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我有多个表的这个查询，我得到了以下结果。然后我想按 t.id 分组。我知道我可以简单地使用 group by t.id 但如何选择 t.id 的最新行，即 url tw5.jpg 和 created_time 10000004

SELECT p.url,t.name,t.num_photo,t.id
FROM photos AS p
LEFT JOIN tag_maps AS tm ON p.id = tm.photo_id
LEFT JOIN tags AS t ON t.id = tm.tag_id

url                       created_time    name      num_photo  id
assets/img/tags/tw1.jpg   1000001         my house  1           1
assets/img/tags/tw2.jpg   1000002         dog       1           2
assets/img/tags/tw3.jpg   1000003         taiwan    2           3
assets/img/tags/tw5.jpg   1000004         taiwan    2           3

这是我在使用 group by t.id 时得到的

url                       created_time    name      num_photo  id
assets/img/tags/tw1.jpg   1000001         my house  1           1
assets/img/tags/tw2.jpg   1000002         dog       1           2
assets/img/tags/tw3.jpg   1000003         taiwan    2           3

这就是我想要的

url                       created_time    name      num_photo  id
assets/img/tags/tw1.jpg   1000001         my house  1           1
assets/img/tags/tw2.jpg   1000002         dog       1           2
assets/img/tags/tw5.jpg   1000004         taiwan    2           3

已编辑

照片表

id    url                      created_time
1     assets/img/tags/tw1.jpg  1000001
2     assets/img/tags/tw2.jpg  1000002
3     assets/img/tags/tw3.jpg  1000003
4     assets/img/tags/tw5.jpg  1000004

标签表

id name     num_photo
1  my house 1
2  dog      1
3  taiwan   2

标记地图

id   tag_id   photo_id
1    1        1
2    2        2
3    3        3
4    3        4
4

1 回答 1

1

编辑#2：没有看到每个表或表结构中的示例数据，我猜测以下将起作用：

SELECT p.url,
  created_time,
  t.name,
  t.num_photo,
  t.id
FROM photos p
LEFT JOIN tag_maps AS tm 
    ON p.id = tm.photo_id
LEFT JOIN tags AS t 
    ON t.id = tm.tag_id
INNER JOIN
(
  select max(created_time) MaxDate, t.id
  FROM photos p
  LEFT JOIN tag_maps AS tm 
      ON p.id = tm.photo_id
  LEFT JOIN tags AS t 
      ON t.id = tm.tag_id
  group by t.id
) d
  on p.created_time = d.MaxDate
  and t.id = d.id;

请参阅带有演示的 SQL Fiddle

或者另一种编写方式是使用从表中返回max(photo_id)by并在连接中使用该结果的子查询：tag_Idtag_maps

SELECT p.url,
  created_time,
  t.name,
  t.num_photo,
  t.id
FROM photos p
INNER JOIN
(
  select max(photo_id) photo_id, tag_id
  from tag_maps
  group by tag_id
) AS tm 
    ON p.id = tm.photo_id
LEFT JOIN tags AS t 
    ON t.id = tm.tag_id

请参阅带有演示的 SQL Fiddle

结果是：

|                     URL | CREATED_TIME |     NAME | NUM_PHOTO | ID |
----------------------------------------------------------------------
| assets/img/tags/tw1.jpg |      1000001 | my house |         1 |  1 |
| assets/img/tags/tw2.jpg |      1000002 |      dog |         1 |  2 |
| assets/img/tags/tw5.jpg |      1000004 |   taiwan |         2 |  3 |

编辑＃1，因为您的 url 不同并且您想要max(id)，那么您应该可以使用：

SELECT p.url,
  t.name,
  t.num_photo,
  Max(t.id) id
FROM photos p
LEFT JOIN tag_maps AS tm 
    ON p.id = tm.photo_id
LEFT JOIN tags AS t 
    ON t.id = tm.tag_id
group by t.name, t.num_photo

OP：您可以使用子查询：

SELECT p.url,
    t.name,
    t.num_photo 
FROM
(
    select MAX(created_time) created_time, url, id
    from photos
    group by url, id
)  AS p
LEFT JOIN tag_maps AS tm 
    ON p.id = tm.photo_id
LEFT JOIN tags AS t 
    ON t.id = tm.tag_id

如果表中的id值photos对于每一行都是唯一的，那么您可能需要使用以下内容：

SELECT p1.url,
    t.name,
    t.num_photo 
FROM photos p1
inner join
(
    select MAX(created_time) created_time, url
    from photos
    group by url
)  AS p2
    on p1.url = p2.url
    and p1.created_time = p2.created_time
LEFT JOIN tag_maps AS tm 
    ON p1.id = tm.photo_id
LEFT JOIN tags AS t 
    ON t.id = tm.tag_id

于 2013-01-14T16:39:41.247 回答