我正在努力准备一个明智的优点列表。我正在使用此mysql query
:
SELECT *
FROM results
ORDER BY qid ASC,
marks DESC
结果是:
但我需要的是这样的(看看标记列,我需要获得qid
行,按marks
排序):
请有人帮助我。
更新
这是result.sql
文件,用于在您的电脑中创建表格。
-- 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 */;
答案 0 :(得分:16)
这正是您所需要的:
SELECT *
FROM results
ORDER BY SUBSTRING( qid
FROM 1
FOR 1 ) ASC , marks DESC
答案 1 :(得分:1)
我真的不明白你的问题,但查看结果,我注意到记录按qid
排序,其中EN
是结果列表中的第一个。
尝试一下,
SELECT *
FROM tableName
ORDER BY CASE WHEN qid LIKE 'EN%' THEN 0
WHEN qid LIKE 'P1%' THEN 1
ELSE 2
END ASC,
marks DESC
答案 2 :(得分:0)
这可能会对您有所帮助:
SELECT *
FROM results
ORDER BY SUBSTRING(qid FROM 1 FOR 9) ASC,
marks DESC