尝试构建hibernate条件查询时出现的问题，该查询通过与联接表实体和复合键的多对多关联进行导航

Question

你能帮我解决这个问题吗？

以下是相关代码摘录：

帐户：

@Entity
@Table(name = "account")
public class Account {

    private Long accountId;
    private List<AccountAttribut> accountAttributes = new LinkedList<AccountAttribut>();
    private Person person;
    private Zielsystem zielsystem;

    public Account() {
    }
    // ...
}

@Entity
@Table(name="account_attribut")
@AssociationOverrides({
    @AssociationOverride(name="aa_pk.account", joinColumns = @JoinColumn(name="account_id")),
    @AssociationOverride(name="aa_pk.attribut", joinColumns = @JoinColumn(name="attribut_id"))
})
public class AccountAttribut {}
private AccountAttributPk aa_pk = new AccountAttributPk();

@Column(name="wert")
private String wert;

@EmbeddedId
public AccountAttributPk getAa_pk() {
    return aa_pk;
}

public void setAa_pk(AccountAttributPk aa_pk) {
    this.aa_pk = aa_pk;
}

@Transient
public Account getAccount() {
    return getAa_pk().getAccount();
}

public void setAccount(Account account) {
    getAa_pk().setAccount(account);
}

@Transient
public Attribut getAttribut() {
    return getAa_pk().getAttribut();
}

public void setAttribut(Attribut attribut) {
    getAa_pk().setAttribut(attribut);
}

public String getWert() {
    return wert;
}

public void setWert(String wert) {
    this.wert = wert;
}

public boolean equals(Object o) {
     if (this== o) return true;
     if (o ==null|| getClass() != o.getClass()) return false;

     AccountAttribut that = (AccountAttribut) o;

     if (getAa_pk() !=null?!getAa_pk().equals(that.getAa_pk()) : that.getAa_pk() !=null) return false;

     return true;
}

 public int hashCode() {
     return (getAa_pk() !=null? getAa_pk().hashCode() : 0);
 }
}

@Embeddable
public class AccountAttributPk implements Serializable {

    private static final long serialVersionUID = -1551814445010872872L;
    private Account account;
    private Attribut attribut;

    @ManyToOne
    public Account getAccount() {
        return account;
    }

    public void setAccount(Account account) {
        this.account = account;
    }

    @ManyToOne
    public Attribut getAttribut() {
        return attribut;
    }

    public void setAttribut(Attribut attribut) {
        this.attribut = attribut;
    }

    public boolean equals(Object o) {
        if (this == o)
            return true;
        if (o == null || getClass() != o.getClass())
            return false;

        AccountAttributPk that = (AccountAttributPk) o;

        if (account != null ? !account.equals(that.account) : that.account != null)
            return false;
        if (attribut != null ? !attribut.equals(that.attribut)
                : that.attribut != null)
            return false;

        return true;
    }

    public int hashCode() {
        int result;
        result = (account != null ? account.hashCode() : 0);
        result = 31 * result + (attribut != null ? attribut.hashCode() : 0);
        return result;
    }
}

List<Account> queried_accounts = HibernateUtils.criteriaList(
   session.createCriteria(Account.class).
      createCriteria("accountAttributes").
         createCriteria("aa_pk.attribut").
            add(Restrictions.isNotNull("name"))
);

例外情况：

org.hibernate.exception.sqlgrammarexception：无法提取org.hibernate.exception.internal.sqlexceptiontypedelegate.convert（sqlexceptiontypedelegate.java:80)（org.hibernate.exception.internal.standardsqlexceptionconverter.convert（standardsqlexceptionconverter.java:49)上的ResultSet

我尝试了以下几种解决方案：

List<Account> queried_accounts = HibernateUtils.criteriaList(
   session.createCriteria(Account.class).
      createCriteria("accountAttributes","accAttrAlias").
       createCriteria("accAttrAlias.aa_pk","accAttrAAPKAlias").
         createCriteria("accAttrAAPKAlias.attribut", "attrAlias").
            add(Restrictions.isNotNull("attrAlias.name"))
);

Java错误日志：org.hibernate.QueryException：不能直接在组件上创建条件html" target="_blank">对象。创建一个拥有实体的条件，并使用一个带点的属性访问组件属性:AccountAttributes.aa_pk

List<Account> queried_accounts = HibernateUtils.criteriaList(
   session.createCriteria(Account.class).
      createCriteria("accountAttributes","accAttrAlias").
         createCriteria("accAttrAlias.aa_pk.attribut", "attrAlias").
            add(Restrictions.isNotNull("attrAlias.name"))
);

Java错误日志：org.hibernate.exception.sqlgrammarexception：无法提取ResultSet

Hibernate错误日志：“Hibernate：选择this_.account_id作为account_1_3_4_,this_.person_id作为person_i2_3_4_,this_.zielsystem_id作为zielsystem_3_3_4_,accattrali1_.account_id作为account_3_4__,accattrali1_.attribut_id作为attribut_2_4_0_,accattrali1_.attribut_id作为attribut_2_4_0_,accattrali1_.attribut_wert作为s2_.name不为null Feb 20 2014 11:25:11 PM org.hibernate.engine.jdbc.spi.sqlexceptionhelper logExceptions警告：SQL错误：1054,SqlState:42s22 Feb 20 2014 11:25:11 PM org.hibernate.engine.jdbc.spi.sqlexceptionhelper logExceptions错误：“WHERE子句”中的列“attralias2_.name”未知“

Answer 1

标准应该与以下变量类似

List<Account> queried_accounts = HibernateUtils.criteriaList(
   session.createCriteria(Account.class).
      createCriteria("accountAttribute","accAttrAlias").
         createCriteria("accAttrAlias.ATTRIBUTE_VARIABLE_NAME", "attrAlias").
            add(Restrictions.isNotNull("attrAlias.name"))
);

以下是标准文档

更新

List<Account> queried_accounts = HibernateUtils.criteriaList(
   session.createCriteria(Account.class).
      createCriteria("accountAttribute","accAttrAlias").
       createCriteria("accAttrAlias.aa_pk","accAttrAAPKAlias").
         createCriteria("accAttrAAPKAlias.attribut", "attrAlias").
            add(Restrictions.isNotNull("attrAlias.name"))
);

List<Account> queried_accounts = HibernateUtils.criteriaList(
   session.createCriteria(Account.class).
      createCriteria("accountAttribute","accAttrAlias").
         createCriteria("accAttrAlias.aa_pk.attribut", "attrAlias").
            add(Restrictions.isNotNull("attrAlias.name"))
);