Онлайн по отпред

niksdevil · 2 Август 2011

Искам 1-во да излизат потребителите който са в таблица : active_users, и след това да излизат по на зад другите потребители където са извън тази таблица ето кода ми :

Код:

<?php 
$page=$_GET['page'];
$limit = 20;
$limit1 = $page * $limit;
$q = mysql_query("SELECT * FROM `users` WHERE `sex`='m' AND `pic`!='i/noavatar.gif ' LIMIT $limit1, $limit");
$i=1;
while($row=mysql_fetch_array($q)) {
?>
<td>
<ul class="profiles_list"> 
<li> 
<div class="thumb_header"> 
<div> 
<a href="<?php echo $row['username']; ?>"><img src="http://saita/online-img/msg_small.gif" alt="Прати писмо на този абонат" title="Изпрати съобщение на <?php echo $row['username']; ?>" border="0" /></a>
</div> 
<div> 

<img src="http://saita/online-img/online_small.gif" alt="<?php echo $row['username']; ?> е налиния" border="0" /> 
 
</div> 
<div> 
<a href="http://saita/<?php echo $row['username']; ?>" target="_blank" title="Отвори профила на <?php echo $row['username']; ?> в нов прозорец"><img src="http://osaita/online-img/blank_small.gif" alt="Отвори профила на <?php echo $row['username']; ?> в нов прозорец" border="0" /></a> 
</div> 
</div> 
<div> <a href="http://saita/<?php echo $row['username']; ?>" title="<?php echo $row['username']; ?>"><img src="<?php echo $row['pic'];?>" class="profileImg" alt="<?php echo $row['username']; ?>" onmouseover="this.style.borderColor='#80bad5'" onmouseout="this.style.borderColor=''" /></a> 
</div> 
<div class="thumb_footer"><a href="http://saita/<?php echo $row['username']; ?>" title="<?php echo $row['username']; ?>"><?php echo $row['username']; ?></a></div> 
</li>
 
</center>
</td>
<?php	
if($i==5 or $i==10 or $i==15) {echo "</tr><tr>"; }
$i=$i+1;
}
?>

Ev0luti0N · 2 Август 2011

Как проверяваш дали потребителят е онлайн или не е? :idea:

niksdevil · 2 Август 2011

ой боже

Ето как вадя потребителите

Код:

<?php 
$i = 1;
$page=$_GET['page'];
$limit = 20;
$limit1 = $page * $limit;
$q = "SELECT `".TBL_ACTIVE_USERS."`.`username`, `users`.`pic`, `users`.`sex` FROM ".TBL_ACTIVE_USERS." INNER JOIN `users` ON `".TBL_ACTIVE_USERS."`.`username`=`users`.`username` ORDER BY `".TBL_ACTIVE_USERS."`.`timestamp` DESC,`".TBL_ACTIVE_USERS."`.`username` LIMIT $limit1, $limit";
$result = $database->query($q);
$num_rows = mysql_numrows($result);
if(!$result || ($num_rows < 0)){
   echo "Error displaying info";
}
else if($num_rows > 0){
   for($i=0; $i<$num_rows; $i++){
      $uname = mysql_result($result,$i,"username");
      $pic = mysql_result($result,$i,"pic");
      $sex = mysql_result($result,$i,"sex");
      if($sex=='m') {
	  ?>
<td><a class="thumbs_img_cnt"href="<?php echo $uname; ?>"><img src="<?php echo $pic;?>" width="120" height="160" onmouseover="this.style.borderColor='#blue'" onmouseout="this.style.borderColor='#dcdcdc'"></a><center><font color="#13B1E1" size="2"><?php echo $uname; ?></font></center></td>
<?php } else { ?>
<td><a class="thumbs_img_cnt" href="<?php echo $uname; ?>"><img src="<?php echo $pic;?>" width="120" height="160" onmouseover="this.style.borderColor='#red'" onmouseout="this.style.borderColor='#dcdcdc'"></a><center><font color="#E22BAB" size="2"><?php echo $uname; ?></font></center></td>
<?php } if($i==4 or $i==9 or $i==14) { echo "</tr><tr>"; }
   }
}
?>

