Will pay if you can stop this site stealing my site content.

This site justinbieber.xona2.net is stealing 100% content of my website. I need help to stop this. few of his pages rank higher than mine on google. At this point i feel discourage to update my blog since other people take advantage from it.

I dont mind if he take 3 or 5 articles, but the whole site? 300 articles including all the images and videos. I believe he is using autoblogging software.

other scam site he has such as;

kesha.xona2.net
britneyspears.xona2.net
ladygaga.xona2.net
and many more.

need ur help now, thanks.

 

Which site is he copying from?

 

justinbieberzone com

he can put watermarks to the images automatically too.

 

have you tried turning off your RSS feeds. As it is possible he is using your feeds to grab your posts...

 

I already changed from full RSS to summary RSS. If i turn off my rss feed, google will not crawl my site, right? I thought rss feed is a must in SEO.

 

Yes it is, but you can try stopping it for a few days so you can see if he is manually copying your site posts, or it is an automatic feed creation software that copies your post. Either way, that info would be useful to stop the perpetrator from stealing more of your posts...

 

It's automatic. I've been watching his blog for 2 weeks. Everytime i post new post, its automatically in his blog within second. And my blog receive trackback right away.

 

The best thing you can do at the moment is rewrite your RSS url writing engine, use your htaccess file to do this. If all else fails hire someone who can hack through his account...

 

lol

anyways:
step 1. contact him, explain the situation.
step 2. contact his hosting provider, explain the situation.
step 3. contact his registrar, explain the situation.
step 4. contact google, explain the situation.
step 5. contact icann, explain the situation.

do you see the general pattern here?

EDIT:
whois for the domain: http://reports.internic.net/cgi/whois?whois_nic=xona2.net&type=domain

 

Do you have express permission to use/display the photos on your site?

 

Why not just ban his server's IPs? That'd be the most quick thing to do. Obviously if he's using autoblog software through WordPress, it's most likely PHP-based software requesting your site from the hosted server. Most a test post, check your logs for the IP, ban it. Then post another test post to see if they're banned. Should work. You could also easily add a ban code to the config file of the site:

$banned = array("24.199.203.189","24.199.203.188"); // Array of banned IPs
if (in_array($_SERVER['REMOTE_ADDR'],$banned)) {
 echo "Permission from IP denied.  Halting execution.";
 exit;
}

Code (markup):

 

if he is good, and wants to copy your content, no matters what you do, he will, so ur best option is to watermark ur images on different places, and i think u doing it already

 

i was thinking, a different approach could be to try to bother him, like, disable for a moment ur blog being seen online by search engines, add a post with wrong information, maybe some porn images, linking to porn urls, wait until his site copies it and then delete it, or better than banning his IP, serve different content to his IP, more difficult, but it could be more difficult for him to find out what's going on

 

says the guy with a link to 'ultimate content fetcher' in his sig =)

 

The situation is really puzzle kind of. I doubt how the coder should have done programming such plugin to steal the content.

 

try insert links of your site in the begin/middle/end of every post
then you'll sure rank higher in google result

 

All link i inserted to the post turn into text in his website. I dont know why

 

I might wanna try putting porn stuff. But I worry because most visitor are little girls and they online 24/7.

 

do one thing.. if you know his ip just do mess with some php like if ip == blah { show p*rn content } else { normal content }

 

Just do a ip block in your htaccess for his ip/domain.

Something like this in your htaccess will do the trick

order allow,deny
deny from 127.0.0.1
deny from 127.0.0.2
deny from 127.0.0.3
deny from domainname.com
allow from all

Code (markup):