Puika1 Posted July 25, 2009 Report Posted July 25, 2009 kas man jādara,lai šajā scriptā uzliktu ka lietotājus neizvelk visus,bet kādu noteiktu grupu,piemēram admin? <?php define('WWW','http://kaut kads webs/'); //Tavas web lapas adrese include_once('forums/conf_global.php'); //Celš uz foruma config failu $link = mysql_connect($INFO['sql_host'],$INFO['sql_user'],$INFO['sql_pass']); mysql_query('use '.$INFO['sql_database'].';'); $page = '<div id="userlist">'; $result = mysql_query('select t1.id,t1.members_display_name,t2.avatar_location from '.$INFO['sql_tbl_prefix'].'members t1 join '.$INFO['sql_tbl_prefix'].'member_extra t2 on (t1.id = t2.id);'); if (!empty($result)) { while ($r = mysql_fetch_array($result)) { if ($r[2] == "") { $avatars = "forums/uploads/default.jpg" ; } else { $avatars = "forums/uploads/$r[2]" ; } $page .='<p> <a href="'.WWW.'forums/index.php?showuser='.$r[0].'"> <img class="avatar" src="'. WWW .''. $avatars .'" alt="'.$r[1].'" /> '.$r[1].'</a> </p>'; } } $page .='</div>'; mysql_close($link); echo $page; ?> Quote
Puika1 Posted July 25, 2009 Author Report Posted July 25, 2009 airbus to es aptuveni zināju bet kur jāliek? Quote
airbus Posted July 27, 2009 Report Posted July 27, 2009 Zem si $result = mysql_query('select t1.id,t1.members_display_name,t2.avatar_location from '.$INFO['sql_tbl_prefix'].'members t1 join '.$INFO['sql_tbl_prefix'].'member_extra t2 on (t1.id = t2.id);'); ieliec if($row->mgroup == "GRUPAS ID") { Quote
Puika1 Posted July 27, 2009 Author Report Posted July 27, 2009 es jau liku meta ara mistisku eror :) Quote
kechums Posted July 27, 2009 Report Posted July 27, 2009 Priekš kam tur if'u? Kverijā var pielikt tādu zvēru kā WHERE. airbus, ja viņš tieši tā, kā tu teici izdarīs, tad pieļaus divas kļūdas vienlaicīgi, tas if's būtu jāliek while ciklā nevis pirms tā, un otrā lieta, tur ir mysql_fetch_array, nevis object. Quote
Puika1 Posted July 27, 2009 Author Report Posted July 27, 2009 ko tad man īsti darīt ar to "WHERE"? Quote
anonīms Posted July 27, 2009 Report Posted July 27, 2009 $result = mysql_query('select t1.id,t1.members_display_name,t2.avatar_location from '.$INFO['sql_tbl_prefix'].'members WHERE grupa = tava_super_grupat 1 join '.$INFO['sql_tbl_prefix'].'member_extra t2 on (t1.id = t2.id);'); Quote
Puika1 Posted July 28, 2009 Author Report Posted July 28, 2009 anonīms es mazliet nesapratu šo vietu 'members WHERE grupa = tava_super_grupat 1 join '.$INFO['sql_tbl_prefix'] ka man te rakstit "tava_super_grupat 1 join " ? Quote
airbus Posted July 28, 2009 Report Posted July 28, 2009 Grupas ID, piemēram 'members WHERE grupa = 11 '.$INFO['sql_tbl_prefix'] Es tā saprātu Quote
airbus Posted July 28, 2009 Report Posted July 28, 2009 Nu protams ka neradīs... Jo tavā datubazē nav tadas tabulas grupa ir jaatrod attiecīga tabula un jaieliek. Quote
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