Question

所以我特别密集我确定 - 但我正在努力学习PHP中的简单（mysql）SQL语句。

我有两张桌子：

entry包含名称＆amp;比赛条目的细节。
votes包含时间戳，唯一ID和entry_id。 entry_id与条目id匹配的每一行代表该人参赛作品的一票。

粗糙的Db结构是：

条目表：

|id|name|email|created|deleted|   [last two cols being datetime stamps]

投票表：

|id|entry_id|created|deleted|

我的尝试（各种重写，可能太累了！）导致只返回一行。

我的最新尝试：

$sql = 'SELECT
        s_id,
        count(sv.s_id) as count
    FROM
        'vote AS sv
    LEFT JOIN
        'entry AS se on sv.entry_id = se.id
    WHERE
        sv.deleted = "0000-00-00 00:00:00"
    AND
        se.deleted = "0000-00-00 00:00:00"
    ORDER BY
        count DESC
    LIMIT
        10';

有人可以告诉我如何最好地实现这个目标吗？

我的目标是返回前十个条目（通过投票数），并在返回数据中包含计数。

谢谢！

史蒂夫

Answer 1

怎么样

SELECT e.id, COUNT(v.id) AS numVotes
FROM entry AS e
JOIN vote AS v ON v.entry_id = e.id
WHERE e.deleted = "0000-00-00 00:00:00"
AND v.deleted = "0000-00-00 00:00:00"
GROUP BY e.id
ORDER BY numVotes DESC
LIMIT 10

Answer 2

select
        e.*,
        cnt
from
        entry e,
(
        select
                entry_id,   
                count(*) cnt
        from
                votes
        where
                deleted = "0000-00-00 00:00:00"
        group by
                entry_id
        order by
                count(*) desc
        limit 10
) c  
where
        c.entry_id = e.id and
        e.deleted = "0000-00-00 00:00:00"
order by
        cnt desc

SQL - 根据单独表中相关条目的计数选择记录

2 个答案: