对于GROUP BY中的每一行，MySQL使用COUNT获取MAX值

Question

假设我有一张桌子

           pages_urls
+----+---------+---------------+
| id | site_id | download_date |
+----+---------+---------------+
|  1 |       1 | 2012-01-01    |
|  2 |       1 | 2012-12-31    |
|  3 |       2 | 2012-01-01    |
|  4 |       2 | 2012-12-31    |
+----+---------+---------------+

对于我想要选择的每个site_id：

1. 上次下载日期
2. 此网站表格中的记录数

我尝试了这个查询：

SELECT
    pages_urls.site_id,
    max_table.download_date,
    COUNT(*)
FROM
    pages_urls
LEFT JOIN
(
SELECT
    site_id,
    MAX(download_date) AS download_date
FROM
    pages_urls AS max_pages_urls
WHERE
    max_pages_urls.site_id=site_id
) AS max_table
ON
    pages_urls.site_id=max_table.site_id
GROUP BY
    site_id;

但是我得到了这个而不是期望的结果：

            desired result           ║            my query result         
+---------+---------------+----------║---------+---------------+----------+
| site_id | download_date | COUNT(*) ║ site_id | download_date | COUNT(*) |
+---------+---------------+----------║---------+---------------+----------+
|       1 | 2012-12-31    |        2 ║       1 | 2012-12-31    |        2 |
|       2 | 2012-12-31    |        2 ║       2 | NULL          |        2 |
+---------+---------------+----------║---------+---------------+----------+
                                     ║

fiddle with table and query

如何获得必要的信息？

Answer 1

SELECT site_id, MAX(download_date) AS download_date, count(*)
FROM pages_urls
GROUP BY site_id;

SQLFiddle demo