- html - 出于某种原因,IE8 对我的 Sass 文件中继承的 html5 CSS 不友好?
- JMeter 在响应断言中使用 span 标签的问题
- html - 在 :hover and :active? 上具有不同效果的 CSS 动画
- html - 相对于居中的 html 内容固定的 CSS 重复背景?
我正在使用 java 上的 hibernate Jersey 和 Spring 构建 RESTful 服务,并意识到我的实体之间的关系应该不包括类的实例,而只是标识符(外键)。但在尝试查询实体时我有异常(exception)。我有以下实体:
@Entity
@javax.persistence.Table(name = "manager_user")
public class ManagerUser extends User {
@ManyToOne(targetEntity = ShopAdminUser.class)
private Integer shopAdminUserId;
//...
}
@Entity
@javax.persistence.Table(name = "shop_admin_user")
public class ShopAdminUser extends User {
@Lob
private String contactData;
public String getContactData() {
return contactData;
}
public void setContactData(String contactData) {
this.contactData = contactData;
}
}
@Entity
@Inheritance(strategy= InheritanceType.TABLE_PER_CLASS)
public abstract class User {
@Id
@GeneratedValue(strategy = GenerationType.TABLE)
private Integer id;
private String firstName;
private String lastName;
@Temporal(TemporalType.DATE)
private Date dob;
@Enumerated(EnumType.STRING)
private Gender gender;
@Column(unique = true, nullable = false)
private String email;
@Column(nullable = false)
private String password;
@Lob
private String personalData;
@OneToOne(targetEntity = Photo.class)
private Integer photoId;
//...getters and setters
}
我在下一个方法中遇到异常:
@Transactional(readOnly = true)
public List<ManagerUser> getByShopAdminUserID(Integer id) {
Session session = sessionFactory.getCurrentSession();
List managerUsers = session.createQuery(
"from ManagerUser m where m.shopAdminUserId = :shopAdminUserId")
.setParameter("shopAdminUserId", id).list();
return managerUsers;
}
堆栈跟踪:
Mar 28, 2015 4:10:56 PM com.sun.jersey.spi.container.ContainerResponse mapMappableContainerException
SEVERE: The RuntimeException could not be mapped to a response, re-throwing to the HTTP container
org.hibernate.PropertyAccessException: could not get a field value by reflection getter of com.foodservice.businesslogic.user.ShopAdminUser.id
at org.hibernate.property.DirectPropertyAccessor$DirectGetter.get(DirectPropertyAccessor.java:60)
at org.hibernate.tuple.entity.AbstractEntityTuplizer.getIdentifier(AbstractEntityTuplizer.java:346)
at org.hibernate.persister.entity.AbstractEntityPersister.getIdentifier(AbstractEntityPersister.java:4746)
at org.hibernate.persister.entity.AbstractEntityPersister.isTransient(AbstractEntityPersister.java:4465)
at org.hibernate.engine.internal.ForeignKeys.isTransient(ForeignKeys.java:243)
at org.hibernate.engine.internal.ForeignKeys.getEntityIdentifierIfNotUnsaved(ForeignKeys.java:293)
at org.hibernate.type.EntityType.getIdentifier(EntityType.java:537)
at org.hibernate.type.ManyToOneType.nullSafeSet(ManyToOneType.java:174)
at org.hibernate.param.NamedParameterSpecification.bind(NamedParameterSpecification.java:67)
at org.hibernate.loader.hql.QueryLoader.bindParameterValues(QueryLoader.java:616)
at org.hibernate.loader.Loader.prepareQueryStatement(Loader.java:1901)
at org.hibernate.loader.Loader.executeQueryStatement(Loader.java:1862)
at org.hibernate.loader.Loader.executeQueryStatement(Loader.java:1839)
at org.hibernate.loader.Loader.doQuery(Loader.java:910)
at org.hibernate.loader.Loader.doQueryAndInitializeNonLazyCollections(Loader.java:355)
at org.hibernate.loader.Loader.doList(Loader.java:2554)
at org.hibernate.loader.Loader.doList(Loader.java:2540)
at org.hibernate.loader.Loader.listIgnoreQueryCache(Loader.java:2370)
at org.hibernate.loader.Loader.list(Loader.java:2365)
at org.hibernate.loader.hql.QueryLoader.list(QueryLoader.java:497)
at org.hibernate.hql.internal.ast.QueryTranslatorImpl.list(QueryTranslatorImpl.java:387)
at org.hibernate.engine.query.spi.HQLQueryPlan.performList(HQLQueryPlan.java:236)
at org.hibernate.internal.SessionImpl.list(SessionImpl.java:1264)
at org.hibernate.internal.QueryImpl.list(QueryImpl.java:103)
at com.foodservice.dao.ManagerUserDAO.getByShopAdminUserID(ManagerUserDAO.java:50)
at com.foodservice.dao.ManagerUserDAO$$FastClassBySpringCGLIB$$7447621a.invoke(<generated>)
...
Caused by: java.lang.IllegalArgumentException: Can not set java.lang.Integer field com.foodservice.businesslogic.user.User.id to java.lang.Integer
at sun.reflect.UnsafeFieldAccessorImpl.throwSetIllegalArgumentException(UnsafeFieldAccessorImpl.java:167)
at sun.reflect.UnsafeFieldAccessorImpl.throwSetIllegalArgumentException(UnsafeFieldAccessorImpl.java:171)
at sun.reflect.UnsafeFieldAccessorImpl.ensureObj(UnsafeFieldAccessorImpl.java:58)
at sun.reflect.UnsafeObjectFieldAccessorImpl.get(UnsafeObjectFieldAccessorImpl.java:36)
at java.lang.reflect.Field.get(Field.java:393)
at org.hibernate.property.DirectPropertyAccessor$DirectGetter.get(DirectPropertyAccessor.java:57)
... 120 more
编辑
Hibernate提供的My SQL
Hibernate:
select
manageruse0_.id as id1_0_,
manageruse0_.dob as dob2_0_,
manageruse0_.email as email3_0_,
manageruse0_.firstName as firstNam4_0_,
manageruse0_.gender as gender5_0_,
manageruse0_.lastName as lastName6_0_,
manageruse0_.password as password7_0_,
manageruse0_.personalData as personal8_0_,
manageruse0_.photoId_id as photoId11_0_,
manageruse0_.systemStatus as systemSt9_0_,
manageruse0_.userType as userTyp10_0_,
manageruse0_.shopAdminUserId_id as shopAdmi2_6_,
manageruse0_.state as state1_6_
from
manager_user manageruse0_
where
manageruse0_.shopAdminUserId_id=?
Hibernate:
select
shopadminu0_.id as id1_0_0_,
shopadminu0_.dob as dob2_0_0_,
shopadminu0_.email as email3_0_0_,
shopadminu0_.firstName as firstNam4_0_0_,
shopadminu0_.gender as gender5_0_0_,
shopadminu0_.lastName as lastName6_0_0_,
shopadminu0_.password as password7_0_0_,
shopadminu0_.personalData as personal8_0_0_,
shopadminu0_.photoId_id as photoId11_0_0_,
shopadminu0_.systemStatus as systemSt9_0_0_,
shopadminu0_.userType as userTyp10_0_0_,
shopadminu0_.contactData as contactD1_16_0_,
photo1_.id as id1_9_1_,
photo1_.image as image2_9_1_,
photo1_.name as name3_9_1_
from
shop_admin_user shopadminu0_
left outer join
photo photo1_
on shopadminu0_.photoId_id=photo1_.id
where
shopadminu0_.id=?
