元素类型“id”的内容必须匹配 - hbm xml

Posted 2023-02-24

【中文标题】元素类型“id”的内容必须匹配 - hbm xml

【英文标题】：The content of element type "id" must match - hbm xml

【发布时间】：2015-06-26 02:22:12

【问题描述】：

我知道这个问题经常被问到。

我收到以下映射错误。

我尝试了一些排列和组合。 我已经尝试了一些解决方案，但似乎无法解决这个问题。

有人可以看看我下面的配置吗？

xmls 是有效的。 我使用hibernate 3和mysql作为数据库

错误日志：

log4j:WARN No appenders could be found for logger (org.hibernate.cfg.Environment).
log4j:WARN Please initialize the log4j system properly.
Initial SessionFactory creation failed.org.hibernate.InvalidMappingException: Could not parse mapping document from resource com/sample/Employee.hbm.xml
Exception in thread "main" java.lang.ExceptionInInitializerError
    at com.sample.SessionFactoryUtil.<clinit>(SessionFactoryUtil.java:17)
    at com.sample.TestEmployee.main(TestEmployee.java:13)
Caused by: org.hibernate.InvalidMappingException: Could not parse mapping document from resource com/sample/Employee.hbm.xml
    at org.hibernate.cfg.Configuration.addResource(Configuration.java:523)
    at org.hibernate.cfg.Configuration.parseMappingElement(Configuration.java:1511)
    at org.hibernate.cfg.Configuration.parseSessionFactory(Configuration.java:1479)
    at org.hibernate.cfg.Configuration.doConfigure(Configuration.java:1458)
    at org.hibernate.cfg.Configuration.doConfigure(Configuration.java:1432)
    at org.hibernate.cfg.Configuration.configure(Configuration.java:1352)
    at org.hibernate.cfg.Configuration.configure(Configuration.java:1338)
    at com.sample.SessionFactoryUtil.<clinit>(SessionFactoryUtil.java:12)
    ... 1 more
Caused by: org.hibernate.MappingException: invalid mapping
    at org.hibernate.cfg.Configuration.addInputStream(Configuration.java:463)
    at org.hibernate.cfg.Configuration.addResource(Configuration.java:520)
    ... 8 more
Caused by: org.xml.sax.SAXParseException: The content of element type "id" must match "(meta*,column*,type?,generator?)".
    at com.sun.org.apache.xerces.internal.util.ErrorHandlerWrapper.createSAXParseException(Unknown Source)
    at com.sun.org.apache.xerces.internal.util.ErrorHandlerWrapper.error(Unknown Source)
    at com.sun.org.apache.xerces.internal.impl.XMLErrorReporter.reportError(Unknown Source)
    at com.sun.org.apache.xerces.internal.impl.XMLErrorReporter.reportError(Unknown Source)
    at com.sun.org.apache.xerces.internal.impl.dtd.XMLDTDValidator.handleEndElement(Unknown Source)
    at com.sun.org.apache.xerces.internal.impl.dtd.XMLDTDValidator.endElement(Unknown Source)
    at com.sun.org.apache.xerces.internal.impl.XMLDocumentFragmentScannerImpl.scanEndElement(Unknown Source)
    at com.sun.org.apache.xerces.internal.impl.XMLDocumentFragmentScannerImpl$FragmentContentDriver.next(Unknown Source)
    at com.sun.org.apache.xerces.internal.impl.XMLDocumentScannerImpl.next(Unknown Source)
    at com.sun.org.apache.xerces.internal.impl.XMLNSDocumentScannerImpl.next(Unknown Source)
    at com.sun.org.apache.xerces.internal.impl.XMLDocumentFragmentScannerImpl.scanDocument(Unknown Source)
    at com.sun.org.apache.xerces.internal.parsers.XML11Configuration.parse(Unknown Source)
    at com.sun.org.apache.xerces.internal.parsers.XML11Configuration.parse(Unknown Source)
    at com.sun.org.apache.xerces.internal.parsers.XMLParser.parse(Unknown Source)
    at com.sun.org.apache.xerces.internal.parsers.AbstractSAXParser.parse(Unknown Source)
    at com.sun.org.apache.xerces.internal.jaxp.SAXParserImpl$JAXPSAXParser.parse(Unknown Source)
    at org.dom4j.io.SAXReader.read(SAXReader.java:465)
    at org.hibernate.cfg.Configuration.addInputStream(Configuration.java:460)
    ... 9 more

