79439863

Date: 2025-02-14 15:46:29
Score: 0.5
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I was able to solve this by adding a name property for Firebase initializeApp like stated in this answer.

 await Firebase.initializeApp(
 name: "dev project",
 options: DefaultFirebaseOptions.currentPlatform);

in case it is useful for someone in the future.

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Posted by: user3808307