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Re:Display Online User 5 Months, 1 Week ago
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Does anyone know how I can translate this forum into my language? I am having trouble readin it.
The B&B in Stratford-Upon-Avon
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Re:Display Online User 5 Months, 1 Week ago
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Excelelnt information. I have stumbled this post for other to find.
Anja from B&B in Stratford-Upon-Avon
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Re:Display Online User 1 Month, 4 Weeks ago
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Hi all!
I have solved the problem with a small change in file mod_cb_superthumb.php
Change this code:
case 9 :
$sql = "select u.id, u.username, u.name, c.avatar FROM #__users u, #__comprofiler c, #__comprofiler_members a, #__session AS s WHERE u.id = c.user_id $imagenot $avatarplease and c.user_id = a.memberid and u.username = s.username and c.approved = 1 and u.block = 0 and c.confirmed = 1 and c.banned = 0 group by u.id limit " . $count;
break;
In this:
case 9 :
$sql = "select u.id, u.username, u.name, c.avatar FROM #__users u, #__comprofiler c, #__session AS s WHERE u.id = c.user_id $imagenot $avatarplease and u.username = s.username and c.approved = 1 and u.block = 0 and c.confirmed = 1 and c.banned = 0 group by u.id limit " . $count;
break;
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Re:Display Online User 1 Month, 4 Weeks ago
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Karma: 63
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V3n0m79 wrote:
QUOTE: Hi all!
I have solved the problem with a small change in file mod_cb_superthumb.php
Change this code:
case 9 :
$sql = "select u.id, u.username, u.name, c.avatar FROM #__users u, #__comprofiler c, #__comprofiler_members a, #__session AS s WHERE u.id = c.user_id $imagenot $avatarplease and c.user_id = a.memberid and u.username = s.username and c.approved = 1 and u.block = 0 and c.confirmed = 1 and c.banned = 0 group by u.id limit " . $count;
break;
In this:
case 9 :
$sql = "select u.id, u.username, u.name, c.avatar FROM #__users u, #__comprofiler c, #__session AS s WHERE u.id = c.user_id $imagenot $avatarplease and u.username = s.username and c.approved = 1 and u.block = 0 and c.confirmed = 1 and c.banned = 0 group by u.id limit " . $count;
break;
thanks for your sharing
i think that it'll be help another user
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