Question

我有两张桌子：

用户

id |用户

喜欢

id |老板|日期

这个查询：

$owner_result = $connector->query("SELECT id FROM users");
while($owner = $connector->fetchArray($owner_result))
{
$rs = $connector->query("SELECT count(id) AS num FROM likes 
WHERE date > DATE_SUB(NOW(), INTERVAL 1 WEEK) 
AND owner='$owner[id]' 
ORDER BY num DESC LIMIT 5");
while($rw = $connector->fetchArray($rs)) 

{echo "This owner ($owner[id]) has $rw[num] likes this week";}

}

我想要做的是返回5 users.id 行，其中包含赞表中最大的行数。我的查询只返回 $ owner [id] ，但不是每个人的喜欢。我认为我的查询的效果也非常低，因为我的查询检查了 likes 表中每个 users.id 的数量，但它可能是喜欢< strong>表格不包含一些 users.id 。

任何有关解决我的问题或改进我的查询的建议都将非常受欢迎。谢谢。

Answer 1

您必须在第二个查询中按所有者分组。你不能先计算行数，而不是先将它们分组。

$owner_result = $connector->query("SELECT id FROM users");
while($owner = $connector->fetchArray($owner_result))
{
$rs = $connector->query("SELECT count(id) AS num FROM likes 
WHERE date > DATE_SUB(NOW(), INTERVAL 1 WEEK) 
AND owner='$owner[id]'
Group BY owner 
ORDER BY num DESC LIMIT 5");
while($rw = $connector->fetchArray($rs)) 

{echo "This owner ($owner[id]) has $rw[num] likes this week";}

}

Answer 2

您应该加入这两个表，以便您可以在一个语句中获取数据：

SELECT
    users.id,
    COUNT(likes.id) AS mylikes     -- specify a name for the computed column
FROM users
INNER JOIN likes
ON users.id = likes.owner
WHERE date > DATE_SUB(NOW(), INTERVAL 1 WEEK)
GROUP BY users.id
ORDER BY COUNT(likes.id) DESC      -- MySQL would allow the alias name 
LIMIT 5                            -- because you want the top 5

如果存在平局，那么您将失去记录（或更多）。

Answer 3

我认为单个查询可以获得指定的结果，例如：

SELECT u.id
     , COUNT(t.owner) AS cnt_likes
  FROM users u
  JOIN likes t
    ON t.owner = u.id
 WHERE t.date > DATE_SUB(NOW(), INTERVAL 1 WEEK)
 GROUP BY u.id
 ORDER BY 2 DESC
 LIMIT 5

likes表上的适当索引应该可以提高性能：

... ON likes (owner, date)

或者，这将给出相同的结果，可能具有（稍微）更好的性能：

SELECT u.id
     , t.cnt_likes
  FROM users u
  JOIN ( SELECT s.owner
              , COUNT(1) AS cnt_likes
           FROM likes s
          WHERE s.date > DATE_SUB(NOW(), INTERVAL 1 WEEK)
          GROUP BY s.owner
       ) t
    ON t.owner = u.id
 ORDER BY t.cnt_likes DESC
 LIMIT 5

如果确保owner表中likes列的所有值都是id表中的user值，那么性能会更好。 ..因为您可以避免连接到users表，并从likes表中获取整个结果：

SELECT s.owner
     , COUNT(1) AS cnt_likes
  FROM likes s
 WHERE s.date > DATE_SUB(NOW(), INTERVAL 1 WEEK)
 GROUP BY s.owner
 ORDER BY cnt_likes DESC
 LIMIT 5

但是，该查询没有进行检查以验证owner列返回的值是否存在，如id表的user列中那样。

连接2个mysql查询

3 个答案: