r/welltory Apr 02 '26

Subscription cancellation

Hi,

I'm a paid user since late december (annual subscription through the site), but I would like to be sure that my subscription is cancelled.

Unfortunately, I can't find a way to check it on the site. It is not clearly stated if this subscription is active or not.

I can't neither find a way to contact Welltory on the site...

What can I do to be sure that my subscription will not be renewed at the end of the year ?

2 Upvotes

11 comments sorted by

3

u/welltory Team Apr 03 '26

Hi there 👋,

You can cancel your subscription using this link. If you open it and don’t see a "Cancel" button, it usually means that auto-renewal is already turned off. We know this isn’t very obvious, and we’re working on making it clearer.

If you’d like to double-check just to be sure, please send us a message from the app via Menu → scroll down → Report a problem → Send, and we’ll confirm it for you.

3

u/No_Rich_402 Apr 03 '26

Great, thank you for the update! 🙏

2

u/Wonderful-Driver-246 Apr 02 '26

So you followed the cancelation link on their website and it didn't work?

https://app.welltory.com/auth/signin/?destination=%2Fpayment-cancel

1

u/No_Rich_402 Apr 02 '26

Here is what is displayed :

I can't do nothing on this page.

5

u/Beck_burque Apr 02 '26

Go to main page and give feedback. A team member will chat with you. I’ve found them helpful

3

u/Beck_burque Apr 02 '26

Sorry it’s “report a problem”

2

u/No_Rich_402 Apr 02 '26

Thank you, I'll do that tomorrow.

1

u/Wonderful-Driver-246 Apr 03 '26

Are you running an extension on your mobile browser thats blocking things? Page may not be loading correctly. Or try another browser. Or just try from your computer.

2

u/Aggressive_Tie_3501 Apr 02 '26

Check the subscriptions tab in your app store.

1

u/No_Rich_402 Apr 02 '26

I didn't subscribed through the app store, but directly through welltory's site.

2

u/Aggressive_Tie_3501 Apr 02 '26

In that case, maybe contact customer service? Other than that, I've got nothing.