问题
When creating Firebase invite intent I try to add link to iOS app as described in documentation:
intent = new AppInviteInvitation.IntentBuilder(context.getString(R.string.invitation_title))
.setMessage(context.getString(R.string.invitation_message))
.setOtherPlatformsTargetApplication(
AppInviteInvitation.IntentBuilder.PlatformMode.PROJECT_PLATFORM_IOS,
"1059710961")
.build();
"1059710961" and "mobi.appintheair.wifi" both cause the same error:
AppInviteAgent: Create invitations failed due to error code: 3
AppInviteAgent: Target client ID must include non-empty and valid client ID: 1059710961. (APPINVITE_CLIENT_ID_ERROR)
What is the correct format for this parameter?
回答1:
To get this client ID you have to do the following:
- Register you iOS app in the Firebase Console
- Download
GoogleServices-Info.plist
for iOS app as we downloadgoogle-services.json
for Android - Look in it and found value for the key
CLIENT_ID
(will be something like this123456789012-abababababababababababababababab.apps.googleusercontent.com
) Add it to builder:
intent = new AppInviteInvitation.IntentBuilder(context.getString(R.string.invitation_title)) ............ .setOtherPlatformsTargetApplication( AppInviteInvitation.IntentBuilder.PlatformMode.PROJECT_PLATFORM_IOS, "123456789012-abababababababababababababababab.apps.googleusercontent.com") .build();
回答2:
the client_id is the one in the plist you download from the firebase console for your iOS app
来源:https://stackoverflow.com/questions/37836448/can-not-add-apple-client-id-to-firebase-invites-intent-on-android