我正试图让你看到朋友们的最新活动，但一次只能显示5个。在这5个人中，如果你的时间比其他任何朋友都多，你可以拥有一个人不止一次。就像上面说的"朋友升级了skillName日期"，然后显示最近的五个。我的代码是这样的：
function recentActivityF($user){
include "mysql.ws";
$query = mysql_query("SELECT u.* FROM friends uf inner join users u on uf.friend = u.username WHERE uf.user = '$user' ORDER BY u.username") or die(mysql_error());
while($row = mysql_fetch_array($query)){
$friends = $row['username'];
//$query2 = mysql_query("SELECT * FROM recentactivity WHERE user LIKE '".$friends."' ORDER BY time DESC LIMIT 1") or die(mysql_error());
$query2 = mysql_query("SELECT u.* FROM recentactivity uf inner join friends u on uf.user = u.friend WHERE uf.user LIKE '".$friends."' ORDER BY uf.time DESC LIMIT 1") or die(mysql_error());
$query3 = mysql_query("SELECT * FROM skills WHERE playerName LIKE '".$friends."' LIMIT 1") or die(mysql_error());
while($row3 = mysql_fetch_array($query3)){
while($row2 = mysql_fetch_array($query2)){
$date = $row2['time'];
$skillRow = $row2['skill']+1;
$skill = findType($skillRow)."xp";
$skillName = findType($skillRow);
echo'Friend: '.$friends.' Levelled Up '.$skillName.' Date: '.$date.'';
}
}
}
}
结果应该是：
1.)Friend: mark Levelled Up Attack date: 28-Aug-2012 09:48
2.)Friend: matt Levelled Up Strength date: 28-Aug-2012 09:47
3.)Friend: kevin Levelled Up Attack date: 28-Aug-2012 09:46
4.)Friend: mark Levelled Up Strength date: 28-Aug-2012 09:45
5.)Friend: april Levelled Up Magic date: 28-Aug-2012 09:44
我的结果是：
1.)Friend: mark Levelled Up Attack date: 28-Aug-2012 09:48
2.)Friend: mark Levelled Up Strength date: 28-Aug-2012 09:45
3.)Friend: matt Levelled Up Strength date: 28-Aug-2012 09:47
4.)Friend: kevin Levelled Up Attack date: 28-Aug-2012 09:46
5.)Friend: april Levelled Up Magic date: 28-Aug-2012 09:44
6.)Friend: april Levelled Up Strength date: 28-Aug-2012 09:45
SQL:
CREATE TABLE `friends` (
`user` varchar(32) NOT NULL,
`friend` varchar(32) NOT NULL
) ENGINE=MyISAM DEFAULT CHARSET=latin1;
INSERT INTO `friends` (`user`, `friend`) VALUES
('Cls Prod', 'pro skiller'),
('Cls Prod', 'polo303'),
('Cls Prod', 'pjtips123'),
('Cls Prod', 'tommy0581'),
('Cls Prod', 'sageofmali');
CREATE TABLE `users` (
`username` varchar(255) NOT NULL DEFAULT '',
`password` varchar(255) DEFAULT NULL,
`rights` varchar(255) DEFAULT NULL,
`online` tinyint(1) NOT NULL,
`friends` text,
`donator` double DEFAULT NULL,
`time` bigint(20) NOT NULL,
PRIMARY KEY (`username`)
) ENGINE=MyISAM DEFAULT CHARSET=latin1;
INSERT INTO `users` (`username`, `password`, `rights`, `online`, `friends`, `donator`, `time`) VALUES
('Cls Prod', 'password', '0', 0, NULL, NULL, 0);
CREATE TABLE `recentactivity` (
`user` varchar(255) NOT NULL,
`skill` int(11) NOT NULL,
`time` varchar(255) NOT NULL,
`killed` varchar(255) NOT NULL
) ENGINE=MyISAM DEFAULT CHARSET=latin1;
INSERT INTO `recentactivity` (`user`, `skill`, `time`, `killed`) VALUES
('Cls Prod', 2, '27-Aug-2012 16:06:40', '');
因此，评论的$query2显示每个朋友的1个，但会显示5个以上，我想将其限制为只显示5个最近的朋友活动。有什么帮助吗？
我注意到你自己正在进行技能到技能的名称翻译，所以这应该足够了：
SELECT f.user AS username, a.time, a.skill
FROM
friends f
JOIN recentactivity a ON(a.user = f.user)
WHERE
f.friend = $username
GROUP BY a.user -- only one event per friend
ORDER BY a.time DESC
LIMIT 5
请在此处查看演示：http://sqlfiddle.com/#！2/66f53/6
编辑：这就是你需要做的全部：
function recentActivityF($user){
include "mysql.ws";
$query = <<<EOD
SELECT f.user AS username, a.time, a.skill
FROM
friends f
JOIN recentactivity a ON(a.user = f.user)
WHERE
f.friend = $user
GROUP BY a.user -- only one event per friend
ORDER BY a.time DESC
LIMIT 5
EOD;
mysql_query($query);
while($row = mysql_fetch_array($query)) {
$friend_name = $row['username'];
$date = $row['time'];
$skillRow = $row['skill']+1;
$skillName = findType($skillRow);
echo'Friend: '.$friend_name.' Levelled Up '.$skillName.' Date: '.$date.'';
}
p.S.标准堆栈溢出免责声明：mysql扩展已弃用，请使用PDO或mysqli。
使用sql限制
基本用法是：
SELECT * FROM table LIMIT 5
如果您想返回记录6-10:
SELECT * FROM table LIMIT 5,5
请给我一点时间浏览一下您的代码，以给出一个好的示例。在你的代码中，你需要在哪里使用它并不明显
仔细查看代码后，似乎不需要第一个查询。你应该能够做到：
SELECT u.*
FROM recentactivity uf
INNER JOIN friends u ON uf.user = u.friend
WHERE u.user = $user
ORDER BY uf.time DESC
LIMIT 5
您可能需要选择更多的字段才能获得所需的所有信息，但这应该可以获得给定用户的recentactivity上的所有记录，按最新记录排序，并仅限于前5名。