Yes, it's very possible. Haven't tested it, but it should be something
like this:

$cursor = -1;
while( $cursor !== 0 ) {
    $info = $oauth->get( 'statuses/followers', array( 'cursor' =>
$cursor ) );
    if( $oauth->http_code === 200 && !isset( $info->error ) ) {
        // Count or whatever here
        $cursor = $info->next_cursor;
    }
}

If you just want to count all of your followers, why not do users/
show? That contains a followers_count variable in it.

On Jun 25, 4:18 pm, Rick <rickstuivenb...@gmail.com> wrote:
> Thank you for your reply.
>
> I don't use next_cursor or whatsoever. I just use the $followers
> variable to use count it with $totaal = count($followers); so I can
> use it in my code. The cursor code I posted before is the only thing I
> use to try and get information.
>
> Is it even possible to get all the followers with the abrahams twitter
> library?
>
> On 25 jun, 15:11, Sam Wierema <samwier...@gmail.com> wrote:
>
>
>
> > You should not increment your cursor, because Twitter returns a cursor
> > for you. And if cursor is 0, it means that there are no more pages (-1
> > + 1 = 0).
>
> > Check your $followers variable that you got from the first call. It
> > should be called something like next_cursor.
>
> > On Jun 25, 2:26 pm, Rick <rickstuivenb...@gmail.com> wrote:
>
> > > Howdy!
>
> > > I am currently making my application OAuth compatible from basic auth,
> > > currently I have a issue I need to get resolved in order to switch to
> > > the new OAuth method.
>
> > > I am getting my followers from: $oauth->get('statuses/followers'); but
> > > that only gives me the last 100 followers, I trought it would be easy
> > > to do it this way:
>
> > > $cursor = -1;
> > > $followers = $oauth->get('statuses/followers', array('cursor' =>
> > > $cursor));
> > > // Other code here..
>
> > > $cursor++;
> > > $followers = $oauth->get('statuses/followers', array('cursor' =>
> > > $cursor));
>
> > > Is it possible to get all my users, because this code does not work
> > > and my inspiration was wrong.
>
> > > Hopefully you can help me.
>
> > > Regards,
>
> > > Rick

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