var_dump(sfContext::getInstance()->getRequest()- >getUriPrefix(),sfContext::getInstance()->getRequest()- >getRelativeUrlRoot()); string 'http://localhost' string '/project/trunk/web'
$uri = sfContext::getInstance()->getRequest()- >getUriPrefix().sfContext::getInstance()->getRequest()- >getRelativeUrlRoot(); then I can just add the folder I want to the end of that. I haven't been able to find another way. I assume this is prolly the best way? Pretty gross code, is there anything out there that will give me the url? I haven't uploaded this to be production server yet, but if anyone else knows of a way to do this please let me know. Thanks, f1g On Aug 15, 11:49 pm, Joshua Estes <f1gm...@gmail.com> wrote: > I have a few different projects that are in sub directories. I'm > trying to generate a url with an image in it. So the base url would > behttp://www.site.com/project/web > > The image is in the sf_upload_dir > (http://www.site.com/project/web/uploads/image.png) > > How do I get the base URL with no script name? > iehttp://www.site.com/project/web > > Thanks, > f1g -- If you want to report a vulnerability issue on symfony, please send it to security at symfony-project.com You received this message because you are subscribed to the Google Groups "symfony users" group. To post to this group, send email to symfony-users@googlegroups.com To unsubscribe from this group, send email to symfony-users+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/symfony-users?hl=en
