Question

我很难构建条件查询以从ID实体的Role实体获取“permissions”属性。权限属性属于Set类型，因此我创建了一个连接并从中进行选择，但查询失败，无效语法异常报告“where子句中的未知列'2L'”

生成错误的条件查询是以这种方式构建的：

EntityManager entityManager = getEntityManager();
CriteriaBuilder criteriaBuilder = entityManager.getCriteriaBuilder();
CriteriaQuery<Object> criteriaQuery = criteriaBuilder.createQuery();

Class<?> queryScopeClass = temp.pack.commons.user.Role.class;

Root<?> from = criteriaQuery.from(queryScopeClass);

Path<?> idAttrPath = from.get("id");
// also tried criteriaBuilder.equal(attributePath, new Long(2))
Predicate predicate = criteriaBuilder.equal(attributePath, criteriaBuilder.literal(new Long(2)))
criteriaQuery.where(predicate);


Path<?> attributePath = from.get("permissions");
PluralAttributePath<?> pluralAttrPath = (PluralAttributePath<?>)attributePath;
PluralAttribute<?, ?, ?> pluralAttr = pluralAttrPath.getAttribute();


Join<?, ?> join = from.join((SetAttribute<Object,?>)pluralAttr);

TypedQuery<Object> typedQuery = entityManager.createQuery(criteriaQuery.select(join));
return (List<P>)typedQuery.getResultList();

当我执行该行以实际返回结果时，抛出以下异常：

javax.persistence.PersistenceException: org.hibernate.exception.SQLGrammarException: could not execute query
    at org.hibernate.ejb.AbstractEntityManagerImpl.convert(AbstractEntityManagerImpl.java:1235)
    at org.hibernate.ejb.AbstractEntityManagerImpl.convert(AbstractEntityManagerImpl.java:1168)
    at org.hibernate.ejb.QueryImpl.getResultList(QueryImpl.java:250)
    at org.hibernate.ejb.criteria.CriteriaQueryCompiler$3.getResultList(CriteriaQueryCompiler.java:260)
    at temp.pack.dao.impl.DefaultDAOService.getProperties(DefaultDAOService.java:628)
...
Caused by: org.hibernate.exception.SQLGrammarException: could not execute query
    at org.hibernate.exception.SQLStateConverter.convert(SQLStateConverter.java:92)
    at org.hibernate.exception.JDBCExceptionHelper.convert(JDBCExceptionHelper.java:66)
    at org.hibernate.loader.Loader.doList(Loader.java:2452)
    at org.hibernate.loader.Loader.listIgnoreQueryCache(Loader.java:2192)
    at org.hibernate.loader.Loader.list(Loader.java:2187)
    at org.hibernate.hql.classic.QueryTranslatorImpl.list(QueryTranslatorImpl.java:936)
    at org.hibernate.engine.query.HQLQueryPlan.performList(HQLQueryPlan.java:196)
    at org.hibernate.impl.SessionImpl.list(SessionImpl.java:1258)
    at org.hibernate.impl.QueryImpl.list(QueryImpl.java:102)
    at org.hibernate.ejb.QueryImpl.getResultList(QueryImpl.java:241)
    ... 20 more
Caused by: com.mysql.jdbc.exceptions.jdbc4.MySQLSyntaxErrorException: Unknown column '2L' in 'where clause'
    at sun.reflect.NativeConstructorAccessorImpl.newInstance0(Native Method)
    at sun.reflect.NativeConstructorAccessorImpl.newInstance(Unknown Source)
    at sun.reflect.DelegatingConstructorAccessorImpl.newInstance(Unknown Source)
    at java.lang.reflect.Constructor.newInstance(Unknown Source)
    at com.mysql.jdbc.Util.handleNewInstance(Util.java:409)
    at com.mysql.jdbc.Util.getInstance(Util.java:384)
    at com.mysql.jdbc.SQLError.createSQLException(SQLError.java:1054)
    at com.mysql.jdbc.MysqlIO.checkErrorPacket(MysqlIO.java:3562)
    at com.mysql.jdbc.MysqlIO.checkErrorPacket(MysqlIO.java:3494)
    at com.mysql.jdbc.MysqlIO.sendCommand(MysqlIO.java:1960)
    at com.mysql.jdbc.MysqlIO.sqlQueryDirect(MysqlIO.java:2114)
    at com.mysql.jdbc.ConnectionImpl.execSQL(ConnectionImpl.java:2696)
    at com.mysql.jdbc.PreparedStatement.executeInternal(PreparedStatement.java:2105)
    at com.mysql.jdbc.PreparedStatement.executeQuery(PreparedStatement.java:2264)
    at com.jolbox.bonecp.PreparedStatementHandle.executeQuery(PreparedStatementHandle.java:179)
    at org.hibernate.jdbc.AbstractBatcher.getResultSet(AbstractBatcher.java:208)
    at org.hibernate.loader.Loader.getResultSet(Loader.java:1869)
    at org.hibernate.loader.Loader.doQuery(Loader.java:718)
    at org.hibernate.loader.Loader.doQueryAndInitializeNonLazyCollections(Loader.java:270)
    at org.hibernate.loader.Loader.doList(Loader.java:2449)
    ... 27 more

