mysql错误资源ID＃3？

Question

当我尝试从数据库中回显列计数器的值时，我收到资源ID＃3错误。我想只获得一个值。任何想法我该怎么做？

$Page = $_SERVER['PHP_SELF'];
$num = preg_replace("/[^0-9]/", '', $Page);


$query = "SELECT * FROM hitscounter WHERE page='$num';";
$res = mysql_query($query);
if (mysql_num_rows($res) > 0) {

mysql_query("UPDATE hitscounter SET counter=counter+1 Where page='$num'");
$views = mysql_query("SELECT counter FROM hitscounter WHERE page=555");
    mysql_fetch_array($views, MYSQL_NUM);
    mysql_free_result($views);
    echo $views;
}

Answer 1

您应该使用result从[{1}}返回的mysql_fetch_array，就像这样：

$result = mysql_fetch_array($views, MYSQL_NUM);
print_r($result);

Answer 2

你的if条件应该是这样的

$result = mysql_fetch_array($res);
if (count($result) > 0){
echo 'do something';
}