这个问题在这里已有答案:
试图在我的测试网站上做一个赞美区域
$result = mysql_query("SELECT message, username FROM compliments order by date DESC");
while($r = mysql_fetch_array($result))
{
$Name = $r["username"];
$Message = $r["message"];
}
但是,获取所有消息,
我想限制到最近的10
所以我试过这个..
$result = mysql_query("SELECT TOP(10) message, username FROM compliments order by date ASC");
while($r = mysql_fetch_array($result))
{
$Name = $r["username"];
$Message = $r["message"];
}
抛出错误
警告:mysql_fetch_array()要求参数1为资源,在第52行的/var/www/vhosts/localhost/httpdocs/readcompliments.php中给出布尔值
我怎么能成功地做到这一点?
编辑:
现在用
SELECT message, username FROM compliments order by date ASC LIMIT 10
它的工作原理!但问题是,它是最老的10!我该怎么做才能获得最新的10个?
如果你使用Mysql
qazxsw poi用于sql Server,qazxsw poi用于Mysql
TOP
代替
LIMIT
编辑
如果你想获得最近的10,你可以使用SELECT message, username FROM compliments order by date ASC LIMIT 10
SELECT TOP(10) message, username FROM compliments order by date ASC
将LIMIT添加到您的查询中:
order by date DESC
不要忘记MySQL已弃用,请将脚本更改为MySQLi
警告MySQL在PHP 5.5.0中已弃用,并且在PHP 7.0.0中已被删除。相反,应使用
SELECT message, username FROM compliments order by date DESC LIMIT 10
或"SELECT message, username FROM compliments order by date ASC LIMIT 10;"
扩展。
MySQLi