0

PHPでリーグアプリケーションを開発しています。ラダー ビュー ページにアクセスすると、そのラダーからすべてのスクワッドを選択し、経験 (league_experience) で並べ替えるクエリがあります。現在の分隊のランクを見つけるようにクエリを変更したいと思います。

$query_squads = "
            SELECT
                s.squad_id AS squad_id, s.ladder_id, s.team_id AS team_id,
                x.experience_id, x.squad_id, SUM(x.value) as total_exp
            FROM league_squads AS s
            LEFT JOIN league_experience AS x ON (s.squad_id = x.squad_id)
            WHERE s.ladder_id = ".$ladder_id."
            GROUP BY s.squad_id, s.ladder_id, s.team_id, x.experience_id, x.squad_id
            ORDER BY total_exp DESC
            ";

これが私のテーブルです

--
-- Table structure for table `league_experience`
--

CREATE TABLE IF NOT EXISTS `league_experience` (
  `experience_id` int(15) NOT NULL,
  `squad_id` int(15) NOT NULL,
  `value` int(15) NOT NULL,
  `date_earned` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP ON UPDATE CURRENT_TIMESTAMP,
  `description` varchar(255) NOT NULL,
  PRIMARY KEY (`experience_id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1;

--
-- Dumping data for table `league_experience`
--

INSERT INTO `league_experience` (`experience_id`, `squad_id`, `value`, `date_earned`, `description`) VALUES
(1, 1, 500, '2013-09-03 07:10:59', 'For being ballers.'),
(2, 2, 250, '2013-09-03 07:10:52', 'For being awesome.');

-- --------------------------------------------------------

--
-- Table structure for table `league_squads`
--

CREATE TABLE IF NOT EXISTS `league_squads` (
  `squad_id` int(15) NOT NULL AUTO_INCREMENT,
  `team_id` int(15) NOT NULL,
  `ladder_id` int(15) NOT NULL,
  `date_joined` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP ON UPDATE CURRENT_TIMESTAMP,
  `status` tinyint(1) NOT NULL,
  `last_rank` tinyint(5) NOT NULL,
  PRIMARY KEY (`squad_id`)
) ENGINE=InnoDB  DEFAULT CHARSET=latin1 AUTO_INCREMENT=3 ;

--
-- Dumping data for table `league_squads`
--

INSERT INTO `league_squads` (`squad_id`, `team_id`, `ladder_id`, `date_joined`, `status`, `last_rank`) VALUES
(1, 1, 1, '2013-09-03 08:16:27', 0, 1),
(2, 2, 1, '2013-09-03 08:16:25', 0, 2);
4

1 に答える 1

1 
SELECT
                s.squad_id AS squad_id, s.ladder_id, s.team_id AS team_id,
                x.experience_id, x.squad_id, SUM(x.value) as total_exp,
                @i:=@i+1 AS rank
            FROM league_squads AS s
            LEFT JOIN league_experience AS x ON (s.squad_id = x.squad_id),
            (SELECT @i:=0) AS foo
            WHERE s.ladder_id = 1
            GROUP BY s.squad_id, s.ladder_id, s.team_id, x.experience_id, x.squad_id
            ORDER BY total_exp DESC

サンプルフィドル

于 2013-09-03T08:46:22.463 に答える