sendLog method

Future<void> sendLog(
  1. LogPayload payload
)

Implementation

Future<void> sendLog(LogPayload payload) async {
  try {
    logDebug('Sending error log:', payload.toJson());
    final url = Uri.parse('$sdkApiUrl/users/sdk/record-logs/');
    final response = await http.post(
      url,
      headers: {'Content-Type': 'application/json'},
      body: jsonEncode(payload.toJson()),
    );
    if (response.statusCode != 200) {
      logError('Failed to send error log:', response.body);
    }
  } catch (e) {
    logError('Error sending log:', e);
  }
}