Ev0luti0N · 2 Август 2011

PHP:

<?php 
$page=$_GET['page'];
$limit = 20;
$limit1 = $page * $limit;
$q = mysql_query("SELECT `users`.* FROM `users` INNER JOIN ".TBL_ACTIVE_USERS." ON `".TBL_ACTIVE_USERS."`.`username`=`users`.`username` WHERE `users`.`sex`='m' AND `users`.`pic`!='i/noavatar.gif ' ORDER BY `".TBL_ACTIVE_USERS."`.`timestamp` DESC,`".TBL_ACTIVE_USERS."`.`username` LIMIT $limit1, $limit");
$i=1;
while($row=mysql_fetch_array($q)) {
?>
<td>
<ul class="profiles_list"> 
<li> 
<div class="thumb_header"> 
<div> 
<a href="<?php echo $row['username']; ?>"><img src="http://saita/online-img/msg_small.gif" alt="Прати писмо на този абонат" title="Изпрати съобщение на <?php echo $row['username']; ?>" border="0" /></a>
</div> 
<div> 

<img src="http://saita/online-img/online_small.gif" alt="<?php echo $row['username']; ?> е налиния" border="0" /> 
 
</div> 
<div> 
<a href="http://saita/<?php echo $row['username']; ?>" target="_blank" title="Отвори профила на <?php echo $row['username']; ?> в нов прозорец"><img src="http://osaita/online-img/blank_small.gif" alt="Отвори профила на <?php echo $row['username']; ?> в нов прозорец" border="0" /></a> 
</div> 
</div> 
<div> <a href="http://saita/<?php echo $row['username']; ?>" title="<?php echo $row['username']; ?>"><img src="<?php echo $row['pic'];?>" class="profileImg" alt="<?php echo $row['username']; ?>" onmouseover="this.style.borderColor='#80bad5'" onmouseout="this.style.borderColor=''" /></a> 
</div> 
<div class="thumb_footer"><a href="http://saita/<?php echo $row['username']; ?>" title="<?php echo $row['username']; ?>"><?php echo $row['username']; ?></a></div> 
</li>
 
</center>
</td>
<?php   
if($i==5 or $i==10 or $i==15) {echo "</tr><tr>"; }
$i=$i+1;
}
?>

Така може би? :idea:

niksdevil · 2 Август 2011

Тъка става а картинката да се сменя за другите потребители ?

<img src="http://saita/online-img/online_small.gif" alt="<?php echo $row['username']; ?> е налиния" border="0" />

на

<img src="http://saita/online-img/offline_small.gif" alt="<?php echo $row['username']; ?> е налиния" border="0" />

Ev0luti0N · 2 Август 2011

PHP:

<?php 
$page=$_GET['page'];
$limit = 20;
$limit1 = $page * $limit;
$q = mysql_query("SELECT `users`.*, IF (".TBL_ACTIVE_USERS.".`username`=`users`.`username`, 1, 0) AS online FROM `users` INNER JOIN ".TBL_ACTIVE_USERS." ON `".TBL_ACTIVE_USERS."`.`username`=`users`.`username` WHERE `users`.`sex`='m' AND `users`.`pic`!='i/noavatar.gif ' ORDER BY `".TBL_ACTIVE_USERS."`.`timestamp` DESC,`".TBL_ACTIVE_USERS."`.`username` LIMIT $limit1, $limit");
$i=1;
while($row=mysql_fetch_array($q)) {
?>
<td>
<ul class="profiles_list"> 
<li> 
<div class="thumb_header"> 
<div> 
<a href="<?php echo $row['username']; ?>"><img src="http://saita/online-img/msg_small.gif" alt="Прати писмо на този абонат" title="Изпрати съобщение на <?php echo $row['username']; ?>" border="0" /></a>
</div> 
<div> 

