I installed the Twitter api (passport-twitter) package yesterday. It does successfully do the Oauth1.0a process with Twitter, and gets the expected key and secret back. HOWEVER, then the package goes on to call https://api.twitter.com/1.1/users/show.json . That endpoint doesn't work with free API access on Twitter, as best as I can tell. It might work with the $100/month plan, but that doesn't fit this project's scope.
Twitter's documentation on tiers is not clear - it says sign in with Twitter is free, which is /technically/ true, unless you use the user's email, and um... who doesn't? It might be nice to document that somewhere.
I can't find in Twitter's 2.0 documentation whether the endpoint that might return email addresses will do so for the free tier. Does anyone know? I could redo the passport-strategy to use the v2.0 API, but given the current downward trend for Twitter, I'm not sure this is worth it. (And all their 'test our API online' stuff is offline, which doesn't make testing super easy, either!)
Hi All,
I installed the Twitter api (passport-twitter) package yesterday. It does successfully do the Oauth1.0a process with Twitter, and gets the expected key and secret back. HOWEVER, then the package goes on to call https://api.twitter.com/1.1/users/show.json . That endpoint doesn't work with free API access on Twitter, as best as I can tell. It might work with the $100/month plan, but that doesn't fit this project's scope.
Twitter's documentation on tiers is not clear - it says sign in with Twitter is free, which is /technically/ true, unless you use the user's email, and um... who doesn't? It might be nice to document that somewhere.
I can't find in Twitter's 2.0 documentation whether the endpoint that might return email addresses will do so for the free tier. Does anyone know? I could redo the passport-strategy to use the v2.0 API, but given the current downward trend for Twitter, I'm not sure this is worth it. (And all their 'test our API online' stuff is offline, which doesn't make testing super easy, either!)