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URL.openStream()

Roger Shields
Greenhorn

Joined: Nov 13, 2006
Posts: 4
I'm trying to read the contents of the reverse lookup page from 411.com at
http://www.411.com/10668/search/ReversePhone?phone=8001234567. (Replace 8001234567 with any number.

But anyway, I get a java.io.IOException with the message of
"Server returned HTTP response code: 403 for URL: http://www.411.com/10668/search/ReversePhone?phone=8001234567".

But if you enter the url into your browser it brings up a web page.

Now my question, I'm wondering if anyone has any idea of what I can do to actually get the content of the page?
Jesus Angeles
Ranch Hand

Joined: Feb 26, 2005
Posts: 2061
Have you tried HttpURLConnection?
Roger Shields
Greenhorn

Joined: Nov 13, 2006
Posts: 4
Yes, tried both. I can enter most other URLs, like http://www.google.com and it works fine. My best guess is that it is a problem with the user agent or something.
Henry Wong
author
Sheriff

Joined: Sep 28, 2004
Posts: 19073
    
  40

An error code 403 is basically an "access denied" error code. It means that there is something with the request that the site doesn't like. Anyway, I took a quick look... and I guess there is a check to only work for supported browsers. So if you fake it out with something like this:



It seems to work.

Henry
[ January 20, 2007: Message edited by: Henry Wong ]

Books: Java Threads, 3rd Edition, Jini in a Nutshell, and Java Gems (contributor)
 
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subject: URL.openStream()