0

我是休眠的新手，我遇到了一个问题。我已经阅读了hibernate网站上的所有入门指南等，但我仍然无法提出解决方案。

我有这样的课：

public class ResultTree {
String attrName;
Map<String, ResultTree> valueMap;
String classValue;
int caseQuant;
Set<Map<String, String>> otherRules;

public String getAttrName() {
    return attrName;
}
public void setAttrName(String attrName) {
    this.attrName = attrName;
}
public Map<String, ResultTree> getValueMap() {
    return valueMap;
}
public void setValueMap(Map<String, ResultTree> valueMap) {
    this.valueMap = valueMap;
}
public String getClassValue() {
    return classValue;
}
public void setClassValue(String classValue) {
    this.classValue = classValue;
}
public int getCaseQuant() {
    return caseQuant;
}
public void setCaseQuant(int caseQuant) {
    this.caseQuant = caseQuant;
}
public Set<Map<String, String>> getOtherRules() {
    return otherRules;
}
public void setOtherRules(Set<Map<String, String>> otherRules) {
    this.otherRules = otherRules;
}

}

像这样的类的 hbm.xml 应该如何看待？我可以自由地创建任何数据结构。

谢谢你的帮助，MM

4

2 回答 2

1

在 Ranna 的解决方案的帮助下，我设法通过将其划分为两个单独的类来对类进行建模：

public class ResultTree {
private Long id;
private String attrName;
private Map<String, ResultTree> valueMap;
private String classValue;
private int caseQuant;
private Set<Rule> otherRules;
}

和

public class Rule {
private Long id;
private Map<String, String> terms;
private ResultTree tree;
private String classValue;
}

hbm.xml 具有以下形式：

    <?xml version="1.0"?>
<!DOCTYPE hibernate-mapping PUBLIC
    "-//Hibernate/Hibernate Mapping DTD 3.0//EN"
    "http://www.hibernate.org/dtd/hibernate-mapping-3.0.dtd">
<hibernate-mapping package="lib.experiment.result">
    <class name="ResultTree" table="RESULT_TREE">
        <id name="id" column="RESULT_TREE_ID" type="long" />
        <property name="attrName" type="string" column="ATTR_NAME" />
        <property name="classValue" type="string" column="CLASS_VALUE" />
        <property name="caseQuant" type="int" column="CASE_QUANT" />
        <map name="valueMap" table="RESULT_TREE_LEAF" lazy="false">
             <key column="RESULT_TREE_ID"/>
             <map-key column="ATTR_VALUE" type="string"/>
             <many-to-many class="ResultTree" />
        </map>
        <set name="otherRules" table="RULE" lazy="false">
            <key column="RESULT_TREE_ID"/>
            <one-to-many class="Rule"/>
        </set>
    </class>
    </hibernate-mapping>

和

    <?xml version="1.0"?>
<!DOCTYPE hibernate-mapping PUBLIC
    "-//Hibernate/Hibernate Mapping DTD 3.0//EN"
    "http://www.hibernate.org/dtd/hibernate-mapping-3.0.dtd">
<hibernate-mapping package="lib.experiment.result">
    <class name="Rule" table="RULE">
        <id name="id" column="RULE_ID" type="long" />
        <property name="classValue" column="CLASS" type="string" />
        <map name="terms" table="RULE_TERM" lazy="false">
             <key column="RULE_ID"/>
             <map-key column="ATTR_NAME" type="string"/>
             <element column="ATTR_VALUE" type="string"/>
        </map>
        <many-to-one name="tree" class="ResultTree" lazy="false">
            <column name="RESULT_TREE_ID"/>
        </many-to-one>
    </class>
    </hibernate-mapping>

非常感谢您的帮助！

于 2012-05-06T23:45:16.993 回答
0

希望这会对你有所帮助。

<?xml version="1.0"?>
<!DOCTYPE hibernate-mapping PUBLIC
"-//Hibernate/Hibernate Mapping DTD 3.0//EN"
"http://hibernate.sourceforge.net/hibernate-mapping-3.0.dtd">
<hibernate-mapping>
<class name="com.ResultTree" table="result_treeid">
<meta attribute="class-description">This class contains student details.</meta>
<id name="id" type="long" column="id">
<generator class="native" />
</id>
<property name="attrName" type="string" length="100" not-null="true" column="attr_name" />
<property name="classValue" type="string" length="100" not-null="true" column="class_value" />
<property name="caseQuant" type="bigint" not-null="true" column="case_quant" />
<map role="valueMap" table="value_map">
     <key column="id"/>
     <map-key column="keyname" type="string"/>
     <element column="valuename" type="ResultTree"/>
</map>
<map role="otherRules" table="other_rules">
     <key column="id"/>
     <map-key column="keyname" type="string"/>
     <element column="valuename" type="string"/>
</map>
</class>
</hibernate-mapping>
于 2012-05-06T16:38:27.077 回答