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Home/ Questions/Q 3496476
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Editorial Team
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Editorial Team
Asked: May 18, 20262026-05-18T12:15:10+00:00 2026-05-18T12:15:10+00:00

So I have a table that I’ve defined as an entity in hibernate like

  • 0

So I have a table that I’ve defined as an entity in hibernate like this:

@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;
}

}

When I tried to write a simple query against this table:

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);
}

I got the following error:

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

Why can’t hibernate figure out what dbgroupid is on my class?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-18T12:15:11+00:00Added an answer on May 18, 2026 at 12:15 pm

    It’s probably because your getter (and setter) is not following the javabeans convention. It should be:

    public long getDbgroupId() {
        return dbgroupid;
    }
    

    What I’d suggest is – name your fields, and then use your IDE to generate setters and getters. It will follow the convention. (Another thing, that is a matter of preference, but in my opinion makes a class easier to read – annotate your fields, not getters)

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