And I thank that is a reasonable expectation. Just make sure you're not unfollowing hundreds of people per day or something so it doesn't raise Twitter red flag.
Instead maintain a list of inactive followers and slowing remove them and/or temporarily block them. Even looking at my personal Twitter account, only connected with a few hundred, I too want to stop following folks that are active OR at least have them NOT count against my following limit. Makes no sense to do so. And it would be great to detect these folks. Leon ________________________________ From: TylerC <tyle...@gmail.com> To: Twitter Development Talk <firstname.lastname@example.org> Sent: Sun, October 25, 2009 10:23:04 PM Subject: [twitter-dev] Re: Check when a friendship was created Well really I would just like to unfollow people I have followed for a while that haven't followed me back lol... It's pretty lame to me that this isnt already a feature but its cool I found my own way to implement the idea. Thanks for nothing :) On Oct 25, 7:31 pm, Zac Bowling <zbowl...@gmail.com> wrote: > What you describe is a very spamy tactic people use to slowly grow > their follower counts without having having dramatically higher > following counts then follower counts. I HATE when people do that > because I get follows (usually from "marketing" or "social media > experts" pushing their personal brand and don't really care so much > about what I have to say and are just shooting for quantity rather > then quality). > > IIRC correctly, it could be against the TOS too. > > Zac Bowling > > On Sun, Oct 25, 2009 at 4:13 PM, TylerC <tyle...@gmail.com> wrote: > > > I am seeking the need to know when a friendship or when I have > > followed a given user. Basically, I want to scan my account for people > > I have been following, enter a given number of days, and unfollow them > > if they have not followed me back in that timespan. I have it all > > worked out expect for checking when I followed a given person. > > > Is there no way to do this via the API, from the looks of it I would > > have to do this artificially with a database or something right? > > > Thanks!