Java Hibernate引发HibernateQueryException:无法解析属性_Java_Hibernate_Spring

Java Hibernate引发HibernateQueryException:无法解析属性

java hibernate spring

Java Hibernate引发HibernateQueryException:无法解析属性,java,hibernate,spring,Java,Hibernate,Spring,所以我有一个表，我在hibernate中将它定义为一个实体，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(

所以我有一个表，我在hibernate中将它定义为一个实体，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;
}

}

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);
}

当我试图针对该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);
}

为什么hibernate不能找出我的类上的dbgroupid是什么？

可能是因为您将它标记为“dbgroupid”，而不是“dbgroupid”？

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;
}

它尝试针对对象解析条件中给定的名称。因此，在typesByPropertyPath映射中，我找到了以下值：

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的大小写与我的条件中的大小写不匹配。所以我将其从dbgroupid更改为dbgroupid，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;
}

}

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);
}

public集合getPreferencesForDBGroup（长dbgroupId）{
最终DetachedCriteria=DetachedCriteria.forClass（Preference.class）
.add（Restrictions.eq（“DBGroupId”，DBGroupId））
.setResultTransformer（DistinctRootEntityResultTransformer.INSTANCE）；
返回getHibernateTemplate（）.findByCriteria（条件）；
}

现在它起作用了

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

public long getDbgroupId() {
    return dbgroupid;
}

我的建议是——命名字段，然后使用IDE生成setter和getter。它将遵循公约。（另一件事，这是一个偏好的问题，但在我看来，使类更易于阅读-注释字段，而不是getter）

这是因为我在getXXX方法中声明了它吗？这就是它得到资本化的地方吗？@Ralph-是的，这是不被禁止的。但我把他的域名作为一个起点，因此我建议修改getter。可能是另一种情况。但是你是对的，开头多个大写字母是很棘手的。你是对的-候补小姐是你的答案吗-所以我删除了我的评论。我有一个类似的问题，我使用了全大写字母，没有找到正确的getter:

.add（Restrictions.eq（“DBGROUPID”，DBGROUPID））

与其让IDE生成getter和setter，不如更好地使用