Добро пожаловать, Гость!
Как соеденить
Общие вопросы | Как соеденить
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Как соеденить
21 Фев 2014Подскажыте как соеденить список онлайн з гостями штоб получилось одне число
Комментарии:
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Mark, не работает... во мой
<div class="onli">
<center>
<a href="/online.php" title="онлайн" style="color:#cdcecf; text-decoration: none">
<font color="#fee300" size="2"><font color="black">Онлайн </font>
<font color="black">
<?=mysql_result(mysql_query("SELECT COUNT(*) FROM `user` WHERE `date_last` > ".(time()-600).""), 0)?>
</font>
</a>
<font color="black" size="2"> (</font>
<font color="black">+
<?=mysql_result(mysql_query("SELECT COUNT(*) FROM `guests` WHERE `date_last` > ".(time()-600)." AND `pereh` > '0'"), 0)?>
</font>
<font color="black" size="2"> гостей )</font></font>
</center>
</div>
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f4rt, echo '<div class="title"><a href="/online.php">Онлайн: '.(mysql_result(mysql_query("SELECT COUNT(*) FROM `user` WHERE `date_last` > ".(time()-1500).""), 0)+mysql_result(mysql_query("SELECT COUNT(*) FROM `guests` WHERE `date_last` > ".(time()-1500)." AND `pereh` >
'0'"), 0)).'</a></div>';
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$onlin = mysql_result(mysql_query
("SELECT COUNT(*) FROM `user`
WHERE `date_last` > ".(time
()-1500).""), 0);
$guest = mysql_result
(mysql_query("SELECT COUNT(*)
FROM `guests` WHERE `date_last`
> ".(time()-1500)." AND `pereh` >
'0'"), 0);
$online = $onlin + $guest;
echo "<div class='title'>";
echo "<a href='/online.php'>Онлайн: ";
echo $online;
echo "</div>";