Employee.hbm.xml

<?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.sample.Employee" table="employee">
   <id name="idEmployee" column="ID">  
        <column name="uid" not-null="true"/>
        <generator class="native" />         
   </id>
   <property name="empName">
       <column name="empName" length="16" not-null="true" />
   </property>
   <property name="sex">
        <column name="sex" length="16" not-null="true" />
   </property>
</class>
</hibernate-mapping>

Employee class:
package com.sample;
public class Employee 

    private Long idEmployee;
    private String empName;
    private String sex;
    public Long getIdEmployee() 
        return idEmployee;
    
    public void setIdEmployee(Long idEmployee) 
        this.idEmployee = idEmployee;
    
    public String getEmpName() 
        return empName;
    
    private String address;
    private int age;


    public void setEmpName(String empName) 
        this.empName = empName;
    
    public String getSex() 
        return sex;
    
    public void setSex(String sex) 
        this.sex = sex;
    
    public String getAddress() 
        return address;
    
    public void setAddress(String address) 
        this.address = address;
    
    public int getAge() 
        return age;
    
    public void setAge(int age) 
        this.age = age;
    

CREATE TABLE `employee` (
`idEmployee` int(11) NOT NULL AUTO_INCREMENT,
`empName` varchar(45) NOT NULL,
`sex` varchar(45) NOT NULL,
`age` int(11) DEFAULT NULL,
`address` varchar(45) DEFAULT NULL,
 PRIMARY KEY (`idEmployee`)
 );

【问题讨论】：

不相关，但请找到另一个教程来学习 Hibernate。您正在像 10 多年前一样使用它，那时还没有注释。如今，注释用于定义映射。 XML 已过时。

当然。将检查最新的教程 :) 嘿，现在我知道以前是怎么做的 ;) 谢谢！！！

【参考方案1】：

我认为您错过了属性“地址”和“年龄”的映射。 你为什么不做？并检查您使用的长度标签，为什么它们都等于 16？ 试试这个：

<hibernate-mapping>
<class name="com.sample.Employee" table="employee">
   <id name="idEmployee" column="ID">  
        <generator class="native" />         
   </id>
   <property name="empName" column="empName"></property>
   <property name="sex" column="sex"></property>
   <property name="address" column="address"></property>
   <property name="age" column="age"></property>
</class>
</hibernate-mapping>

【参考方案2】：

异常说：

Caused by: org.xml.sax.SAXParseException: The content of element type "id" must match "(meta*,column*,type?,generator?)".

您的 ID 类型不匹配。在Employee 类上，您的idEmployee 变量类型为Long，在.hbm.xml 文件中，您的ID 属性映射具有native 生成器，这意味着ID 默认为Integer 类型。所以你应该在你的 ID 属性上使用 type="long" 注释。或者只是尝试使用不同类型的 ID 生成器。在 .hbm.xml 文件中的 ID 注释上，您使用两个名称作为 ID 列。您正在使用column="ID" 和column="uid"。 .hbm.xml 文件的示例：

<hibernate-mapping>
     <class name="com.sample.Employee" table="employee">
        <id name="idEmployee" type="long">
            <column name="uid" not-null="true"/>
            <generator class="native" />
        </id>
        <property name="empName">
            <column name="empName" length="16" not-null="true" />
        </property>
        <property name="sex">
            <column name="sex" length="16" not-null="true" />
        </property>
    </class>
</hibernate-mapping>

或者尝试使用SequenceStyleGenerator。生成器，它通过给定的序列名称为每个表创建单独的 ID 序列。生成的 id 的默认类型是Long。 .hbm.xml 文件示例：

 <hibernate-mapping>
    <class name="com.sample.Employee" table="employee">
        <id name="idEmployee" column="ID">
            <generator class="org.hibernate.id.enhanced.SequenceStyleGenerator">
                <param name="optimizer">none</param>
                <param name="increment_size">1</param>
                <param name="sequence_name">seq_employee_id</param>
            </generator>
        </id>
        <property name="empName">
            <column name="empName" length="16" not-null="true" />
        </property>
        <property name="sex">
            <column name="sex" length="16" not-null="true" />
        </property>
    </class>
 </hibernate-mapping>

希望对你有帮助...☺