Question

我们有一个Hibernate-gnerated域对象的拼凑而成，例如

@Entity
@Table(name = "events_t", schema = "public")
public class EventsT implements java.io.Serializable {    
    private int id;
    private RecallsT recallsT; // another table
}

使用Spring Data，我无法做到

RecallsT recallsT = recallsDAO.findById(recallId);

我被迫做了

Optional<RecallsT> recallsT = recallsDAO.findById(recallId);

但这引入了另一个问题：现在我不能再使用我的Hibernate对象，因为这不会起作用：

eventsT.setRecallsT(recallsT);

现在错误将是它不适合＆＃34;可选＆lt; ...＆gt;＆＃34;将对象转换为普通对象。正如我在Hibernate实体中所展示的那样，setter采用直接的普通对象，因为我们生成了域对象的传统方式。

怎么办？

Answer 1

您可以改为编写

<img class="myImg" src="https://cdn.theatlantic.com/assets/media/img/mt/2017/06/GettyImages_675371680/lead_960.jpg?1498239007" alt="Trolltunga, Norway" width="300" height="200">

<img class="myImg" src="https://cdn.theatlantic.com/assets/media/img/mt/2017/06/GettyImages_675371680/lead_960.jpg?1498239007" alt="Trolltunga, Norway" width="300" height="200">

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可选表示可能缺少数据，并且具有使用此包装器的方法。有关正确使用可选is here.

的更多信息

Answer 2

你也可以写：

  private Todo findTodoEntryById(Long id) {
        Optional<Todo> todoResult = repository.findOne(id);
        return todoResult.orElseThrow(() -> new TodoNotFoundException(id));
}

如何使用Spring Data的Optional <object>返回

2 个答案: