When using openInputStream on the content-resolver, the returned input-
stream is not as 'flexible' as a FileInputStream. For example,
FileInputStreams can be retried if something goes wrong. The one
returned by openInputStream can not.
To get the fully qualified path-name to the image-file (jpg/png),
query the Uri of the image and obtain the value of the DATA ("_data")
column. This value is a String, the fully-qualified path to the image
file. Then you can create a FileInputStream from this file.
On Nov 17, 4:08 am, MrChaz <mrchazmob...@googlemail.com> wrote:
> 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- Hide quoted text -
>
> - Show quoted text -
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