java - Hibernate抛出HibernateQueryException : could not resolve property

因此，我有一个这样的表，它已定义为 hibernate 状态的实体，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;
}

}

当我尝试针对此表编写一个简单查询时:

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);
}

我收到以下错误:

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

为什么不能 hibernate 找出类上的dbgroupid？

最佳答案

可能是因为您的getter(和setter)没有遵循javabeans约定。它应该是:

public long getDbgroupId() {
    return dbgroupid;
}

我的建议是-命名您的字段，然后使用您的IDE生成 setter 和 getter 。它将遵循惯例。 (另一件事，这是一个优先事项，但我认为这使一类课更易于阅读-为您的字段加上注释，而不是获取方法)

关于java - Hibernate抛出HibernateQueryException : could not resolve property，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/4441837/