pinacoladaxb

I'm having some trouble with a query I'm trying to make. I can't figure out what's wrong here. Can anyone see the issue? sqltest.php source: <?php $q = 'test'; $con = mysql_connect('localhost', 'pauliuko', 'password') or die(mysql_error()); mysql_select_db('pauliuko_atomicriot', $con) or die(mysql_error()); $query = mysql_query("SELECT COUNT(*) FROM 'ac_users' WHERE 'username' = '$q'") or die(mysql_error()); list($number)=mysql_fetch_row($query); echo $number; mysql_close($con); ?> sqltest.php displays: 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 ''ac_users' WHERE 'username' = 'test'' at line 1 I double checked my database and there is a table named 'ac_users' with a column named 'username'.

I'm trying to write a script where a user types a query into a field, and the script finds a paragraph or two about that topic from the internet. Could this be done by having the script do an "I'm feeling lucky" google search and then return the readable content from the html file? :-\ How on earth would I go about doing this? Thanks for any help.

I don't want to do anything malicious here. I'm re-uploading some of my old YouTube videos in higher quality, and I do not want to lose certain stats from the old videos. I was wondering if it would be possible to transfer ratings, views, and/or comments from the old videos to the new videos, using either the Zend Gdata library or a modded version of it. Is that possible? Zend Gdata documentation: http://framework.zend.com/manual/en/zend.gdata.html PHP Developer's Guide: http://code.google.com/apis/youtube/2.0/developers_guide_php.html It seems to me that Google is opening itself up here for people to potentially screw around with its databases...

I need to install an extension for my php site, but since I am using a web hosting service (hostmonster), I do not have access to the directory that contains all the PHP extensions. I DO, however, have access to my php.ini file. How can I install this extension (dbg.so)?

Thanks. I found the source of the problem in an .htaccess file

This is a weird problem, but the solution should be simple. I just need to know what server configuration file to edit (I'm running apache.) Anyways... If you go to my server's homepage (http://98.111.193.184/ ) and hover over any link, you'll see in the bottom-left corner of your browser where the link links to. However, when you click on it, you'll notice it takes you to to an entirely different server (see the url you're on). Does anyone know what the problem could be?

No, gammerz. This isnt funny. The site's over. I've had my fun. This has gotten way too serious.

Ok, everyone, I deleted the site because of all the trouble it seems to have caused. Sorry for everything. Oh, and if you added my to any such spam lists I kindly ask that you take my ip off, if possible.

Ahh so much to reply to Yes, I know you can "easily" read the server header, but in order to do that you'd have to view the raw html of the email and know what to look for. Besides, no one checks every email they get to see if it is really coming from where it says it is. As for the spam issue, that shouldn't be a problem unless I use this for spamming. I've just used it to mess with my friends and I always tell them later. But I do take your point seriously, so I'll remove the site the instant someone abuses it, and I'll know if they do. I don't appreciate you calling me a fake programmer. I'm as real as you, just less mature so I make sites like these. First off: What was funny?, and second: Why would he be angry? You asked for assistance - albeit malicious - and I provided a possible answer. Sorry I accidentally switched the replies around. That message was directed towards gammerz. So sorry. I don't see how this qualifies as chainletter. I personally hate chainletters so much I look the other way when I get chain emails about kidnapped children, etc. So if this is a valid issue tell me how to prevent it.

@jordanwb That's hilarious! But I told God on you and he's angry. @gemmerz Ya I suppose, but I'm looking for a more hidden approach if possible.

I wrote this a while ago and it was a lot of fun to use: http://pauliukonis.com/miscfiles/email.php It can send person 'A' an email and make it look like it was sent by person 'B'. Stupid, and simple, I know. But now I'm curious: Is there a way that I can make it so that when person 'A' replies, it is sent to person 'B' AND the "pranker?" Thanks.

Yes, but take a look at this. All ports are closed except 21761: Warning: fsockopen() [function.fsockopen]: unable to connect to 173.71.166.97:7777 (Connection timed out) in /home/pauliuko/public_html/atomicriot/portcheck.php on line 6 7777, Warning: fsockopen() [function.fsockopen]: unable to connect to 173.71.166.97:21761 (Connection timed out) in /home/pauliuko/public_html/atomicriot/portcheck.php on line 6 21761, Warning: fsockopen() [function.fsockopen]: unable to connect to 173.71.166.97:13000 (Connection timed out) in /home/pauliuko/public_html/atomicriot/portcheck.php on line 6 13000, Warning: fsockopen() [function.fsockopen]: unable to connect to 173.71.166.97:13139 (Connection timed out) in /home/pauliuko/public_html/atomicriot/portcheck.php on line 6 13139, Warning: fsockopen() [function.fsockopen]: unable to connect to 173.71.166.97:27900 (Connection timed out) in /home/pauliuko/public_html/atomicriot/portcheck.php on line 6 27900, Isn't it weird that there's a space after the port that was open? Hmm...

I am making an application that can only work if the user has forwarded a couple ports to their computer. I need a php page that will check if those ports are open and unblocked by the user's windows firewall. Here's what I have: <?php $ip=$_SERVER['REMOTE_ADDR'];//Gets user's IP address $port = array(765,6546,6500,6515,7777,13000,13139,27900,3783,6667,28900,29900,29901);//ports that should be forwarded and unblocked $loopindex=0; while ($loopindex<count($port)){//loops through the ports and checks if each is open $checkport = fsockopen($ip, $port[$loopindex], $errnum, $errstr, 2); if(!$checkport) echo $port[$loopindex].",";//echos a list of ports that are not open $loopindex++; } ?> But when I run the script it says all of those ports are not responding. I'm no expert here, so I don't know what I did wrong. Help me please.

Hey thanks for the replies, but I just restarted my server and it works fine now. Also, don't bother connecting to it; you're not supposed to see anything.

I own the domain atomicriot.com, which is hosted on hostmonster. I am running a server at home as well, which I want to use only for my online games that I make. My game server is opened on port 6121, on my ip (71.60.197.242). My goal is to allow people to connect to that server by typing in "server.atomicriot.com" rather than "71.60.197.242." In order to do this, I contacted hostmonster and they changed the A Record for server.atomicriot.com to 71.60.197.242. Now my problem is that I cannot seem to connect to the server when typing in "server.atomicriot.com." am getting an error that says that the connection was attempted on something that is not a socket. In addition, here's what happens in my Terminal Application (macs only) when I type in "ping server.atomicriot.com": PING server.atomicriot.com (71.60.197.242): 56 data bytes 36 bytes from c-71-60-197-242.hsd1.pa.comcast.net (71.60.197.242): Destination Host Unreachable Vr HL TOS Len ID Flg off TTL Pro cks Src Dst 4 5 00 5400 afcb 0 0000 40 01 b1ea 10.0.1.197 71.60.197.242 Can someone please help me correct this problem? Thanks. -Adam Note - this post is an extension of THIS TOPIC, which I marked as solved a while ago. You don't need to read, however, to help me.