Question

好的，我有两个MYSQL表：

CREATE TABLE `sessions` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `userID` int(11) NOT NULL,
  `runTime` int(11) NOT NULL,
  `monstersKilled` int(11) NOT NULL,
  `profit` int(11) NOT NULL,
  `tasks` int(11) NOT NULL,
  `xpGain` int(11) NOT NULL,
  `lastupdate` int(11) NOT NULL,
  PRIMARY KEY (`id`)
);

和

CREATE TABLE  `users` (
  `userid` int(11) NOT NULL AUTO_INCREMENT,
  `user` text NOT NULL,
  `pass` text NOT NULL,
  `paypal` text NOT NULL,
  `active` tinyint(1) NOT NULL DEFAULT '0',
  `strikes` int(11) NOT NULL DEFAULT '0',
  `session` text NOT NULL,
  `brand` text NOT NULL,
  `identifier` text NOT NULL,
  PRIMARY KEY (`userid`)
) ENGINE=MyISAM AUTO_INCREMENT=651 DEFAULT CHARSET=latin1;

如您所见，会话具有指向用户ID的链接。现在我想制作一个总分高的PHP文件，但凭借我对PHP的一点知识和对MYSQL的了解不多，我不知道如何开始。总得分将按总运行时间排序。

实施例：在用户中，我有用户＃1（cody）＆amp;用户＃2（乔）。现在，在会议中，我有3个会议：

id, userID, runTime, monstersKilled, profit, tasks, xpGain, lastupdate
12, 1, 27, 14, 6200, 0, 5050, 1282325410
19, 1, 18, 1, 277, 1, 168, 1278897756
1968, 2, 195, 433, 111345, 4, 73606, 1280993244

打印输出应该是：

Place, username, Total Run Time, Total Monsters Killed, Total profit, Total tasks, Total Exp Gain
1. Joe, 195, 433,11345,4,73606
2. Cody, 55, 15, 1, 5218

Answer 1

使用：

  SELECT u.user,
         SUM(s.runtime) AS total_run_time,
         SUM(s.monsterskilled) AS total_monsters_killed,
         SUM(s.profit) AS total_profit,
         SUM(s.tasks) AS total_tasks, 
         SUM(s.xpgain) AS total_xp_gained
    FROM USERS u
    JOIN SESSION s ON s.userid = u.userid
GROUP BY u.user
ORDER BY total_run_time DESC

Answer 2

此脚本应该为您输出整齐的表中的高分。感谢OMG Ponies的SQL查询。

<?php
$dbhost = 'localhost';  // Database Host
$dbuser = 'user';       // Database User
$dbpass = 'password';   // Database Password
$dbname = 'database';   // Database Name

$db = mysql_connect($dbhost, $dbuser, $dbpass);
mysql_select_db($dbname, $db);

$limit = 10; // The number of users to show in the top highscores.  Set to 0 to show all

$sql = 'SELECT u.user,
         SUM(s.runtime) AS total_run_time,
         SUM(s.monsterskilled) AS total_monsters_killed,
         SUM(s.profit) AS total_profit,
         SUM(s.tasks) AS total_tasks, 
         SUM(s.xpgain) AS total_xp_gained
        FROM USERS u
        JOIN SESSION s ON s.userid = u.userid
        GROUP BY u.user
        ORDER BY total_run_time DESC';

if ($limit > 0) {
    $sql .= ' LIMIT '.$limit;
}

$get_scores = mysql_query($sql);
$scores = array();

$rank = 0;
while($score = mysql_fetch_array($get_scores)) {
    ++$rank;
    $scores[] = '<td>'.$rank.'</td><td>'.$score['user'].'</td><td>'.$score['total_run_time'].'</td><td>'.$score['total_monsters_killed'].'</td><td>'.$score['total_profit'].'</td><td>'.$score['total_tasks'].'</td><td>'.$score['total_xp_gained'].'</td>';
}

echo '<table><tbody>';
echo '<tr><td>Place</td><td>Username</td><td>Total Run Time</td><td>Total Monsters Killed, Total profit</td><td>Total tasks</td><td>Total Exp Gain</td></tr><tr>';
echo implode('</tr><tr>', $scores);
echo '</tr></tbody></table>';
?>

SQL高分

2 个答案: