• Post Reply Bookmark Topic Watch Topic
  • New Topic
programming forums Java Mobile Certification Databases Caching Books Engineering Micro Controllers OS Languages Paradigms IDEs Build Tools Frameworks Application Servers Open Source This Site Careers Other Pie Elite all forums
this forum made possible by our volunteer staff, including ...
Marshals:
  • Campbell Ritchie
  • Tim Cooke
  • paul wheaton
  • Paul Clapham
  • Ron McLeod
Sheriffs:
  • Jeanne Boyarsky
  • Liutauras Vilda
Saloon Keepers:
  • Tim Holloway
  • Carey Brown
  • Roland Mueller
  • Piet Souris
Bartenders:

opening a url

 
Ranch Hand
Posts: 57
  • Mark post as helpful
  • send pies
    Number of slices to send:
    Optional 'thank-you' note:
  • Quote
  • Report post to moderator
Dear java guru's
I was opening a url to a page like this.
url = new URL("http://www.webappcabaret.com/abc/report/successmail.jsp");
try {
in = new BufferedInputStream( url.openStream() );
} catch ( IOException e ) {
System.out.println( "Couldn't retrieve data from url!" );
}
But it is giving exception
So when I ask the provider about this they just said
"You CANNOT open a webappcabaret from within webappcabaret"
Is there any other way to do this?
Payal
 
Saloon Keeper
Posts: 28765
211
Android Eclipse IDE Tomcat Server Redhat Java Linux
  • Mark post as helpful
  • send pies
    Number of slices to send:
    Optional 'thank-you' note:
  • Quote
  • Report post to moderator
Put a
System.out.println("Exception was '" + e.getMessage() + "'" );
In your catch clause. You may find out something useful.
 
With a little knowledge, a cast iron skillet is non-stick and lasts a lifetime.
reply
    Bookmark Topic Watch Topic
  • New Topic