我正在尝试以下JPQL,但它失败了:
Query query = em.createQuery("SELECT u FROM User u WHERE 'admin' MEMBER OF u.roles");
List users = query.query.getResultList();
我得到以下异常:
ERROR [main] PARSER.error(454) | <AST>:0:0: unexpected end of subtree
java.lang.IllegalArgumentException: org.hibernate.hql.ast.QuerySyntaxException: unexpected end of subtree [SELECT u FROM com.online.data.User u WHERE 'admin' MEMBER OF u.roles] ERROR [main] PARSER.error(454) | <AST>:0:0: expecting "from", found '<ASTNULL>'
...
...
Caused by: org.hibernate.hql.ast.QuerySyntaxException: unexpected end of subtree [SELECT u FROM com.online.data.User u WHERE 'admin' MEMBER OF u.roles]
我有Spring 3.0.1.RELEASE,Hibernate 3.5.1-Final和maven来粘合依赖项。
用户类:
@Entity
public class User {
@Id
@Column(name = "USER_ID")
@GeneratedValue(strategy = GenerationType.IDENTITY)
private long id;
@Column(unique = true, nullable = false)
private String username;
private boolean enabled;
@ElementCollection
private Set<String> roles = new HashSet<String>();
...
}
Spring配置:
<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:context="http://www.springframework.org/schema/context"
xmlns:tx="http://www.springframework.org/schema/tx" xmlns:p="http://www.springframework.org/schema/p" xmlns:aop="http://www.springframework.org/schema/aop"
xsi:schemaLocation="
http://www.springframework.org/schema/beans
http://www.springframework.org/schema/beans/spring-beans-3.0.xsd
http://www.springframework.org/schema/context
http://www.springframework.org/schema/tx/spring-context-3.0.xsd
http://www.springframework.org/schema/tx
http://www.springframework.org/schema/tx/spring-tx-3.0.xsd
http://www.springframework.org/schema/aop
http://www.springframework.org/schema/aop/spring-aop-3.0.xsd">
<!-- Reading annotation driven configuration -->
<tx:annotation-driven transaction-manager="transactionManager" />
<bean class="org.springframework.dao.annotation.PersistenceExceptionTranslationPostProcessor" />
<bean class="org.springframework.orm.jpa.support.PersistenceAnnotationBeanPostProcessor" />
<bean id="dataSource" class="org.apache.commons.dbcp.BasicDataSource" destroy-method="close">
<property name="driverClassName" value="${jdbc.driverClassName}" />
<property name="url" value="${jdbc.url}" />
<property name="username" value="${jdbc.username}" />
<property name="password" value="${jdbc.password}" />
<property name="maxActive" value="100" />
<property name="maxWait" value="1000" />
<property name="poolPreparedStatements" value="true" />
<property name="defaultAutoCommit" value="true" />
</bean>
<bean id="transactionManager" class="org.springframework.orm.jpa.JpaTransactionManager">
<property name="entityManagerFactory" ref="entityManagerFactory" />
<property name="dataSource" ref="dataSource" />
</bean>
<bean id="entityManagerFactory" class="org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean">
<property name="dataSource" ref="dataSource" />
<property name="jpaVendorAdapter">
<bean class="org.springframework.orm.jpa.vendor.HibernateJpaVendorAdapter">
<property name="showSql" value="true" />
<property name="databasePlatform" value="${hibernate.dialect}" />
</bean>
</property>
<property name="loadTimeWeaver">
<bean class="org.springframework.instrument.classloading.InstrumentationLoadTimeWeaver" />
</property>
<property name="jpaProperties">
<props>
<prop key="hibernate.hbm2ddl.auto">update</prop>
<prop key="hibernate.current_session_context_class">thread</prop>
<prop key="hibernate.cache.provider_class">org.hibernate.cache.NoCacheProvider</prop>
<prop key="hibernate.show_sql">true</prop>
<prop key="hibernate.format_sql">false</prop>
<prop key="hibernate.show_comments">true</prop>
</props>
</property>
<property name="persistenceUnitName" value="punit" />
</bean>
<bean id="JpaTemplate" class="org.springframework.orm.jpa.JpaTemplate">
<property name="entityManagerFactory" ref="entityManagerFactory" />
</bean>
</beans>
的persistence.xml
<?xml version="1.0" encoding="UTF-8"?>
<persistence xmlns="http://java.sun.com/xml/ns/persistence" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://java.sun.com/xml/ns/persistence">
<persistence-unit name="punit" transaction-type="RESOURCE_LOCAL" />
</persistence>
pom.xml maven依赖项。
<dependency>
<groupId>org.hibernate</groupId>
<artifactId>hibernate</artifactId>
<version>${hibernate.version}</version>
<type>pom</type>
</dependency>
<dependency>
<groupId>org.hibernate</groupId>
<artifactId>hibernate-core</artifactId>
<version>${hibernate.version}</version>
</dependency>
<dependency>
<groupId>org.hibernate</groupId>
<artifactId>hibernate-annotations</artifactId>
<version>${hibernate.version}</version>
</dependency>
<dependency>
<groupId>org.hibernate</groupId>
<artifactId>hibernate-entitymanager</artifactId>
<version>${hibernate.version}</version>
</dependency>
<dependency>
<groupId>commons-dbcp</groupId>
<artifactId>commons-dbcp</artifactId>
<version>1.2.2</version>
<type>jar</type>
</dependency>
<dependency>
<groupId>org.springframework.security</groupId>
<artifactId>spring-security-web</artifactId>
<version>${spring.version}</version>
</dependency>
<dependency>
<groupId>org.springframework.security</groupId>
<artifactId>spring-security-config</artifactId>
<version>${spring.version}</version>
</dependency>
<dependency>
<groupId>org.springframework.security</groupId>
<artifactId>spring-security-taglibs</artifactId>
<version>${spring.version}</version>
</dependency>
<dependency>
<groupId>org.springframework.security</groupId>
<artifactId>spring-security-acl</artifactId>
<version>${spring.version}</version>
</dependency>
<dependency>
<groupId>javax.annotation</groupId>
<artifactId>jsr250-api</artifactId>
<version>1.0</version>
</dependency>
<properties>
<!-- Application settings -->
<spring.version>3.0.1.RELEASE</spring.version>
<hibernate.version>3.5.1-Final</hibernate.version>
运行单元测试以检查配置,我能够运行其他JPQL查询,我唯一无法运行的是IS EMPTY,成员条件。
完整的单元测试如下:
TestIntegration
@RunWith(SpringJUnit4ClassRunner.class)
@ContextConfiguration(locations = { "/spring/dataLayer.xml"})
@Transactional
@TransactionConfiguration
public class TestUserDaoImplIntegration {
@PersistenceContext
private EntityManager em;
@Test
public void shouldTest() throws Exception {
try {
//WORKS
Query query = em.createQuery("SELECT u FROM User u WHERE 'admin' in elements(u.roles)");
List users = query.query.getResultList();
} catch (Exception e) {
e.printStackTrace();
throw e;
}
try {
//DOES NOT WORK
