Саке, картинка вниз уходит
Добавлено 15.07.15 в 12:27:19:
echo "<div class='nav2'>";
echo "<img src='/style/icons/foto.png' alt='*' /> ";
echo "<a href='/foto/$ank[id]/'>Фотографии</a> ";
echo "(" . mysql_result(mysql_query("SELECT COUNT(*) FROM `gallery_foto` WHERE `id_user` = '$ank[id]'"
,0) . "<br />";
вот так не могу.
echo '<div class="nav2"><a href="/foto/'.$ank['id'].'"><img src="/style/icons/foto.jpg"> Фотографии ('.mysql_num_rows(mysql_query('SELECT `id` FROM `gallery_foto` WHERE `id_user` = "'.$ank['id'].'"')).')</a></div>';
<?php
echo "<a class='nav2' style='display: block;' href='/foto/$ank[id]/'> <img src='/style/icons/foto.png' alt='*' /> Фотографии (".mysql_result(mysql_query("SELECT COUNT(*) FROM `gallery_foto` WHERE `id_user` = '$ank[id]'",0)." </a>";
?>
Netscape, можешь дать код без тега php не могу скопировать
echo "<a class='nav2' style='display: block;' href='/foto/$ank[id]/'> <img src='/style/icons/foto.png' alt='*' /> Фотографии (".mysql_result(mysql_query("SELECT COUNT(*) FROM `gallery_foto` WHERE `id_user` = '$ank[id]'",0)." </a>";
Netscape, а тут попа
$k_f = mysql_result(mysql_query("SELECT COUNT(id) FROM `frends_new` WHERE `to` = '$ank[id]' LIMIT 1", 0);
$k_fr = mysql_result(mysql_query("SELECT COUNT(*) FROM `frends` WHERE `user` = '$ank[id]' AND `i` = '1'", 0);
$res = mysql_query("select `frend` from `frends` WHERE `user` = '$ank[id]' AND `i` = '1'";
echo '<div class="nav2">';
echo '<img src="/style/icons/druzya.png" alt="*" /> ';
echo '<a href="/user/frends/?id=' . $ank['id'] . '">Друзья</a> (' . $k_fr . '</b>/';
$i = 0;
while ($k_fr = mysql_fetch_array($res))
{
if (mysql_result(mysql_query("SELECT COUNT(*) FROM `user` WHERE `id` = '$k_fr[frend]' && `date_last` > '".(time()-600)."'",0) != 0)
$i++;
}
echo "<span style='color:green'><a href='/user/frends/online.php?id=".$ank['id']."'>$i</a></span>";
if ($k_f>0 && $ank['id'] == $user['id'])echo " <a href='/user/frends/new.php'><font color='red'>+$k_f</font></a>";
echo "</div>";