角色和权限类已使用JPA和一些Hibernate注释进行映射，如下所示：

public abstract class Role implements Serializable {

 /**
  * The id of this role. Internal use only.
  * 
  * @since 1.0
  */
 @Id @GeneratedValue
 protected long id;

 /**
  * Set of permissions granted to this role.
  * 
  * @since 1.0
  */
 @OneToMany(cascade = { CascadeType.PERSIST, CascadeType.MERGE }, mappedBy="sourceRole")
 protected Set<Permission> permissions = new HashSet<Permission>();

...

}

public class Permission implements Serializable {
 private static final long serialVersionUID = 1L;

 /**
  * The id of this permission. Used internally for persistence.
  * 
  * @since 1.0
  */
 @Id @GeneratedValue
 @Column(name = "PERMISSION_ID")
 protected long id;

 /**
  * The group to which the owner of this permission is being granted permission to.
  * 
  * @since 1.0
  */
 @ManyToOne(cascade = { CascadeType.PERSIST, CascadeType.MERGE })
 @JoinColumn(name = "TARGET_ROLE_ID")
 @ForeignKey(name = "FK_TARGET_GROUP_PERMISSION_ID",
   inverseName = "FK_PERMISSION_ID_TARGET_GROUP")
 protected Group targetGroup;

 /**
  * The role that has been granted this permission.
  * 
  * @since 1.0
  */
 @ManyToOne(cascade = { CascadeType.PERSIST, CascadeType.MERGE })
 @JoinColumn(name = "SOURCE_ROLE_ID")
 @ForeignKey(name = "FK_SOURCE_GROUP", inverseName = "FK_GROUP_PERMISSIONS")
 private Role sourceRole;

...

}

此条件查询有什么问题？

我希望调用typedQuery.getResultList（）来返回只包含一个元素的集合列表：id = 2的角色的权限对象集合。这是尝试选择（并初始化）只有来自id为2的对象角色的“permissions”集合。

我是标准查询的新手，我很难找到它的错误。

跟进： 这是Hibernate试图执行的查询：

select permission1_.PERMISSION_ID as PERMISSION1_12_,
    permission1_.IS_REQUIRED as IS2_12_,
    permission1_.SOURCE_ROLE_ID as SOURCE3_12_,
    permission1_.TARGET_ROLE_ID as TARGET4_12_
from (
        select ROLE_ID,
        NAME,
        DESCRIPTION,
        IS_ACTION,
        LABEL,
        null as FIRST_NAME,
        null as LAST_NAME,
        null as PASSWORD_HASH,
        1 as clazz_ from GROUPS
    union
        select ROLE_ID,
            NAME,
            null as DESCRIPTION,
            null as IS_ACTION,
            null as LABEL,
            FIRST_NAME,
            LAST_NAME,
            PASSWORD_HASH,
            2 as clazz_ from USERS
    )
role0_ inner join PERMISSIONS permission1_ on role0_.ROLE_ID=permission1_.SOURCE_ROLE_ID
    where (role0_.ROLE_ID=2L )

显然问题在于2L。从MySQL的角度看，2L实际上是一个字符串或一列，因为它上面有L并且没有引用。

这是Hibernate中的错误吗？它看起来像我。 L不应该在生成的SQL Query中...

任何人都对这发生了什么有任何想法？在我看来，这将是一个无法被其他人忽视的主要错误，所以我假设有一个很好的改变，我做错了。

谢谢！

爱德华

Answer 1

您可以尝试使用用于在条件中创建连接的两个表的别名。这也可能是问题，因为hibernate只为HQL，s和实体加载创建别名，但标准依赖于用户创建相同的。

Answer 2

通过将id属性数据类型从long替换为Long来解决问题。这个简单的更改使得Hibernate在查询中开始放置没有L的2，并且不再发生故障。

任何人都可以告诉为什么这个JPA条件查询在执行时会生成无效的SQL语句（Hibernate）以及如何修复它？

2 个答案: