【发布时间】:2015-11-27 13:27:57
【问题描述】:
我有一个让我发疯的错误。当我尝试创建一个新对象时,Hibernate 返回一个错误:
object 引用了一个未保存的瞬态实例 - 在刷新之前保存瞬态实例: EquipmentType;嵌套异常是 org.hibernate.TransientObjectException: 对象引用了一个未保存的瞬态实例 - 在此之前保存瞬态实例
我有以下 Spring 服务:
命令服务
public class CommandService implements ICommandService {
@Override
public Command createCommand(String name, String eqpName) {
Command cmd = new Command();
Equipment eqp = getEquipment(eqpName);
cmd.setEquipment(eqp);
cmd.setName(name);
return cmd;
}
@Override
public Equipment getEquipment(String eqpName) {
IEquipmentService eqpService = (IEquipmentService) SPRING_CONTEXT.getBean("EquipmentService");
Equipment eqp = eqpService.findEquipmentByName(eqpName);
EquipmentType eqpType = new EquipmentType();
eqpType.setName("MyType");
eqp.setType(eqpType);
Model model = new Model();
model.setName("MyModel");
eqp.setModel(model);
eqpService.saveEquipment(eqp);
return eqp;
}
}
设备服务
public class EquipmentService implements IEquipmentService {
private DAO dao; // This DAO is autowired
@Override
public void saveEquipment(Equipment eqp) {
completeModel(eqp);
completeEqpType(eqp);
((EquipmentDAO) dao).merge(eqp);
}
private void completeModel(Equipment eqp) {
IModelService modelService = (IModelService) SPRING_CONTEXT.getBean("ModelService");
Model result = modelService.findModelByName(eqp.getModel().getName()); // Fails here
eqp.setModel(result);
}
private void completeEqpType(Equipment eqp) {
IEquipmentTypeService eqpTypeService = (IEquipmentTypeService) SPRING_CONTEXT.getBean("EquipmentTypeService");
EquipmentType result = eqpTypeService.findEquipmentTypeByName(eqp.getType().getName());
eqp.setType(result);
}
}
请注意,IEquipmentService 和 ICommandService 的所有方法都配置为事务性的。
我的对象Equipment 由Model 和EquipmentType 组成。
这些对象引用了数据库中已经创建的对象,这就是我不想再次保存它们的原因。
这里是休眠配置:
<class name="Equipment" table="equipment" lazy="false">
<id name="equipmentId" type="java.lang.Integer">
<column name="EQUIPMENT_ID" />
<generator class="native" />
</id>
<many-to-one name="type" class="EquipmentType" fetch="join" not-null="true">
<column name="TYPE_ID" not-null="true" />
</many-to-one>
<many-to-one name="model" class="Model" fetch="join" not-null="true">
<column name="MODEL_ID" not-null="true" />
</many-to-one>
</class>
EDIT:根据要求,这里是equals(..)和hashcode(..)的方法
对于设备
@Override
public int hashCode() {
final int prime = 31;
int result = 1;
result = prime * result + ((equipmentId == null) ? 0 : equipmentId.hashCode());
result = prime * result + ((type == null) ? 0 : type.hashCode());
result = prime * result + ((model == null) ? 0 : model.hashCode());
return result;
}
@Override
public boolean equals(Object obj) {
if (obj == this) {
return true;
}
if (obj instanceof Equipment) {
Equipment eqp = (Equipment) obj;
return new EqualsBuilder().append(this.equipmentId, eqp.getEquipmentId()).isEquals();
}
return false;
}
对于 Model 和 EquipmentType,这些方法没有明确定义。
您能帮我弄清楚为什么当我调用createCommand(..) 时会返回此错误吗?
【问题讨论】:
-
你是如何为你的模型类定义 equals() 和 hashcode() 方法的?你能添加那些有问题的细节吗?
-
@mohit 我编辑了我的原始帖子。请注意,
EqualsBuilder由 Apache Commons 提供。