java java.lang.string_无法将java.lang.String字段设置为java.lang.String

我目前正在开发带有套接字服务器的小型MMO应用程序.我使用的数据库是PostgreSQL,而我使用的是Hibernate ORM.

当请求一个用户拥有的所有头像时,我偶然发现了一个例外.

我参与了3个班级的学习,分别是：

> GameServerClient

>数据库

>数据库查询

当用户(客户端应用程序)通过套接字向服务器发送请求时,将调用一个方法,该方法应返回所有头像的JsonString.

但是,使用来自UserOwnsAvatar的HQL查询(其中user =：username)并将结果放入UserOwnsAvatar对象的ArrayList中,它将返回无法设置的java.lang.String字段nl.marcusink.mmo.server.database.table.User .username到java.lang.String

完整的stackTrace是：

org.hibernate.property.access.spi.PropertyAccessException: Error accessing field [private java.lang.String nl.marcusink.mmo.server.database.table.User.username] by reflection for persistent property [nl.marcusink.mmo.server.database.table.User#username] : Mjollnir94

at org.hibernate.property.access.spi.GetterFieldImpl.get(GetterFieldImpl.java:43)

at org.hibernate.tuple.entity.AbstractEntityTuplizer.getIdentifier(AbstractEntityTuplizer.java:223)

at org.hibernate.persister.entity.AbstractEntityPersister.getIdentifier(AbstractEntityPersister.java:4594)

at org.hibernate.type.EntityType.toLoggableString(EntityType.java:505)

at org.hibernate.internal.util.EntityPrinter.toString(EntityPrinter.java:87)

at org.hibernate.engine.spi.QueryParameters.traceParameters(QueryParameters.java:281)

at org.hibernate.engine.query.spi.HQLQueryPlan.performList(HQLQueryPlan.java:194)

at org.hibernate.internal.SessionImpl.list(SessionImpl.java:1268)

at org.hibernate.internal.QueryImpl.list(QueryImpl.java:87)

at nl.marcusink.mmo.server.database.Database$Queries.avatarsRequest(Database.java:134)

at nl.marcusink.mmo.server.connection.GameServerClient.run(GameServerClient.java:91)

at java.lang.Thread.run(Thread.java:745)

Caused by: java.lang.IllegalArgumentException: Can not set java.lang.String field nl.marcusink.mmo.server.database.table.User.username to java.lang.String

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.access.spi.GetterFieldImpl.get(GetterFieldImpl.java:39)

... 11 more

查询代码为：

Query query = session.createQuery("from UserOwnsAvatar where user = :username");

query.setParameter("username", username);

ArrayList ownedAvatars = (ArrayList) query.list();

最后一行是错误的原因,有什么想法吗？

编辑

@Id

@ManyToOne(targetEntity = User.class)

@JoinColumn(name = "username", nullable = false)

private User user;

@Id

@OneToOne(targetEntity = Avatar.class)

@JoinColumn(name = "avatar", nullable = false, unique = true)

private Avatar avatar;

这里的用户名等于User对象的用户名,即：

@Id

@Column(name = "username", unique = true, nullable = false)

private String username;

解决方法:

您将必须设置完整(或仅具有相关字段的对象),而不是该对象的一个​​特定值.

我了解的是您正在尝试在设置参数时设置字符串,但是列的类型为User. Hibernate试图在String上调用getUsername方法,这就是错误的原因.

因此,将您的代码更改为如下所示：

User user = getSomeUser();

Query query = session.createQuery("from UserOwnsAvatar where user.username = :username");

query.setParameter("username", user);

标签：sockets,server,hibernate,java

来源： https://codeday.me/bug/20191119/2036948.html