正影响加入。如何使用示例中的连接选择上一个而不是第一个记录？

Answer 1

我希望你能发现这个有用的东西：

SELECT a.id,
       a.title,
       a.body,
       p.id    AS prev_id,
       p.title AS prev_title,
       n.id    AS next_id,
       n.title AS next_title
FROM   posts a
       INNER JOIN
       (
            SELECT 
                _a.id AS RefID,
                MIN(_a.id - _p.id) AS MinDistPrev,
                MIN(_n.id - _a.id) AS MinDistNext
            FROM   
                posts _a
                LEFT JOIN posts _p
                      ON _p.id < _a.id
                LEFT JOIN posts _n
                      ON _n.id > _a.id
            WHERE  
                _a.id = ?
            GROUP BY
                _a.id
       ) AS _PrevNextDist
        ON _PrevNextDist.RefID = a.ID
       LEFT JOIN posts p
              ON p.id < a.id
                 AND a.id - p.id = _PrevNextDist.MinDistPrev
       LEFT JOIN posts n
              ON n.id > a.id
                 AND n.id - a.id = _PrevNextDist.MinDistNext
LIMIT  1

另外，在原始代码中添加ORDER BY p.id DESC也解决了问题

SELECT a.id,
       a.title,
       a.body,
       p.id    AS prev_id,
       p.title AS prev_title,
       n.id    AS next_id,
       n.title AS next_title
FROM   posts a
       LEFT JOIN posts p
              ON p.id < a.id
       LEFT JOIN posts n
              ON n.id > a.id
WHERE  a.id = ?
ORDER BY p.id DESC
LIMIT  1

在join语句中选择上一个和下一个记录

1 个答案: