I can login to my app then share something. This needs an OPENED session state. However, when I am not logged in, then I want to share something, I need to open the session. I am using a ViewPager so e.g. when I go from one page to another and this code
Session.openActiveSession(getActivity(), true, new StatusCallback() {
@Override
public void call(Session session, SessionState state, Exception exception) {
}
});
is in the beginning of the code, then the session becomes active, and I get automatically logged in, which is wrong! That's why I put this code block into an onClickListener, so I only want to open the session if I click the share button in my app:
if (session != null && session.isOpened()) {
publishFeedDialog();
}
else {
Session.openActiveSession(getActivity(), true, new StatusCallback() {
@Override
public void call(Session session, SessionState state, Exception exception) {
}
});
publishFeedDialog();
}
private void publishFeedDialog() {
session = Session.getActiveSession();
Log.i("TAG", session.getState() + ""); //OPENING
WebDialog feedDialog = (
new WebDialog.FeedDialogBuilder(getActivity(),
Session.getActiveSession(),
params))
.setOnCompleteListener(new OnCompleteListener() {
@Override
public void onComplete(Bundle values,
FacebookException error) {
if (error == null) {
final String postId = values.getString("post_id");
if (postId != null) {
} else {
// User clicked the Cancel button
}
} else if (error instanceof FacebookOperationCanceledException) {
} else {
// Generic, ex: network error
}
}
})
.build();
feedDialog.show();
}
The error:
Attempted to use a session that was not open.
So I open the session in vain, because it is still OPENING when the WebDialog should appear.
Please help.