<img src="http://saita/online-img/<?php ($row['online'] == 1 ? 'online_small' : 'offline_small'); ?>.gif" alt="<?php echo $row['username']; ?> е налиния" border="0" /> 
 
</div> 
<div> 
<a href="http://saita/<?php echo $row['username']; ?>" target="_blank" title="Отвори профила на <?php echo $row['username']; ?> в нов прозорец"><img src="http://osaita/online-img/blank_small.gif" alt="Отвори профила на <?php echo $row['username']; ?> в нов прозорец" border="0" /></a> 
</div> 
</div> 
<div> <a href="http://saita/<?php echo $row['username']; ?>" title="<?php echo $row['username']; ?>"><img src="<?php echo $row['pic'];?>" class="profileImg" alt="<?php echo $row['username']; ?>" onmouseover="this.style.borderColor='#80bad5'" onmouseout="this.style.borderColor=''" /></a> 
</div> 
<div class="thumb_footer"><a href="http://saita/<?php echo $row['username']; ?>" title="<?php echo $row['username']; ?>"><?php echo $row['username']; ?></a></div> 
</li>
 
</center>
</td>
<?php   
if($i==5 or $i==10 or $i==15) {echo "</tr><tr>"; }
$i=$i+1;
}
?>

Така може би? :idea:

niksdevil · 2 Август 2011

Тъка не вади нито една от 2-те картинки

Просто ако е в таблицата active_users да слага онлайн ако е извън таблицата оффлайн

Ev0luti0N · 2 Август 2011

niksdevil каза:
Тъка не вади нито една от 2-те картинки

Просто ако е в таблицата active_users да слага онлайн ако е извън таблицата оффлайн

Да, на теория е лесно.. Ама на практика не мога да го измисля в момента..

PHP:

<?php 
$page=$_GET['page'];
$limit = 20;
$limit1 = $page * $limit;
$q = mysql_query("SELECT `users`.*, ".TBL_ACTIVE_USERS.".`username` AS user FROM `users` INNER JOIN ".TBL_ACTIVE_USERS." ON `".TBL_ACTIVE_USERS."`.`username`=`users`.`username` WHERE `users`.`sex`='m' AND `users`.`pic`!='i/noavatar.gif ' ORDER BY `".TBL_ACTIVE_USERS."`.`timestamp` DESC,`".TBL_ACTIVE_USERS."`.`username` LIMIT $limit1, $limit");
$i=1;
while($row=mysql_fetch_array($q)) {
?>
<td>
<ul class="profiles_list"> 
<li> 
<div class="thumb_header"> 
<div> 
<a href="<?php echo $row['username']; ?>"><img src="http://saita/online-img/msg_small.gif" alt="Прати писмо на този абонат" title="Изпрати съобщение на <?php echo $row['username']; ?>" border="0" /></a>
</div> 
<div> 

<img src="http://saita/online-img/<?php ($row['username'] == $row['user'] ? 'online_small' : 'offline_small'); ?>.gif" alt="<?php echo $row['username']; ?> е налиния" border="0" /> 
 
</div> 
<div> 
<a href="http://saita/<?php echo $row['username']; ?>" target="_blank" title="Отвори профила на <?php echo $row['username']; ?> в нов прозорец"><img src="http://osaita/online-img/blank_small.gif" alt="Отвори профила на <?php echo $row['username']; ?> в нов прозорец" border="0" /></a> 
</div> 
</div> 
<div> <a href="http://saita/<?php echo $row['username']; ?>" title="<?php echo $row['username']; ?>"><img src="<?php echo $row['pic'];?>" class="profileImg" alt="<?php echo $row['username']; ?>" onmouseover="this.style.borderColor='#80bad5'" onmouseout="this.style.borderColor=''" /></a> 
</div> 
<div class="thumb_footer"><a href="http://saita/<?php echo $row['username']; ?>" title="<?php echo $row['username']; ?>"><?php echo $row['username']; ?></a></div> 
</li>
 
