unregister push on logout #1309
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| @@ -106,6 +106,7 @@ public class NotifyClient { | |||
| try await identityClient.unregister(account: account) | |||
| notifyWatcherAgreementKeysProvider.removeAgreement(account: account) | |||
| try notifyStorage.clearDatabase(account: account) | |||
| Task { try await pushClient.unregister()} | |||
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What if the unregister fails (e.g. no internet?) Will this flow be tried again or will the push registration still be present?
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good point, I will make it throw
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@chris13524 what code push api returns for already unregistered client on /unregister ?
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It gives the same response code regardless if the client was already unregistered or not: 200
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I will make it throw
I still think there is an issue. If it throws, what does caller do with that? If simply display an error message, the question to me is when will it do the unregister again?
Looks to me that you are clearing the notify storage database (and other things) before you unregister the push client. So this would mean if the way the unregister takes place is based on the notify storage, then the unregister won't happen again. I'm assuming that logout should be an atomic operation to prevent the receiving of push notifications it should not.
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| do { | ||
| let (_, response) = try await URLSession.shared.data(for: request) | ||
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| guard let httpResponse = response as? HTTPURLResponse, httpResponse.statusCode == 200 else { |
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200
Instead of checking for a specific status code, can you check if it is any 2xx status code? We may want e.g. change to 204 (no content) and all 2xx codes mean success
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Description
Resolves # (issue)
How Has This Been Tested?
Due Dilligence