Php 如何计算SQL中最常见的值?
目前,我有以下代码:Php 如何计算SQL中最常见的值?,php,mysql,count,mysql-num-rows,Php,Mysql,Count,Mysql Num Rows,目前,我有以下代码: $forummost = mysql_query(" SELECT door, COUNT(door) AS doorCount FROM forum_posts GROUP BY door ORDER BY COUNT(door) DESC "); $forummostc = mysql_num_rows($forummost); $forummostf = mysql_fetch_assoc($forummost); 有人能解释一下为什么my
$forummost = mysql_query("
SELECT door,
COUNT(door) AS doorCount
FROM forum_posts
GROUP BY door
ORDER BY COUNT(door)
DESC
");
$forummostc = mysql_num_rows($forummost);
$forummostf = mysql_fetch_assoc($forummost);
有人能解释一下为什么mysql的行数($forummost)不起作用吗
提前谢谢
mysql\u num\u rows
统计查询返回的行数。您的查询是每个门的计数是多少
,因此每个门只返回一行。在mysql中只选择该列的不同值。如前所述,在您的情况下,它会选择尽可能多的行,只要有不同的door
值。试试这个,我已经测试过了,它是有效的
<?php
$link = mysql_connect('localhost','root','');
if (!$link)
{
die('Could not connect to MySQL: ' . mysql_error());
}
echo 'Connection OK'."</br>";
$forummost = mysql_query("
SELECT door,
COUNT(door) AS doorCount
FROM forum_posts
GROUP BY door
ORDER BY COUNT(door)
DESC
");
$forummostc = mysql_num_rows($forummost);
if ($forummostc == 0) {
echo "No rows found, nothing to print so am exiting";
exit;
}
while ($row = mysql_fetch_assoc($forummost)) {
echo $row["door"];
echo $row["doorCount"];
}
mysql_free_result($forummost);
mysql_close($link);
?>
您得到或输出了什么错误?我得到“227”,但当我在PHPMyAdmin SQL中选择同一个用户时,我得到:“1158”返回数组中的行数是不同门的数目。@Mossawi:您能编辑您的问题并在那里发布预期和实际的输出吗?