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link to build.gradle instead of app id

This commit is contained in:
Damodar Lohani 2021-07-06 13:38:20 +05:45
parent 1629d8d19a
commit 2f28bc112f

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@ -11,7 +11,7 @@ If you are building your Flutter application for multiple devices, you have to f
For **iOS** add your app name and Bundle ID, You can find your Bundle Identifier in the General tab for your app's primary target in Xcode.
#### Android
For **Android** add your app <u>name</u> and <u>package name</u>, Your package name is generally the <a href="https://github.com/appwrite/playground-for-flutter/blob/0fdbdff98384fff940ed0b1e08cf14cfe3a2be3e/android/app/build.gradle#L41" target="_blank" rel="noopener">applicationId</a> in your app-level build.gradle file. By registering your new app platform, you are allowing your app to communicate with the Appwrite API.
For **Android** add your app <u>name</u> and <u>package name</u>, Your package name is generally the **applicationId** in your app-level <a href="https://github.com/appwrite/playground-for-flutter/blob/0fdbdff98384fff940ed0b1e08cf14cfe3a2be3e/android/app/build.gradle#L41" target="_blank" rel="noopener">build.gradle</a> file. By registering your new app platform, you are allowing your app to communicate with the Appwrite API.
#### Linux
For **Linux** add your app <u>name</u> and <u>package name</u>, Your package name is generally the **name** in your <a href="https://github.com/appwrite/playground-for-flutter/blob/0fdbdff98384fff940ed0b1e08cf14cfe3a2be3e/pubspec.yaml#L1" target="_blank" rel="noopener">pubspec.yaml<a> file. If you cannot find the correct package name, run the application in linux, and make any request with proper exception handling, you should get the application id needed to add in the received error message.