</center>
</td>
<?php   
if($i==5 or $i==10 or $i==15) {echo "</tr><tr>"; }
$i=$i+1;
}
?>

Така? :idea:

niksdevil · 2 Август 2011

Пак

Ev0luti0N · 2 Август 2011

PHP:

<?php 
$page=$_GET['page'];
$limit = 20;
$limit1 = $page * $limit;
$q = "SELECT `".TBL_ACTIVE_USERS."`.`username`, `users`.username as uName, `users`.`pic`, `users`.`sex` ".TBL_ACTIVE_USERS.", (SELECT COUNT(*) FROM `".TBL_ACTIVE_USERS."` WHERE username=uName) as is_active  FROM  `".TBL_ACTIVE_USERS."` WHERE INNER JOIN `users` ON `".TBL_ACTIVE_USERS."`.`username`=`users`.`username`  ORDER BY is_active DESC LIMIT $limit1, $limit";
$q = mysql_query($q) or die (mysql_error());
$i=1;
while($row=mysql_fetch_array($q)) {
?>
<td>
<ul class="profiles_list"> 
<li> 
<div class="thumb_header"> 
<div> 
<a href="<?php echo $row['username']; ?>"><img src="http://saita/online-img/msg_small.gif" alt="Прати писмо на този абонат" title="Изпрати съобщение на <?php echo $row['username']; ?>" border="0" /></a>
</div> 
<div> 

<img src="http://saita/online-img/<?php ($row['is_active'] == 1 ? 'online_small' : 'offline_small'); ?>.gif" alt="<?php echo $row['username']; ?> е налиния" border="0" /> 
 
</div> 
<div> 
<a href="http://saita/<?php echo $row['username']; ?>" target="_blank" title="Отвори профила на <?php echo $row['username']; ?> в нов прозорец"><img src="http://osaita/online-img/blank_small.gif" alt="Отвори профила на <?php echo $row['username']; ?> в нов прозорец" border="0" /></a> 
</div> 
</div> 
<div> <a href="http://saita/<?php echo $row['username']; ?>" title="<?php echo $row['username']; ?>"><img src="<?php echo $row['pic'];?>" class="profileImg" alt="<?php echo $row['username']; ?>" onmouseover="this.style.borderColor='#80bad5'" onmouseout="this.style.borderColor=''" /></a> 
</div> 
<div class="thumb_footer"><a href="http://saita/<?php echo $row['username']; ?>" title="<?php echo $row['username']; ?>"><?php echo $row['username']; ?></a></div> 
</li>
 
</center>
</td>
<?php   
if($i==5 or $i==10 or $i==15) {echo "</tr><tr>"; }
$i=$i+1;
}
?>

niksdevil · 2 Август 2011

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'INNER JOIN `users` ON `active_users`.`username`=`users`.`username` ORDER BY is_' at line 1

Ev0luti0N · 2 Август 2011

PHP:

<?php 
$page=$_GET['page'];
$limit = 20;
$limit1 = $page * $limit;
$q = "SELECT `".TBL_ACTIVE_USERS."`.`username`, `users`.`username` as uName, `users`.`pic`, `users`.`sex`, (SELECT COUNT(*) FROM `".TBL_ACTIVE_USERS."` WHERE username=uName) as is_active  FROM  `".TBL_ACTIVE_USERS."` WHERE INNER JOIN `users` ON `".TBL_ACTIVE_USERS."`.`username`=`users`.`username`  ORDER BY is_active DESC LIMIT $limit1, $limit";
$q = mysql_query($q) or die (mysql_error());
$i=1;
while($row=mysql_fetch_array($q)) {
?>
<td>
<ul class="profiles_list"> 
<li> 
<div class="thumb_header"> 
<div> 
<a href="<?php echo $row['username']; ?>"><img src="http://saita/online-img/msg_small.gif" alt="Прати писмо на този абонат" title="Изпрати съобщение на <?php echo $row['username']; ?>" border="0" /></a>
</div> 
<div> 

