MySQL查询排序两列，一个ASC另一个DESC

Question

我正在尝试准备一个主题明智的优点列表。我正在使用这个 mysql 查询:

SELECT * 
FROM results 
ORDER BY qid ASC,
    marks DESC

结果是:

但我需要的是这样的(查看标记列，我需要得到相同的 qid 行，按 marks 排序):

请任何人帮助我。

更新: 这是 result.sql 文件，用于在您的电脑中创建表。

-- phpMyAdmin SQL Dump
-- version 3.5.2.2
-- http://www.phpmyadmin.net
--
-- Host: 127.0.0.1
-- Generation Time: Mar 09, 2013 at 05:40 PM
-- Server version: 5.5.27
-- PHP Version: 5.4.7

SET SQL_MODE="NO_AUTO_VALUE_ON_ZERO";
SET time_zone = "+00:00";


/*!40101 SET @OLD_CHARACTER_SET_CLIENT=@@CHARACTER_SET_CLIENT */;
/*!40101 SET @OLD_CHARACTER_SET_RESULTS=@@CHARACTER_SET_RESULTS */;
/*!40101 SET @OLD_COLLATION_CONNECTION=@@COLLATION_CONNECTION */;
/*!40101 SET NAMES utf8 */;

--
-- Database: `ges_omeca`
--

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

--
-- Table structure for table `results`
--

CREATE TABLE IF NOT EXISTS `results` (
  `exam_id` int(11) NOT NULL AUTO_INCREMENT,
  `sid` varchar(50) NOT NULL,
  `qid` varchar(100) NOT NULL,
  `corrects` int(3) NOT NULL,
  `total_qs` int(3) NOT NULL,
  `marks` float NOT NULL,
  `date_time` datetime NOT NULL COMMENT 'DateTime when user submits the answer script.',
  PRIMARY KEY (`exam_id`)
) ENGINE=InnoDB  DEFAULT CHARSET=latin1 AUTO_INCREMENT=14 ;

--
-- Dumping data for table `results`
--

INSERT INTO `results` (`exam_id`, `sid`, `qid`, `corrects`, `total_qs`, `marks`, `date_time`) VALUES
(1, 'guest', 'EN_(Set-B)_(07.02.13)', 37, 40, 36.25, '2023-02-13 01:10:00'),
(2, 'guest', 'EN_(Set-B)_(07.02.13)', 11, 40, 10.25, '2013-02-23 01:56:58'),
(3, 'guest', 'P1_(Set-D)_(10.02.13)', 2, 100, 36.25, '2013-02-23 03:42:57'),
(4, 'guest', 'P1_(Set-B)_(09.02.13)', 5, 40, 5, '2013-02-23 03:46:59'),
(5, 'guest', 'EN_(Set-A)_(07.02.13)', 1, 40, 0.25, '2013-02-23 04:46:59'),
(6, 'guest', 'EN_(Set-A)_(07.02.13)', 6, 40, 5.5, '2013-02-23 04:59:59'),
(7, 'guest', 'P1_(Set-D)_(10.02.13)', 10, 100, 9.25, '2013-02-24 08:57:17'),
(8, 'guest', 'P1_(Set-B)_(09.02.13)', 5, 40, 5, '2013-02-24 01:23:50'),
(9, 'guest', 'EN_(Set-D)_(07.02.13)', 0, 40, -0.5, '2013-02-25 12:45:33'),
(10, 'guest', 'EN_(Set-D)_(07.02.13)', 2, 40, 1.5, '2013-02-25 01:45:38'),
(11, 'guest', 'P1_(Set-B)_(09.02.13)', 2, 40, 2, '2013-02-25 04:06:28'),
(12, 'guest', 'EN_(Set-C)_(07.02.13)', 5, 40, 4.5, '2013-02-25 04:42:27'),
(13, 'guest', 'P1_(Set-C)_(10.02.13)', 6, 40, 6, '2013-02-25 05:00:57');

/*!40101 SET CHARACTER_SET_CLIENT=@OLD_CHARACTER_SET_CLIENT */;
/*!40101 SET CHARACTER_SET_RESULTS=@OLD_CHARACTER_SET_RESULTS */;
/*!40101 SET COLLATION_CONNECTION=@OLD_COLLATION_CONNECTION */;

Answer 1

这正是你所需要的:

SELECT * 
FROM results
ORDER BY SUBSTRING( qid
FROM 1 
FOR 1 ) ASC , marks DESC