subzeroid / instagrapi

🔥 The fastest and powerful Python library for Instagram Private API 2024
https://hikerapi.com/p/bkXQlaVe
MIT License
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Getting Feed URL after success upload Feed #1406

Closed ardinusawan closed 1 year ago

ardinusawan commented 1 year ago

Is your feature request related to a problem? Please describe.

in this function .photo_upload or .album_upload is any way to getting upload url from the return?

There is media.dict() and there is attributes thumbnail url and video url. I want it has feed url. Is it possible?

Describe the solution you'd like

media = cl.photo_upload() media.feed_url

Describe alternatives you've considered

Or anyway to getting url from media_id or media_pk?

Additional context

nope

adw0rd commented 1 year ago

I think you need media.code

Or send your PR with improvements

ardinusawan commented 1 year ago

media.code is what I need, I just need to build url from it. Thanks for fast response!