表格：患者 ( id, 姓名, id_status, ...) -> FK 到 pattient_statuspattient_status (id, 描述) -> 目標(biāo)表我需要的只是在 pattient.class 中獲取 pattient_status.description，因?yàn)槲业?GET 方法需要 JSON 返回上的此信息。代碼：@Entity@Table(name="cad_paciente")public class Paciente {... (other columns)@OneToOne    @JoinColumn(insertable=false, updatable=false, name = "id_status_paciente", referencedColumnName = "id")    private StatusPaciente status;public String getStatusPaciente(){        return status.getStatus();    }----@Entity@Table(name="cad_status_paciente")public class StatusPaciente {    @Id    @GeneratedValue(strategy = GenerationType.IDENTITY)    private Long id;    @Column(name="ds_status")    @Size(max=50)    private String status;這正確列出了我的信息，但在 POST 方法上，JPA 正確保存但返回消息：Resolved [org.springframework.http.converter.HttpMessageNotWritableException: Could not write JSON: (was java.lang.NullPointerException); nested exception is com.fasterxml.jackson.databind.JsonMappingException: (was java.lang.NullPointerException) (through reference chain: com.spin.spincare.model.Paciente["statusPaciente"])]我應(yīng)該怎么辦？