Question

SELECT
    deal_woot.*,
    site.woot_off,
    site.name AS site_name
FROM deal_woot
INNER JOIN site ON site.id = site_id
WHERE site_id IN (2, 3, 4, 5, 6)
GROUP BY site_id
ORDER BY deal_woot.id DESC
LIMIT 5

我想在分组之前先订购，我该如何做到这一点？

Answer 1

使用以下子查询：SELECT *,COUNT(*) FROM (SELECT * from actions order by date DESC) AS actions GROUP BY ip;

Answer 2

由于您按site_id进行分组，因此只会返回1 deal_woot行。请尝试按MAX()排序，这将为每id返回最高site_id。

SELECT
    deal_woot.*,
    site.woot_off,
    site.name AS site_name
FROM deal_woot
INNER JOIN site ON site.id = site_id
WHERE site_id IN (2, 3, 4, 5, 6)
GROUP BY site_id
ORDER BY MAX(deal_woot.id) DESC
LIMIT 5

注意：由于它是UB，实际上会返回deal_woot行，请尝试查询您的查询：

SELECT
    deal_woot.*,
    site.woot_off,
    site.name AS site_name
FROM site JOIN (
    SELECT site_id, MAX(deal_woot.id) MaxID
    FROM deal_woot
    WHERE site_id IN (2, 3, 4, 5, 6)
    GROUP BY site_id
  ) sg ON site.id = sg.site_id
  JOIN deal_woot
    ON site.id = deal_woot.site_id AND deal_woot.id = sg.MaxID
ORDER BY sg.MaxID DESC
LIMIT 5

Answer 3

您正在site_id上执行GROUP BY，而您正在使用deal_woot.id安排记录。

即使你在Group之前进行Order By，输出也会保持不变。

您是否有任何具体要求来做这些事情，因为您所怀疑的与Order by和group by无关。

Answer 4

SELECT
     deal_woot.*,
     site.woot_off,
     site.name AS site_name 
FROM deal_woot 
INNER JOIN site ON site.id = site_id 
WHERE site_id IN (2, 3, 4, 5, 6) 
GROUP BY deal_woot.id DESC, site_id 
LIMIT 5

我认为deal_woot.id是唯一的，分组将基于site_id

GROUP BY查询忽略ORDER BY子句

4 个答案: