# Is it possible to integrate with Etsy OAuth? **URL:** \ **Category:** Get Help\ **Tags:** etsy, oauth\ **Created:** [June 8, 2017, 5:58pm UTC](https://community.auth0.com/t/is-it-possible-to-integrate-with-etsy-oauth/6977 "2017-06-08T17:58:46Z")\ **Posts on this page:** 3\ **Page:** 1
**Author:** ![alex11](https://upload.wikimedia.org/wikipedia/commons/thumb/7/7c/User_font_awesome.svg/768px-User_font_awesome.svg.png) [@alex11](https://community.auth0.com/u/alex11)\ **Post date:** [June 8, 2017, 5:58pm UTC](https://community.auth0.com/t/is-it-possible-to-integrate-with-etsy-oauth/6977/1 "2017-06-08T17:58:46Z")
Since I don’t have the preconfigured option for Etsy connections in the social tab, I was thinking if there is an easy way to implement with Auth0 this connection to Etsy OAuth through a serverless Angular 4 web app? ---
**Author:** ![jmangelo](https://sea1.discourse-cdn.com/auth0/user_avatar/community.auth0.com/jmangelo/32/7923_2.png) [@jmangelo](https://community.auth0.com/u/jmangelo)\ **Post date:** [June 9, 2017, 7:29pm UTC](https://community.auth0.com/t/is-it-possible-to-integrate-with-etsy-oauth/6977/2 "2017-06-09T19:29:40Z")
According to Etsy [documentation](https://www.etsy.com/developers/documentation/getting_started/oauth) they provide an OAuth1 implementation. > The Etsy API uses OAuth 1.0 to give developers access… This means you should be able to follow their documentation and Auth0 on how to add a generic OAuth1 authorization server implementation to Auth0 so that you could then use it as the default social connections. More specifically, you should follow: [Adding a generic OAuth1 Authorization Server to Auth0](https://auth0.com/docs/tutorials/adding-generic-oauth1-connection) ---
**Author:** ![system](https://global.discourse-cdn.com/auth0/original/3X/b/5/b560f266932fcce5d57f09a6595bd50518cae3f3.jpeg) [@system](https://community.auth0.com/u/system)\ **Post date:** [March 2, 2018, 5:54am UTC](https://community.auth0.com/t/is-it-possible-to-integrate-with-etsy-oauth/6977/3 "2018-03-02T05:54:30Z")