Hibernate引发HibernateQueryException:无法解析属性
Hibernate引发HibernateQueryException:无法解析属性
提问于 2010-12-15 00:57:37
所以我有一个在hibernate中定义为实体的表，如下所示：
@Entity
@Table(name = "sec_Preference")
public class Preference {
private long id;
@Column(name = "PreferenceId", nullable = false, insertable = true, updatable = true, length = 19, precision = 0)
@GeneratedValue(strategy = GenerationType.AUTO)
@Id
public long getId() {
return id;
}
public void setId(long id) {
this.id = id;
}
private long systemuserid;
@Column(name = "SystemUserId", nullable = true, insertable = true, updatable = true, length = 19, precision = 0)
@Basic
public long getSystemUserId() {
return systemuserid;
}
public void setSystemUserId(long systemuserid) {
this.systemuserid = systemuserid;
}
private long dbgroupid;
@Column(name = "DBGroupId", nullable = true, insertable = true, updatable = true, length = 19, precision = 0)
@Basic
public long getDBGroupId() {
return dbgroupid;
}
public void setDBGroupId(long dbgroupid) {
this.dbgroupid = dbgroupid;
}
private long externalgroupid;
@Column(name = "ExternalGroupId", nullable = true, insertable = true, updatable = true, length = 19, precision = 0)
@Basic
public long getExternalGroupId() {
return externalgroupid;
}
public void setExternalGroupId(long externalgroupid) {
this.externalgroupid = externalgroupid;
}
private long securityroleid;
@Column(name = "SecurityRoleId", nullable = true, insertable = true, updatable = true, length = 19, precision = 0)
@Basic
public long getSecurityRoleId() {
return securityroleid;
}
public void setSecurityRoleId(long securityroleid) {
this.securityroleid = securityroleid;
}
public void setEnum(com.vitalimages.common.server.security.Preference pref) {
this.preferencekey = pref.name();
}
private String preferencekey;
@Column(name = "PreferenceKey", nullable = false, insertable = true, updatable = true, length = 255, precision = 0)
@Basic
public String getKey() {
return preferencekey;
}
public void setKey(String key) {
this.preferencekey = key;
}
private String preferencevalue;
@Column(name = "PreferenceValue", nullable = true, insertable = true, updatable = true, length = 255, precision = 0)
@Basic
public String getValue() {
return preferencevalue;
}
public void setValue(String value) {
this.preferencevalue = value;
}
}当我试图对这个表编写一个简单的查询时：
public Collection<Preference> getPreferencesForDBGroup(long dbgroupId) {
final DetachedCriteria criteria = DetachedCriteria.forClass(Preference.class)
.add(Restrictions.eq("dbgroupid", dbgroupId))
.setResultTransformer(DistinctRootEntityResultTransformer.INSTANCE);
return getHibernateTemplate().findByCriteria(criteria);
}我得到了以下错误：
org.springframework.orm.hibernate3.HibernateQueryException: could not resolve property: dbgroupid of: com.common.server.domain.sec.Preference; nested exception is org.hibernate.QueryException: could not resolve property: dbgroupid of: com.common.server.domain.sec.Preference为什么hibernate不能弄清楚我的类上的dbgroupid是什么？
回答 3
Stack Overflow用户
回答已采纳
发布于 2010-12-15 01:03:01
这可能是因为你的getter (和setter)没有遵循javabean的约定。它应该是：
public long getDbgroupId() {
return dbgroupid;
}我的建议是-命名您的字段，然后使用IDE生成setter和getter。它将遵循惯例。(另一件事，这是一个偏好问题，但在我看来，让类更容易阅读-注释你的字段，而不是getter)
Stack Overflow用户
发布于 2010-12-15 01:02:36
我在这方面取得了一些进展，但我仍然不明白hibernate的名字是从哪里来的。我对hibernate内部进行了调试，发现了以下类：
org.hibernate.persister.entity.AbstractPropertyMapping在这个类中有一个方法：
public Type toType(String propertyName) throws QueryException {
Type type = (Type) typesByPropertyPath.get(propertyName);
if (type == null) {
throw propertyException(propertyName);
}
return type;
}它尝试根据对象解析标准中给出的名称。因此，在typesByPropertyPath映射中，我发现了以下值：
id -> DBGroupId=org.hibernate.type.LongType@1e96ffd
key -> value=org.hibernate.type.StringType@aa2ee4
value -> value=org.hibernate.type.StringType@aa2ee4
systemUserId -> DBGroupId=org.hibernate.type.LongType@1e96ffd
securityRoleId -> DBGroupId=org.hibernate.type.LongType@1e96ffd
externalGroupId -> DBGroupId=org.hibernate.type.LongType@1e96ffd
DBGroupId -> DBGroupId=org.hibernate.type.LongType@1e96ffd在这里，您可以看到DBGroupId的大小写与我的标准中的大小写不匹配。所以我把它从dbgroupid改为DBGroupId，如下所示：
public Collection<Preference> getPreferencesForDBGroup(long dbgroupId) {
final DetachedCriteria criteria = DetachedCriteria.forClass(Preference.class)
.add(Restrictions.eq("DBGroupId", dbgroupId))
.setResultTransformer(DistinctRootEntityResultTransformer.INSTANCE);
return getHibernateTemplate().findByCriteria(criteria);
}现在它起作用了。
Stack Overflow用户
发布于 2010-12-15 01:00:47
也许是因为您给它贴上了"DBGroupId“的标签，而不是"dbgroupid"？
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原文链接：
https://stackoverflow.com/questions/4441837
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