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我创建了一个对象，它映射了我的数据库中的两个表，即 Dictionary 表和 Token 表。表示这两个表之间连接的对象（类）称为 DictionaryToken。

这是课程：

import java.util.List;

import javax.persistence.CascadeType;
import javax.persistence.Column;
import javax.persistence.Entity;
 import javax.persistence.FetchType;
 import javax.persistence.GeneratedValue;
import javax.persistence.GenerationType;
 import javax.persistence.Id;
import javax.persistence.OneToMany;
 import javax.persistence.Table;

import org.apache.log4j.Logger;


@Entity
@Table(name="dictionary", catalog="emscribedxcode")

公共类 DictionaryToken { 私有静态 Logger LOG = Logger.getLogger(DictionaryToken.class);

private Long _seq;
private String _code;
private String _acute;
private String _gender;
private String _codeType;
private String _papplydate;
private String _capplydate;
private Long _tokenLength;
private List <TokenDictionary> _token;
private int _type;
private String _system;
private String _physicalsystem;
/*
 * type of 0 is a straight line insert type of 1 is a language dictionary
 * entyr type of 2 is a multiple token entry
 */
@Id
@GeneratedValue(strategy = GenerationType.AUTO)
@Column(name = "seq")
public Long getSeq() {
    return _seq;
}
public void setSeq(Long seq_) {
    _seq = seq_;
}

@Column(name = "code")
public String getCode() {
    return _code;
}
public void setCode(String code_) {
    _code = code_;
}

@Column(name = "acute")
public String getAcute() {
    return _acute;
}
public void setAcute(String acute_) {
    _acute = acute_;
}

@Column(name = "gender")
public String getGender() {
    return _gender;
}
public void setGender(String gender_) {
    _gender = gender_;
}

@Column(name = "codetype")
public String getCodeType() {
    return _codeType;
}
public void setCodeType(String codeType_) {
    _codeType = codeType_;
}

@Column(name = "papplydate")
public String getPapplydate() {
    return _papplydate;
}
public void setPapplydate(String papplydate_) {
    _papplydate = papplydate_;
}

@Column(name = "capplydate")
public String getCapplydate() {
    return _capplydate;
}
public void setCapplydate(String capplydate_) {
    _capplydate = capplydate_;
}


@Column(name = "token_length")
public Long getTokenLength() {
    return _tokenLength;
}
public void setTokenLength(Long tokenLength_) {
    _tokenLength = tokenLength_;
}
@OneToMany (mappedBy = "dictionarytoken", targetEntity = TokenDictionary.class, cascade = CascadeType.ALL, fetch = FetchType.EAGER)
public List<TokenDictionary> get_token() {
    return _token;
}
public void set_token(List<TokenDictionary> _token) {
    this._token = _token;
}
public void addToToken(TokenDictionary token){
    this._token.add(token);
}
@Column(name = "type")
public int getType() {
    return _type;
}
public void setType(int _type) {
    this._type = _type;
}
@Column(name = "physicalsystem")
public String get_physicalsystem() {
    return _physicalsystem;
}
public void set_physicalsystem(String _physicalsystem) {
    this._physicalsystem = _physicalsystem;
}
@Column(name = "codingsystem")
public String get_system() {
    return _system;
}
public void set_system(String _system) {
    this._system = _system;
}



   }

这是我的问题。除非我添加条件，否则我可以毫无问题地使用具有此对象的服务执行查询。这是检索条目的方法

public List<DictionaryToken> getDictionaryTokenEntries(String system) {
    Session session = null;
    List<DictionaryToken> dictonaries = new ArrayList<DictionaryToken>();
    try {
        session = HibernateUtils.beginTransaction("emscribedxcode");
           session.createCriteria(Dictionary.class).addOrder(Order.desc("codeType"))

        Criteria criteria = session.createCriteria(DictionaryToken.class);
                  /*******THIS IS THE PROBLEM STATEMENT*************************/
        if (system != null) {
            criteria.add(Restrictions.eq("codingsystem", system));
        }
                  /****************************************************************/
        // dictonaries = criteria.list();
        Order order = Order.asc("seq");
        criteria.addOrder(order);
        dictonaries = criteria.list();
        System.out.println("Dictionaryentries = " + dictonaries.size());

        // System.out.println("Dictionaries entries EVICT start...");
        // for(Dictionary dic : dictonaries){
        // session.evict(dic);
        // }
        // System.out.println("Dictionaries entries EVICT end");
    } catch (HibernateException e_) {
        e_.printStackTrace();
        NTEVENT_LOG.error("Error while getting List of Dictionary entries");
    } finally {
        if (session != null && session.isOpen()) {
            try {
                HibernateUtils.closeSessions();
            } catch (HibernateException e) {
                // TODO Auto-generated catch block
                e.printStackTrace();
            }
        }
    }
    return dictonaries;
}

添加条件时，出现以下错误：

org.hibernate.QueryException：无法解析属性：编码系统：com.artificialmed.domain.dictionary.model.DictionaryToken

我知道这与对象的性质有关，该对象实际上是我的字典类和基础表以及我的令牌类和表之间的连接。

字段编码系统是我的字典类中的一个字段。我想我应该使用别名，但我不知道在当前情况下如何做到这一点。任何帮助将不胜感激。

艾略特

4

1 回答 1

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这是一个新手问题。Hibernate 要求反映表的模型的 getter 和 setter 具有特定格式。getter 必须是 get+，其中 name 是基础表中的字段名。setter 必须是 set+ 其中 name 是基础表的字段名。是的，名称的第一个字母必须大写。

于 2012-10-25T01:26:38.067 回答