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

Can not set java.lang.String field to java.lang.String

提问by Marcel

I am currently developing a small MMO application with a socket server. The database i am using is PostgreSQL and i am using the Hibernate ORM. I stumbled on to a exeption when requesting all the avatars an single user owns.

我目前正在开发一个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，我使用的是 Hibernate ORM。在请求单个用户拥有的所有头像时，我偶然发现了一个例外。

I got 3 classes involed, those are :

我参加了 3 个课程，它们是：

• GameServerClient
• Database
• Database.Queries

• 游戏服务器客户端
• 数据库
• 数据库.查询

When the user (client application) sends a request to the server via the sockets, a method is called which should return a JsonString of the all the Avatars.

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。

How ever, using the HQL query from UserOwnsAvatar where user = :usernameand puting the result in an ArrayList of the UserOwnsAvatar object it returns an Can not set java.lang.String field nl.marcusink.mmo.server.database.table.User.username to java.lang.String

但是，使用 HQL 查询from UserOwnsAvatar where user = :username并将结果放入 UserOwnsAvatar 对象的 ArrayList 中，它会返回一个Can not set java.lang.String field nl.marcusink.mmo.server.database.table.User.username to java.lang.String

the full stackTrace is :

完整的堆栈跟踪是：

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

the query code is :

查询代码是：

Query query = session.createQuery("from UserOwnsAvatar where user = :username");
query.setParameter("username", username);
ArrayList<UserOwnsAvatar> ownedAvatars = (ArrayList<UserOwnsAvatar>) query.list();

The last line is the cause of the error, any ideas?

最后一行是错误的原因，有什么想法吗？

EDIT

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;

The username here is equal to the username of the User object, which is :

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;

回答by SureshS

You will have to set the complete (or object with only relevant fields) instead of one particular value of that object.

您必须设置完整的（或仅具有相关字段的对象）而不是该对象的一个​​特定值。

What I understood is you are trying to set a Stringwhile setting the parameter, but the column is of type User. Hibernate is trying to call the getUsernamemethod on Stringand that's why the error.

我的理解是您正在尝试String设置参数，但该列的类型为User。Hibernate 正在尝试调用该getUsername方法String，这就是错误的原因。

So change your code to something like this:

所以把你的代码改成这样：

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

回答by Forkmohit

user = :username is incorrect. user is not a string field of the entity. Either use a join or use hibernate criteria to create a alias for User and add a restriction, for ex, .add(Restriction.eq("user.username",username).

user = :username 不正确。用户不是实体的字符串字段。使用加入或使用休眠条件为用户创建别名并添加限制，例如，.add(Restriction.eq("user.username",username)。