selenin

Thanks a lot Jacques works great.

Hi Does anybody know the preg_replace for everything else then arabic. Couldn't figure it out. Thanks

Thanks Jacques1 I had to set the header.

Hi I try to get arab back and now I have this can I somehow encode it to get arabic signs thanks selenin

Thank you but is not working on my server, I think I'll give it up because the answer of the xml file is so terrible slow

lol yes it's adult cam site and I have to write an affiliate code Why are you getting it and I don't, I'm really a null

Don't understand this xpath that's what I wrote $cams = new SimpleXMLElement('http://chaturbate.com/affiliates/api/onlinerooms/?format=xml&wm=N6TZA', null, true); foreach ($cams->xpath('chantal') as $username) { if ($username == ""){ echo "not online"; }else{ echo "online"; } }

Don't get it just a blank site any other solution

Hi I want to check if an user is online, if the user is online the object is set otherwise it doesn't exist. I tried it like that but it doesn't work at all. $cams = new SimpleXMLElement('http://chaturbate.com/affiliates/api/onlinerooms/?format=xml&wm=N6TZA', null, true); if (isset($cams->username['chantal'])){ echo "online"; }

Okaj one I got html_entity_decode()

Thanks a lot I made htmlentities($variable, ENT_QUOTES, "UTF-8") works great, I'm so glad Now does anybody know how I make ö and ö to an ö

Hi Does anybody know how I encode this ö to this ö original it's an ö

Hi I need to make this Url: http://www.facebook.com/pages/MySchöppingen24de/309926495718477 to this http://www.facebook.com/pages/MySchöppingen24de/309926495718477 how I encode

Yes the error message is simple, because I'm not getting any results, normally I get 10 Error message Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home/selenin/public_html/artmeo/google.php on line 5

SELECT * FROM `sites` JOIN users ON sites.user = users.login WHERE (`banned` = '0' AND `user`!='{$data->login}') AND sites.user IN (select concat_ws(',', login) from users WHERE `coins`>=`cpc`) AND `id` NOT IN (SELECT `site_id` FROM `plused` WHERE `user_id`='{$data->id}') ORDER BY `cpc` DESC LIMIT 0, 10 I tried it like that but I'm only getting error messages, I really don't understand this JOIN

Do you think that really works I'm getting all the time error messages, it's the first time I use JOIN, I tried as well UNION but the same

$site2 = mysql_query("SELECT * FROM `sites` WHERE (`banned` = '0' AND `user`!='{$data->login}') AND sites.user IN (select concat_ws(',', login) from users WHERE `coins`>=`cpc`) AND `id` NOT IN (SELECT `site_id` FROM `plused` WHERE `user_id`='{$data->id}') ORDER BY `cpc` DESC LIMIT 0, 10"); Yes I know but I want as well order by coins, but coins is from another table, the table users. First it should be ordered by cpc desc and sometimes the cpc are the same the it should be ordered by users.coins

Don't know where to edit my post like that?

Hi I have a question, my script so far $site2 = mysql_query("SELECT * FROM `sites` WHERE (`banned` = '0' AND `user`!='{$data->login}') AND sites.user IN (select concat_ws(',', login) from users WHERE `coins`>=`cpc`) AND `id` NOT IN (SELECT `site_id` FROM `plused` WHERE `user_id`='{$data->id}') ORDER BY `cpc` DESC LIMIT 0, 10"); $ext = mysql_num_rows($site2); Now it's ordered by cpc desc, but I want to order as well by from the table users coins desc. Then the user with more coins is before the other. But sites.cpc is another table then users.coins, is this possible?

Thanks a lot that works, I have to do it like that

nobody I really don't get it

Hi I want the icon of a digg user $apicall = 'http://services.digg.com/2.0/user.getInfo?usernames='.$username.'&type=xml'; $result = simplexml_load_file($apicall); print_r($result); With that code I'll get this: So I want the icon the profile image of the user and I make this: print (string)$result->users->Ericquits->icon; This works, but my question is if there is a solution to get the icon without Ericquits the username? Because if write it little I don't get the icon...

okay got it, had to implode... Thanks

Hi I built a session array like this $notuserchannel = array(); $notuserchannel = $_SESSION['user']; $notuserchannel[] = $_GET['nextsub']; $_SESSION['user'] = $notuserchannel; Now I want mysql that youtube not in array session, I tried it like that: $sql = dbquery("SELECT youtube, subcredits FROM myMembers WHERE subcredits > 0 and id != '".$_SESSION['id']."' and youtube NOT IN ('".$_SESSION['user']."') and '".$_SESSION['id']."' NOT IN (select concat_ws(',', sub_giver) from sub_activity where sub_received = myMembers.id) ORDER BY myMembers.subcredits DESC LIMIT 1"); youtube not in session, but it's not working...

Okaj for the media:thumbnail, I made this: $names = preg_match_all('/<media:thumbnail url=["\'](.+?)["\']/', $page_contents, $matches); echo $matches[1][0]; and I get this: this one I'm close only the ?> in the end is too much