Question

I use Yii framework to develop a website, I want to know how to get the last 5 images orderd by create_time but not within the same album by using Yii active record and by plain SQL.

here’s my albums table:

CREATE TABLE IF NOT EXISTS `tbl_album` (
  `album_id` int(11) NOT NULL AUTO_INCREMENT,
  `album_name` varchar(45) DEFAULT NULL,
  `album_folder_name` int(11) DEFAULT NULL,
  `create_time` datetime DEFAULT NULL,
  `create_user_id` int(11) DEFAULT NULL,
  `update_time` datetime DEFAULT NULL,
  `update_user_id` int(11) DEFAULT NULL,
  PRIMARY KEY (`album_id`)
) ENGINE=InnoDB  DEFAULT CHARSET=utf8 AUTO_INCREMENT=55 ;

and here’s my images table:

CREATE TABLE IF NOT EXISTS `tbl_image` (
  `image_id` int(11) NOT NULL AUTO_INCREMENT,
  `image_name` varchar(100) DEFAULT NULL,
  `image_description` varchar(100) DEFAULT NULL,
  `image_album_id` int(11) NOT NULL,
  `create_time` datetime DEFAULT NULL,
  `create_user_id` int(11) DEFAULT NULL,
  `update_time` datetime DEFAULT NULL,
  `update_user_id` int(11) DEFAULT NULL,
  PRIMARY KEY (`image_id`),
  KEY `fk_image_album` (`image_album_id`)
) ENGINE=InnoDB  DEFAULT CHARSET=utf8 AUTO_INCREMENT=9 ;

please note that I use “create_time” for sorting images.

Again, I need a query to get the last 5 images orderd by create_time but not within the same album by Yii active record and plain SQL.

for example the last 7 images are:

a.jpg in the album 7
b.jpg in the album 5
c.jpg in the album 7
d.jpg in the album 6
e.jpg in the album 3
f.jpg in the album 4
g.jpg in the album 2
h.jpg in the album 1

I need the query result to be like the following:

a.jpg in album 7
b.jpg in album 5
d.jpg in album 6
e.jpg in album 3
f.jpg in album 4

not like the following:

a.jpg in album 7
b.jpg in album 5
c.jpg in album 7
d.jpg in album 6
e.jpg in album 3

Thanks in advance.

Answer 1

Try this:

SELECT
  a.album_name,
  i1.image_name
FROM tbl_album a
INNER JOIN tbl_image i1 ON i1.image_album_id = a.album_id
INNER JOIN
(
   SELECT image_album_id, MIN(image_id) image_id
   FROM tbl_image
   GROUP BY image_album_id
) i2  ON i2.image_album_id = i1.image_album_id 
     AND i1.image_id = i2.image_id
ORDER BY a.album_name DESC
LIMIT 5;

The JOIN with:

   SELECT image_album_id, MIN(image_id) image_id
   FROM tbl_image
   GROUP BY image_album_id

will insure that for each album, the first image will be returned. Therefore you will got only one image for each album, then LIMIT 5 will limit the result set to be only 5 albums.

SQL Fiddle Demo

Note that: The ORDER BY clause will determine which five albums will be returned by the LIMIT 5 clause. So don’t expect that the albums are returned in the way they are stored as the expected result in your question, because records in the table has no specific order. They are stored as a set, and you have to specify an ORDER BY clause in your query to get them in a specific order.

---

Update: IF you are looking for the last created image for each album, use the MAX(creaet_time) instead like so:

SELECT
  a.album_name,
  i1.image_name,
  date_format(i1.create_time, '%Y-%m-%d') create_time
FROM tbl_album a
INNER JOIN tbl_image i1 ON i1.image_album_id = a.album_id
INNER JOIN
(
   SELECT image_album_id, MAX(create_time) LatestCreateTime
   FROM tbl_image
   GROUP BY image_album_id
) i2  ON i2.image_album_id = i1.image_album_id 
     AND i1.create_time = i2.LatestCreateTime
ORDER BY i1.create_time DESC
LIMIT 5;

Updated SQL Fiddle Demo

---

Update 2: For the duplicate values, use the DISTINCT keyword, or you can use the MAX(image_id) instead, since the image_id is autoincremental, like so:

SELECT
  a.album_name,
  i1.image_name,
  date_format(i1.create_time, '%Y-%m-%d') create_time
FROM tbl_album a
INNER JOIN tbl_image i1 ON i1.image_album_id = a.album_id
INNER JOIN
(
   SELECT image_album_id, MAX(image_id) LatestId
   FROM tbl_image
   GROUP BY image_album_id
) i2  ON i2.image_album_id = i1.image_album_id 
     AND i1.image_id = i2.LatestId
ORDER BY i1.create_time DESC
LIMIT 5;

Updated SQL Fiddle Demo