JPA Hibernate-排除类上的未知实体
[英]JPA Hibernate - Unknown Entity on excluded class
问题描述
When I run my test class called UserServiceTest I get this error. 
当我运行名为UserServiceTest的测试类时，出现此错误。 The weird thing is, I am not trying to persist an instance of the UserServiceTest type. 奇怪的是，我不是要持久保存UserServiceTest类型的实例。
The code which it fails on. 它失败的代码。 If I comment out this code the rest of the tests run and I dont get this error. 如果我注释掉此代码，其余的测试将运行，并且不会出现此错误。
User persistedUser = new User() {
{
setUsername("Jan123");
setName("Jan");
setText("test bio");
setLocation("<location>");
setWebsite("<website>");
}
};
entityManager.getTransaction().begin();
entityManager.persist(persistedUser);
entityManager.getTransaction().commit();
java.lang.IllegalArgumentException: Unknown entity: UserServiceTest$1 at org.hibernate.internal.SessionImpl.firePersist(SessionImpl.java:808) at org.hibernate.internal.SessionImpl.persist(SessionImpl.java:789) at UserServiceTest.setUp(UserServiceTest.java:55) at java.base/jdk.internal.reflect.NativeMethodAccessorImpl.invoke0(Native Method) at java.base/jdk.internal.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:62) at java.base/jdk.internal.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:43) at java.base/java.lang.reflect.Method.invoke(Method.java:566) at org.junit.runners.model.FrameworkMethod$1.runReflectiveCall(FrameworkMethod.java:50) at org.junit.internal.runners.model.ReflectiveCallable.run(ReflectiveCallable.java:12) at org.junit.runners.model.FrameworkMethod.invokeExplosively(FrameworkMethod.java:47) at org.junit.internal.runners.statements.RunBefores.evaluate(RunBefores.java:24) at org.junit.internal.runners.statements.RunAf
Here is my persistence.xml. 这是我的persistence.xml。 I checked that it is used by malforming the XML on purpose to see if it throws an XML parsing error, which it does. 我检查了是否故意使XML格式错误，以查看它是否引发XML解析错误。 Tables are created just fine, as I can see in the output of the build and in the SQL Server Management Studio. 正如我在构建输出和SQL Server Management Studio中所看到的那样，表创建得很好。
<?xml version="1.0" encoding="UTF-8"?>
<persistence version="2.1" xmlns="http://xmlns.jcp.org/xml/ns/persistence" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/persistence http://xmlns.jcp.org/xml/ns/persistence/persistence_2_1.xsd">
<persistence-unit name="KwetterPUTest" transaction-type="RESOURCE_LOCAL">
<provider>org.hibernate.jpa.HibernatePersistenceProvider</provider>
<class>Kweet</class>
<class>User</class>
<exclude-unlisted-classes>true</exclude-unlisted-classes>
<properties>
<property name="hibernate.connection.url" value="jdbc:sqlserver://localhost;databaseName=KwetterTest;" />
<property name="hibernate.connection.username" value="sa" />
<property name="hibernate.connection.password" value="admin" />
<property name="hibernate.show_sql" value="true"/>
<property name="javax.persistence.jdbc.driver" value="com.microsoft.sqlserver.jdbc.SQLServerDriver" />
<property name="javax.persistence.schema-generation.database.action" value="drop-and-create"/>
<property name="hibernate.dialect" value="org.hibernate.dialect.SQLServerDialect"/>
<property name="hibernate.hbm2ddl.auto" value="create"/>
</properties>
</persistence-unit>
</persistence>
Edit: Here is the relevant code from the UserServiceTest. 编辑：这是UserServiceTest中的相关代码。 I did a usage search in NetBeans for any instances of UserServiceTest that I perhaps wrongfully instantiated. 我在NetBeans中进行了用法搜索，以查找可能错误地实例化的UserServiceTest的任何实例。 There are none. 没有了。
private static EntityManagerFactory entityManagerFactory;
private static EntityManager entityManager;
private static UserService userService;
private static User persistedUser;
@BeforeClass
public static void setUp() {
entityManagerFactory = Persistence.createEntityManagerFactory("KwetterPUTest");
entityManager = entityManagerFactory.createEntityManager();
userService = new UserService(new UserDAOJPAImpl(entityManager));
persistedUser = new User() {
{
setUsername("Jan123");
setName("Jan");
setText("test bio");
setLocation("<location>");
setWebsite("<website>");
}
};
entityManager.getTransaction().begin();
entityManager.persist(persistedUser);
entityManager.getTransaction().commit();
}
Edit 2: pom file for those who are interested. 编辑2：感兴趣的人的pom文件。 I used the latest 4.x version of Hibernate as well as the latest 5.x and 5.2.2. 我使用了最新的4.x版本的Hibernate以及最新的5.x和5.2.2。 None changed anything. 没有任何改变。
<dependencies> <dependency> <groupId>junit</groupId> <artifactId>junit</artifactId> <version>4.12</version> <scope>test</scope> </dependency> <dependency> <groupId>org.hibernate</groupId> <artifactId>hibernate-core</artifactId> <version>5.4.2.Final</version> <scope>provided</scope> </dependency> <dependency> <groupId>com.microsoft.sqlserver</groupId> <artifactId>mssql-jdbc</artifactId> <version>7.2.1.jre8</version> <scope>test</scope> </dependency> <dependency> <groupId>javax</groupId> <artifactId>javaee-api</artifactId> <version>8.0</version> <type>jar</type> </dependency> </dependencies>
1 个解决方案
解决方案1
1 已采纳 2019-04-12 15:34:10
Why are instantiating the User class like this: 为什么要这样实例化User类：
persistedUser = new User() {
{
setUsername("Jan123");
setName("Jan");
setText("test bio");
setLocation("<location>");
setWebsite("<website>");
}
};
This will creates an inner class of type: UserServiceTest$1 这将创建一个内部类型： UserServiceTest$1
And this class is not recognized by Hibernate as the stack trace says: Hibernate无法识别此类，因为堆栈跟踪显示：
java.lang.IllegalArgumentException: Unknown entity: UserServiceTest$1
You must do it like this: 您必须这样做：
persistedUser = new User();
persistedUser.setUsername("Jan123");
persistedUser.setName("Jan");
persistedUser.setText("test bio");
persistedUser.setLocation("<location>");
persistedUser.setWebsite("<website>");
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