If I remember rightly, the gallery returns you a content uri so you'll want to do something like: getContentResolver ().openInputStream(uri)
On Nov 17, 5:53 am, Abhi <abhishek.r.sha...@gmail.com> wrote: > Hi Dianne, > > I am trying to do the following. > > I have an Image viewer where in the user picks an Image from within > the gallery. The uri to that selected Image is available to me. Now, I > want to use this URI information and send it as a file over a socket > using FileInputStream. Is this a valid syntax to perform the above > action? > > FileInputStream fis = new FileInputStream(uri.getPath()); // > Here uri is the URI of the selected Image > byte[] buffer = new byte[fis.available()]; > fis.read(buffer); > > ObjectOutputStream oos = new ObjectOutputStream > (socket.getOutputStream()); > oos.writeObject(buffer); > > Please help me to move forward. > > Thanks, > > Abhi -- You received this message because you are subscribed to the Google Groups "Android Developers" group. To post to this group, send email to android-developers@googlegroups.com To unsubscribe from this group, send email to android-developers+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/android-developers?hl=en
