假设我有以下架构：
艺术家：
+------------+------------------+------+-----+---------+----------------+
| Field | Type | Null | Key | Default | Extra |
+------------+------------------+------+-----+---------+----------------+
| id | int(10) unsigned | NO | PRI | NULL | auto_increment | |
| name | varchar(255) | YES | UNI | NULL | |
+------------+------------------+------+-----+---------+----------------+
事件：
+------------+------------------+------+-----+---------+----------------+
| Field | Type | Null | Key | Default | Extra |
+------------+------------------+------+-----+---------+----------------+
| id | int(10) unsigned | NO | PRI | NULL | auto_increment |
| date | timestamp | YES | | NULL | |
| artist_id | int(11) | YES | | NULL | |
| venue_id | int(11) | YES | | NULL | |
+------------+------------------+------+-----+---------+----------------+
资产：
+---------------------+------------------+------+-----+---------+----------------+
| Field | Type | Null | Key | Default | Extra |
+---------------------+------------------+------+-----+---------+----------------+
| id | int(10) unsigned | NO | PRI | NULL | a
| event_id | int(11) | YES | | NULL |
| source_asset_title | varchar(255) | YES | | NULL | |
| source_created_time | timestamp | YES | | NULL | |
我希望给定artist.id的每个事件都有 4个资产的结果集，按事件日期排序，例如：
+----------+----------+------------------------------------------------------------------------------------------------------------------------+---------------------+---------------------+
| event_id | asset_id | source_asset_title | event_date | date |
+----------+----------+------------------------------------------------------------------------------------------------------------------------+---------------------+---------------------+
| 1 | 2089 | aba | 2015-12-03 07:00:00 | 2015-12-03 07:00:00 |
| 1 | 2101 | abb | 2015-12-03 07:00:00 | 2011-04-07 15:30:00 |
| 1 | 2102 | abc | 2015-12-03 07:00:00 | 2011-05-22 16:00:00 |
| 1 | 2107 | abd | 2015-12-03 07:00:00 | 2011-06-11 15:00:00 |
| 2 | 2109 | abe | 2011-07-18 15:00:00 | 2011-07-18 15:00:00 |
| 2 | 2113 | abf | 2011-07-18 15:00:00 | 2011-07-24 15:30:00 |
| 2 | 2115 | abg | 2011-07-18 15:00:00 | 2011-08-25 16:00:00 |
| 2 | 2123 | abh | 2011-07-18 15:00:00 | 2011-08-28 16:00:00 |
| 3 | 2126 | abi | 2011-09-01 16:00:00 | 2011-09-01 16:00:00 |
| 3 | 2129 | abj | 2011-09-01 16:00:00 | 2011-09-10 16:00:00 |
| 3 | 2135 | abk | 2011-09-01 16:00:00 | 2011-10-14 16:00:00 |
| 3 | 2147 | abl | 2011-09-01 16:00:00 | 2011-10-22 16:00:00 |
如果每个事件没有一个子查询，我怎么能实现这个目标呢？
我相信这里的模式和结果集使得它与StackExchange上的其他问题的不同之处在于新问题是合适的。
答案 0 :(得分：0)
以下是artist.id = 1的SQL，由events.date
从较新到较旧排序SELECT
events.id as event_id,
assets.id as asset_id,
source_asset_title
FROM assets
INNER JOIN events ON events.id = assets.event_id
INNER JOIN artists ON artists.id = events.artist_id
WHERE artists.id = 1
ORDER BY events.date DESC
答案 1 :(得分：0)
不，没有子查询就无法实现指定的结果集。作为替代方案，请考虑GROUP_CONCAT，但您可能需要增加MySQL的group_concat_max_len的大小以包含您要描述的所有列。这可能是也可能不是一个好主意。