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I'm getting ready to release an iOS app that offers subscriptions that we also offer via our website. Our annual subscription via our website is $89 even. I'm thinking of setting the iOS subscription price to $89.99 ($88.99 just looks kooky to me), but I thought I recalled reading that Apple does not allow you to offer the same product at a lower price via your website.

I've searched through the App Store Review Guidelines and I can't seem to find any reference to that now. Am I just mis-remembering things? (I do know that you are not allowed to reference other stores from within the app.)

So my question is, if I want to offer the in-app product at $89.99, do I have to raise the price of the corresponding product on my website to the same price or higher?

Brian Rak
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    I'm voting to close this question as off-topic because it's about product pricing and not about programming. – rmaddy Sep 03 '18 at 20:04
  • It's true that it's not directly about programming, but it is part of the process that any developer has to go through to get an app offering in-app subscriptions onto the App Store. – Brian Rak Sep 03 '18 at 20:15
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    Questions about Apple's policies and App Store approval are also off topic here. – rmaddy Sep 03 '18 at 20:18
  • You cannot offer the ability to purchase subscriptions from your website in your app. You cannot even tell people that they can do this, so Apple has no visibility of the price you may charge in other locations. It is up to you to whether your customers will be upset if they find out they paid a dollar more in the app. – Paulw11 Sep 03 '18 at 21:25
  • @rmaddy, okay, my bad. Thank you for clarifying. – Brian Rak Sep 03 '18 at 21:26
  • @Paulw11, that's great to hear. Thank you! – Brian Rak Sep 03 '18 at 21:26

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Assuming you are using in-app purchase for subscription in iOS, there are some specific price tiers from which you can choose the price of a subscription and that price will not be compared at other places(like your website). Hope this helps. Best luck for your new application.

BhargavR
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