我做错了什么?帮助我)
最佳答案
我有一个似乎问题。现在我解决了。很简单。你的代码-->
@ManyToOne(targetEntity = ShopAdminUser.class)
private Integer shopAdminUserId;
应该是
@ManyToOne(targetEntity = ShopAdminUser.class)
@JoinColumn(name="shopAdminUserId" references...=""(i forget how to spell))
private ShopAdminUser shopAdminUser;
这也是如此:
@OneToOne(targetEntity = Photo.class)
private Integer photoId;
应该是
@OneToOne(targetEntity = Photo.class)
@JoinColumn(name="photoId" references...=""(i forget how to spell))
private Photo photo;
因为 hibernate 会自动将类转换为 int 或其他类型
关于java - Hibernate 中无法将 java.lang.Integer 字段...User.id 设置为 java.lang.Integer 异常,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/29318362/
在 Tomcat 6/Ubuntu 12.04 上启动 Grails 2.1.0 应用程序时出现以下错误。 Error 500 - Internal Server Error. groovy.lang
在运行 Storm 拓扑时,我收到此错误。拓扑完美运行 5 分钟,没有任何错误,然后失败。我正在使用 Config.TOPOLOGY_TICK_TUPLE_FREQ_SECS as 300 sec i
我有一个 jsp 代码在其中一台机器上运行良好。但是当我复制到另一台机器时,我得到了这个 no such method found 异常。我是 Spring 的新手。有人可以解释我错过了什么吗? 以下
已关闭。此问题需要 debugging details 。目前不接受答案。 编辑问题以包含 desired behavior, a specific problem or error, and the
我的代码在下面给出了一个错误; Exception in thread "main" java.lang.NoSuchMethodError: com/myApp/Client.cypherCBC(L
我正在尝试一个 Restful web 服务示例,所以当我要访问 url 时,我遇到了异常 java.lang.NoSuchMethodError: jersey.repackaged.com.goo
我正在将一个 Spring web 项目转换为一个 Maven 项目,但我收到了这个错误: java.lang.NoSuchMethodError: org.jboss.logging.Logger.
在我的项目中,我有一个像这样的枚举: public enum MyEnum { FIRST(1), SECOND(2); private int value; private MyEnum(int v
我创建了这个简单的示例,用于读取 Linux 正常运行时间: public String getMachineUptime() throws IOException { String[] di
我正在使用 Eclipse,并且正在使用 Java。我的目标是使用 bogoSort 方法对 vector 进行排序在一个 vector (vectorExample)中适应我的 vector 类型,
我正在运行以下查询。它显示一条错误消息。如何解决这个错误? ListrouteList=null; List companyList = session.createS
我有以下模型类: @Entity @Table(name="user_content") @org.hibernate.annotations.NamedQueries({ @org.
我有那个错误。这是我的代码: GmailSettingsService service = new GmailSettingsService(APPLICATION_NAME, DOMAIN_NAME
实际上我在执行我的java程序时遇到了下面提到的错误 Exception in thread "pool-1-thread-1" java.lang.ClassCastException: jav
java.lang.ClassCastException: java.lang.Float cannot be cast to java.lang.String 我在以下代码中遇到此异常: Strin
我正在尝试从 linkedhashset 中检索随机元素。下面是我的代码,但它每次都给我异常。 private static void generateRandomUserId(Set userIds
我已经完成了 Android 中的代码: List spinnerArray = new ArrayList(); for (int i = 0; i item = (LinkedTreeMap)
这个问题已经有答案了: Explanation of ClassCastException in Java (12 个回答) 已关闭 6 年前。 我已经编写了 java 到 Json 的代码,同时从页
这个问题在这里已经有了答案: ClassCastException java.lang.Long cannot be cast to clojure.lang.IFn (4 个答案) 关闭 6 年前
我在运行时遇到问题来编译这段代码,这给我一个错误,java.lang.Integer 无法转换为 Java.lang.Double。如果有人帮助我更正此代码,我将非常高兴 double x; pu
我是一名优秀的程序员,十分优秀!