Jiabin, You would put the video into the "raw" folder, so yeah, it should work.
http://developer.android.com/guide/topics/resources/providing-resources.html On Jun 24, 1:44 pm, Jiabin Qin <[email protected]> wrote: > I searched Google and found several ways , but I am not sure which is > the correct one. > > File mFilePtr = new File(getExternalFilesDir(null), mFilename); > InputStream is = getResources().openRawResource(id); > is.read(mBuffer, 0, bufsize); > > The doubt for this is , in the method: openRawResource(id), on the > developers site, it says: > Open a data stream for reading a raw resource. This can only be used > with resources whose value is the name of an asset files -- that is, > it can be used to open drawable, sound, and raw resources; it will > fail on string and color resources. > id The resource identifier to open, as generated by the appt tool. > > Is it ok to read my video in that way? > > Thanks in advance! > Jiabin > Singapore -- You received this message because you are subscribed to the Google Groups "Android Developers" group. To post to this group, send email to [email protected] To unsubscribe from this group, send email to [email protected] For more options, visit this group at http://groups.google.com/group/android-developers?hl=en
