Resolved Subscription question

[ginofresh]

so I subscription to something on my account and then I unsubscribed to it but it still show up in my subscription list still is there anyway I can remove it or will it go away as the subscription is done?

 

[C DM]

Subscribe where? Where does it still show up?

 

[ginofresh]

Subscribe where? Where does it still show up?

in my subscribing things on my account under subscriptions

 

[C DM]

in my subscribing things on my account under subscriptions

Is this related to iCloud? Where do you go to see this in iOS?

 

[ginofresh]

Is this related to iCloud? Where do you go to see this in iOS?

go to the App Store and then press on your picture in the corner and then go down to subscriptions and there is something there

 

[gwang73]

go to the App Store and then press on your picture in the corner and then go down to subscriptions and there is something there

Subscriptions will stay active until the time expires when you cancel it. It should show you the expiration date under the service. Once it expires, it moves into the expired section.

 

[ginofresh]

Subscriptions will stay active until the time expires when you cancel it. It should show you the expiration date under the service. Once it expires, it moves into the expired section.

and then when does it go away?

 

[gwang73]

and then when does it go away?

I'm not sure it ever goes completely away. I have one showing that I canceled 4 months ago and it still shows under expired.
I also wish that it would go completely away.

 

[papbot]

I'm not sure it ever goes completely away. I have one showing that I canceled 4 months ago and it still shows under expired.
I also wish that it would go completely away.

I think they leave it there on the possibility that you may decide to renew it again. I had one app that actually discontinued its subscription service but remained listed under my subscriptions for about a year or so afterwards. It only disappeared when the app itself was no longer available on the App Store.