对象继承-重构问题-创建灵活服务的可能性

这应该很容易，但是显然我缺乏基本知识。我有一项服务，可以在BugTracking系统中创建票证。创建插件非常简单。有一个界面

public interface BugTracker {
    boolean createAndSendIssue(Issue issue);
    boolean updateAndSendIssue(Issue issue);
    boolean closeIssue(Issue issue);
}

和JIRA的API客户端：

@Component
public class JiraClient implements BugTracker {
    @Override
    public boolean createAndSendIssue(Issue issue) {
        //Logic for using JIRA REST API
        return false;
    }

    @Override
    public boolean updateAndSendIssue(Issue issue) {
        //Logic for using JIRA REST API
        return false;
    }

    @Override
    public boolean closeIssue(Issue issue) {
        //Logic for using JIRA REST API
        return false;
    }
}

Bugzilla的API CLient：

@Component
public class BugzillaClient implements BugTracker {

    @Override
    public boolean createAndSendIssue(Issue issue) {
        //Logic for using BUGZILLA REST API
        return false;
    }

    @Override
    public boolean updateAndSendIssue(Issue issue) {
        //Logic for using BUGZILLA REST API
        return false;
    }

    @Override
    public boolean closeIssue(Issue issue) {
        //Logic for using BUGZILLA REST API
        return false;
    }
}

问题Service正在处理中MVC。它看起来像：

@Service
public class BugTrackerService {
    private final BugzillaClient bugzillaClient;
    private final JiraClient jiraClient;
    private static final String JIRA_TYPE = "Jira";
    private static final String BUGZILLA_TYPE = "Bugzilla";

    @Autowired
    BugTrackerService(BugzillaClient bugzillaClient, JiraClient jiraClient) {
        this.bugzillaClient = bugzillaClient;
        this.jiraClient = jiraClient;
    }

    boolean processIssueTicketRequest(Issue issue){
        if ( issue.getBugTrackerType().bugTrackingSystemType.name.equals(JIRA_TYPE)) {
            return jiraClient.createAndSendIssue(issue);
        } else if (issue.getBugTrackerType().bugTrackingSystemType.name.equals(BUGZILLA_TYPE)){
            return bugzillaClient.createAndSendIssue(issue);
        } else {
            return false;
        }
    }
}

是否可以在BugTrackerService.processIssueTicketRequest没有自动连接所有客户端的情况下创建方法？目前，这不是什么大问题，但是在与越来越多的系统集成之后，该服务将变得复杂。

java spring 重构