Question

我有一个简单的网络应用程序，当用户点击图像上的收藏夹时，数据库会存储user_id和他们正在查看的image_id，表格如下所示：

Favorites
---------------------
-user_id - image_id -
---------------------
-abc     - 123      -
-abc     - 456      -
-def     - 123      -
---------------------

我正在努力寻找前10名最喜欢的图片（全球），即总体上最受欢迎的10张图片。查询只需要找到最常出现的10个image_id值。到目前为止，我已经尝试了一些基于

的内容

SELECT image_id, COUNT(*) FROM favourites GROUP BY image_id LIMIT 100 ORDER DESC

完成此任务的正确查询是什么？

Answer 1

以下查询应该可以解决问题，它与您的代码几乎相同，但最后一点是不同的：

select
    image_id,
    count(*)
from
    favourites
group by
    image_id
order by
    count(*) desc
limit 10

您可能还希望阅读Q&A that I wrote，其中涵盖了很多类似的内容。

编辑：

要回答下面的评论之一，在count(*)语句中使用order by会导致它再次计算吗？

没有

mysql> select * from test2;
+------+-------+-------+
| id   | barry | third |
+------+-------+-------+
|    1 | ccc   |  NULL |
| NULL | d     |     1 |
| NULL | d     |     2 |
| NULL | d     |     3 |
+------+-------+-------+
4 rows in set (0.00 sec)

mysql> explain select barry, max(third) from test2;
+----+-------------+-------+------+---------------+------+---------+------+------+-------+
| id | select_type | table | type | possible_keys | key  | key_len | ref  | rows | Extra |
+----+-------------+-------+------+---------------+------+---------+------+------+-------+
|  1 | SIMPLE      | test2 | ALL  | NULL          | NULL | NULL    | NULL |    4 |       |
+----+-------------+-------+------+---------------+------+---------+------+------+-------+
1 row in set (0.11 sec)

mysql> explain select barry, max(third) from test2 order by barry;
+----+-------------+-------+------+---------------+------+---------+------+------+-------+
| id | select_type | table | type | possible_keys | key  | key_len | ref  | rows | Extra |
+----+-------------+-------+------+---------------+------+---------+------+------+-------+
|  1 | SIMPLE      | test2 | ALL  | NULL          | NULL | NULL    | NULL |    4 |       |
+----+-------------+-------+------+---------------+------+---------+------+------+-------+
1 row in set (0.00 sec)


mysql> explain select barry, max(third) from test2 order by max(third);
+----+-------------+-------+------+---------------+------+---------+------+------+-----------------+
| id | select_type | table | type | possible_keys | key  | key_len | ref  | rows | Extra           |
+----+-------------+-------+------+---------------+------+---------+------+------+-----------------+
|  1 | SIMPLE      | test2 | ALL  | NULL          | NULL | NULL    | NULL |    4 | Using temporary |
+----+-------------+-------+------+---------------+------+---------+------+------+-----------------+
1 row in set (0.00 sec)

您可以从中看到它将数据存储在temporary中并从那里使用它。

Answer 2

试试这个：

SELECT
  image_id, count(image_id)
FROM Favorites
GROUP BY image_id
ORDER BY 2 DESC
LIMIT 10

MySQL查询根据具有最高投票的项目进行选择

2 个答案: