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GROUP_CONCAT with additional grouping?

I have started on a script to generate statistics of how often and when a particular article is called using

SELECT `title`, `page_id`, COUNT(*) AS `total`, 
GROUP_CONCAT(DISTINCT `date_created` 
             ORDER BY `date_created` SEPARATOR ',') dates 
FROM `statistics` 
WHERE `supplier_id` = '27' 
GROUP BY `title` 
ORDER BY `title`;

which produces

[0] => Array
        (
            [0] => Array
                (
                    [total] => 3
                    [dates] => 2011-04-26,2011-04-27
                )

            [statistics] => Array
                (
                    [title] => Title 2
                    [page_id] => 2
                )  
        )

[1] => Array
        (
            [0] => Array
                (
                    [total] => 6
                    [dates] => 2011-04-26,2011-04-27,2011-04-28
                )

            [statistics] => Array
                (
                    [title] => Title 7
                    [page_id] => 7
                )
        )

Good stuff. But how can I add a field similar to dates which adds a breakdown of ids grouped by date_created thus creating something like:

[2] => Array
    (
        [0] => Array
            (
                [total] => 6
                [dates] => 2011-04-26,2011-04-27,2011-04-28
                [total_by_date] => 2,1,3
            )

        [statistics] => Array
            (
                [title] => Title 7
                [page_id] => 7
            )
)

I would like to add an extra GROUP BY along these lines:

GROUP_CONCAT(DISTINCT COUNT(*) AS `total_by_date` 
             GROUP BY `date_created` ORDER BY `date_created` SEPARATOR ',')

but it's not working, how to I fix this?

DUMP:

CREATE TABLE `statistics` (
  `id` int(11) NOT NULL auto_increment,
  `pagetype` varchar(50) character set utf8 collate utf8_unicode_ci NOT NULL,
  `supplier_id` int(11) NOT NULL,
  `page_id` int(11) default NULL,
  `title` varchar(255) character set utf8 collate utf8_unicode_ci NOT NULL,
  `date_created` date NOT NULL,
  PRIMARY KEY  (`id`)
) ENGINE=MyISAM  DEFAULT CHARSET=latin1 AUTO_INCREMENT=920 ;

-- 
-- Dumping data for table `statistics`
-- 

INSERT INTO `statistics` (`id`, `pagetype`, `supplier_id`, `page_id`, `title`, `date_created`) VALUES 
(1, 'newsarticle', 27, 2751, 'Title 1', '2011-04-26'),
(2, 'newsarticle', 27, 2751, 'Title 1', '2011-04-26'),
(3, 'newsarticle', 27, 2751, 'Title 1', '2011-04-27'),
(4, 'newsarticle', 27, 462009, 'Title 2', '2011-04-26'),
(5, 'newsarticle', 27, 462009, 'Title 2', '2011-04-26'),
(6, 'newsarticle', 27, 462009, 'Title 2', '2011-04-27'),
(7, 'newsarticle', 27, 462009, 'Title 2', '2011-04-27'),
(8, 'newsarticle', 27, 462009, 'Title 2', '2011-04-27'),
(9, 'newsarticle', 27, 462009, 'Title 2', '2011-04-28'),
(10, 'newsarticle', 27, 46200, 'Title 3', '2011-04-26'),
(11, 'newsarticle', 27, 46200, 'Title 3', '2011-04-26'),
(12, 'newsarticle', 27, 46200, 'Title 3', '2011-04-26'),
(13, 'newsarticle', 27, 46200, 'Title 3', '2011-04-26'),
(14, 'newsarticle', 27, 46200, 'Title 3', '2011-04-26');
like image 748
Taff Avatar asked Mar 21 '26 07:03

Taff


1 Answers

Not 100% sure this is gonna work, but it's a start:

Do the 2nd GROUP_CONCAT in a sub-query.

SELECT title
  , s1.page_id
  , s2.total_by_date
  , GROUP_CONCAT(DISTINCT s1.date_created
                 ORDER BY s1.date_created SEPARATOR ',') as dates  
FROM statistics s1
INNER JOIN  
  (SELECT page_id
     ,GROUP_CONCAT(/*DISTINCT*/ gc.cnt
     ORDER BY gc.date_created SEPARATOR ',') AS total_by_date
   FROM 
   (
      SELECT count(*) as cnt
        ,date_created
      FROM statistics
      WHERE supplier_id = '27'
      GROUP BY date_created
   ) gc
   GROUP BY gc.date_created
  ) s2 ON (s1.page_id = s2.page_id)      
WHERE supplier_id = '27'  
GROUP BY title  
ORDER BY title; 

Using distinct count(*) would hide items with the same count, which is not I think you want, so I've put it in in commented form.
BTW the ORDER BY title is not needed here, because GROUP BY already sorts on title.

like image 105
Johan Avatar answered Mar 23 '26 20:03

Johan



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