Query query = em.createQuery("SELECT u FROM User u WHERE 'admin' MEMBER OF u.roles");
List users = query.query.getResultList();
} catch (Exception e) {
e.printStackTrace();
throw e;
}
}
}
答案 0 :(得分:10)
您的查询对我来说非常好。为了记录,这是JPA 2.0规范写的关于MEMBER OF
运算符的内容:
4.6.13收集成员表达
使用的语法 collection_member_expression中的比较运算符
MEMBER OF
如下:collection_member_expression ::= entity_or_value_expression [NOT] MEMBER [OF] collection_valued_path_expression entity_or_value_expression ::= single_valued_object_path_expression | state_field_path_expression | simple_entity_or_value_expression simple_entity_or_value_expression ::= identification_variable | input_parameter | literal
此表达式测试指定的值是否为其成员 由...指定的集合 集合值路径表达式。
评估为的表达式 不支持嵌入类型 集合成员表达式。支持 在收集中使用嵌入物 成员表达式可以添加到 此规范的未来版本。
如果收集价值路径 表达式指定一个空的 集合,
MEMBER
的值 OF表达式为FALSE
和值NOT MEMBER OF
表达式是TRUE
。否则,如果是的值 集合成员表达式中的collection_valued_path_expression或entity_or_value_expression为NULL 或未知,价值 集合成员表达式是 未知的。示例:
SELECT p FROM Person p WHERE 'Joe' MEMBER OF p.nicknames
因此,因为我在查询中看不到任何错误,所以我使用EclipseLink 1 测试了您的代码,以下代码片段正常运行:
Query query = em.createQuery("SELECT u FROM User u WHERE 'admin' MEMBER OF u.roles");
List list = query.getResultList();
但是Hibernate EntityManager 3.5.1-Final确实失败了。这听起来像个错误,随意提出a Jira issue。
1 以防万一,我使用了以下Maven配置文件(对于JPA提供商):
<profile>
<id>eclipselink</id>
<repositories>
<repository>
<id>eclipselink</id>
<url>http://www.eclipse.org/downloads/download.php?r=1&nf=1&file=/rt/eclipselink/maven.repo/</url>
</repository>
</repositories>
<dependencies>
<!-- See http://wiki.eclipse.org/EclipseLink/Maven -->
<dependency>
<groupId>org.eclipse.persistence</groupId>
<artifactId>eclipselink</artifactId>
<version>2.0.0</version>
</dependency>
<!-- optional - only needed if you are using JPA outside of a Java EE container-->
<dependency>
<groupId>org.eclipse.persistence</groupId>
<artifactId>javax.persistence</artifactId>
<version>2.0.0</version>
<scope>provided</scope>
</dependency>
</dependencies>
</profile>
这是我的persistence.xml
:
<?xml version="1.0" encoding="UTF-8"?>
<persistence
xmlns="http://java.sun.com/xml/ns/persistence"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://java.sun.com/xml/ns/persistence http://java.sun.com/xml/ns/persistence/persistence_2_0.xsd"
version="2.0">
<persistence-unit name="TestPu" transaction-type="RESOURCE_LOCAL">
<class>com.stackoverflow.q2688144.User</class>
<properties>
<property name="javax.persistence.jdbc.driver" value="org.apache.derby.jdbc.EmbeddedDriver"/>
<property name="javax.persistence.jdbc.url" value="jdbc:derby:testdb;create=true"/>
<property name="eclipselink.target-database" value="DERBY"/>
<property name="eclipselink.ddl-generation" value="drop-and-create-tables"/>
</properties>
</persistence-unit>
</persistence>
在HHH-5209 中报告了更新
答案 1 :(得分:7)
Hibernate Bug #HHH-5209, 解决方法使用以下语法:
select user from User user where :role in elements(user.roles)
答案 2 :(得分:3)
您的查询绝对正确,应该有效。我今天遇到了同样的问题,感谢这篇文章,我能够不停地撞到墙上。
由于@ElementCollection对于具有内在关联的连接元素来说确实是一个很好的快捷方式,因此您仍然可以在查询中使用连接语义。
为了更好地说明这一点,您需要加入您的收藏集,并在其中设置条件。
把它变成更具体的东西...
您的查询:
SELECT u FROM User u WHERE 'admin' MEMBER OF u.roles
可以写成:
SELECT u FROM User u JOIN u.roles r WHERE r = 'admin'
希望这有帮助!
答案 3 :(得分:0)
我认为member of
不能与@ElementCollection
属性一起使用。