Getting SDKERR_UNINITIALIZE error when initializing zoom SDK second time on same machine

Meeting SDK
1

Hello there,

We are trying to build a calling bot that as per our requirement should join the zoom call with multiple different users (SDK Keys and secrets) from the same machine.

However, as soon as we try to launch our application a second time with a different user, it gives us SDKError code SDKERR_UNINITIALIZE

Here is our code block to get an idea of what we are doing:

 ZOOM_SDK_DOTNET_WRAP.CZoomSDKeDotNetWrap.Instance.GetAuthServiceWrap().Add_CB_onAuthenticationReturn(onAuthenticationReturn);
            ZOOM_SDK_DOTNET_WRAP.CZoomSDKeDotNetWrap.Instance.GetAuthServiceWrap().Add_CB_onLoginRet(onLoginRet);
            ZOOM_SDK_DOTNET_WRAP.CZoomSDKeDotNetWrap.Instance.GetAuthServiceWrap().Add_CB_onLogout(onLogout);

            ZOOM_SDK_DOTNET_WRAP.AuthContext param = new ZOOM_SDK_DOTNET_WRAP.AuthContext();
            ZOOM_SDK_DOTNET_WRAP.AuthParam authparam = new ZOOM_SDK_DOTNET_WRAP.AuthParam();
            param.jwt_token = ZoomToken();
            authparam.appKey = ApiKey;
            authparam.appSecret = ApiSecret;

            SDKError err = ZOOM_SDK_DOTNET_WRAP.CZoomSDKeDotNetWrap.Instance.GetAuthServiceWrap().SDKAuth(authparam);
            Logger.Info("App Key " + authparam.appKey);
            Logger.Info(err.ToString());
            if(err != SDKError.SDKERR_SUCCESS)
            {
                Application.Current.Shutdown(0);
            }

We also enabled logs and attached the log file for internal checking. Here is the link to download logs: zoom_sdk_demo.exe_Video_0.log - Google Drive

Is there any restriction on a number of SDK bots joining from the same machine but different users?

2

Welcome, @nitin.k,

Thank you for posting in the Zoom Developer forum and sorry for the missed response. I am happy to help here. As a start, can you share the version of the SDK you are using? Also, please give a detailed description of the steps to reproduce the issue. This information will help us reproduce the behavior on our end.

Let me know if you have any questions about this.