<img src="http://saita/online-img/<?php ($row['is_active'] == 1 ? 'online_small' : 'offline_small'); ?>.gif" alt="<?php echo $row['username']; ?> е налиния" border="0" /> 
 
</div> 
<div> 
<a href="http://saita/<?php echo $row['username']; ?>" target="_blank" title="Отвори профила на <?php echo $row['username']; ?> в нов прозорец"><img src="http://osaita/online-img/blank_small.gif" alt="Отвори профила на <?php echo $row['username']; ?> в нов прозорец" border="0" /></a> 
</div> 
</div> 
<div> <a href="http://saita/<?php echo $row['username']; ?>" title="<?php echo $row['username']; ?>"><img src="<?php echo $row['pic'];?>" class="profileImg" alt="<?php echo $row['username']; ?>" onmouseover="this.style.borderColor='#80bad5'" onmouseout="this.style.borderColor=''" /></a> 
</div> 
<div class="thumb_footer"><a href="http://saita/<?php echo $row['username']; ?>" title="<?php echo $row['username']; ?>"><?php echo $row['username']; ?></a></div> 
</li>
 
</center>
</td>
<?php   
if($i==5 or $i==10 or $i==15) {echo "</tr><tr>"; }
$i=$i+1;
}
?>

niksdevil · 2 Август 2011

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'INNER JOIN `users` ON `active_users`.`username`=`users`.`username` ORDER BY is_' at line 1

пак това вади

Ev0luti0N · 2 Август 2011

PHP:

<?php 
$page=$_GET['page'];
$limit = 20;
$limit1 = $page * $limit;
$q = "SELECT `".TBL_ACTIVE_USERS."`.`username`, `users`.`username` as uName, `users`.`pic`, `users`.`sex`, (SELECT COUNT(*) FROM `".TBL_ACTIVE_USERS."` WHERE username=uName) as is_active  FROM  `".TBL_ACTIVE_USERS."` INNER JOIN `users` ON `".TBL_ACTIVE_USERS."`.`username`=`users`.`username` ORDER BY is_active DESC LIMIT $limit1, $limit";
$q = mysql_query($q) or die (mysql_error());
$i=1;
while($row=mysql_fetch_array($q)) {
?>
<td>
<ul class="profiles_list"> 
<li> 
<div class="thumb_header"> 
<div> 
<a href="<?php echo $row['username']; ?>"><img src="http://saita/online-img/msg_small.gif" alt="Прати писмо на този абонат" title="Изпрати съобщение на <?php echo $row['username']; ?>" border="0" /></a>
</div> 
<div> 

<img src="http://saita/online-img/<?php ($row['is_active'] == 1 ? 'online_small' : 'offline_small'); ?>.gif" alt="<?php echo $row['username']; ?> е налиния" border="0" /> 
 
</div> 
<div> 
<a href="http://saita/<?php echo $row['username']; ?>" target="_blank" title="Отвори профила на <?php echo $row['username']; ?> в нов прозорец"><img src="http://osaita/online-img/blank_small.gif" alt="Отвори профила на <?php echo $row['username']; ?> в нов прозорец" border="0" /></a> 
</div> 
</div> 
<div> <a href="http://saita/<?php echo $row['username']; ?>" title="<?php echo $row['username']; ?>"><img src="<?php echo $row['pic'];?>" class="profileImg" alt="<?php echo $row['username']; ?>" onmouseover="this.style.borderColor='#80bad5'" onmouseout="this.style.borderColor=''" /></a> 
</div> 
<div class="thumb_footer"><a href="http://saita/<?php echo $row['username']; ?>" title="<?php echo $row['username']; ?>"><?php echo $row['username']; ?></a></div> 
</li>
 
</center>
</td>
<?php   
if($i==5 or $i==10 or $i==15) {echo "</tr><tr>"; }
$i=$i+1;
}
?>

niksdevil · 2 Август 2011

пак нито една от 2-те :?

Онлайн по отпред

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