用hibernate怎么实现一个表中多个字段的联合唯一

<? xml version="1.0" encoding="utf-8" ?>

<! DOCTYPE hibernate-mapping

    PUBLIC "-//Hibernate/Hibernate Mapping DTD 3.0//EN"

    "http://hibernate.sourceforge.net/hibernate-mapping-3.0.dtd" >

< hibernate-mapping >

    < class name ="onlyfun.caterpillar.User" table ="user" >

        < composite-id >

            < key-property name ="name"

                          column ="name"

                          type ="java.lang.String" />

            < key-property name ="sex"

                          column ="sex"

                          type ="java.lang.String" />

        </ composite-id >

    </ class >

</ hibernate-mapping >




并且定义的User类要覆盖equals和hashCode方法

Java code

    
    
    
    

     
     
     
     
      
     
     
     
     public
     
     
     
      
     
     
     
     boolean
     
     
     
      equals(Object obj) { 
     
     
     
     if
     
     
     
     (obj 
     
     
     
     ==
     
     
     
      
     
     
     
     this
     
     
     
     ) { 
     
     
     
     return
     
     
     
      
     
     
     
     true
     
     
     
     ; } 
     
     
     
     if
     
     
     
     (
     
     
     
     !
     
     
     
     (obj 
     
     
     
     instanceof
     
     
     
      User)) { 
     
     
     
     return
     
     
     
      
     
     
     
     false
     
     
     
     ; } User user 
     
     
     
     =
     
     
     
      (User) obj; 
     
     
     
     return
     
     
     
      
     
     
     
     new
     
     
     
      EqualsBuilder() .append(
     
     
     
     this
     
     
     
     .name, user.getName()) .append(
     
     
     
     this
     
     
     
     .sex, user.getSex()) .isEquals(); } 
     
     
     
     public
     
     
     
      
     
     
     
     int
     
     
     
      hashCode() { 
     
     
     
     return
     
     
     
      
     
     
     
     new
     
     
     
      HashCodeBuilder() .append(
     
     
     
     this
     
     
     
     .name) .append(
     
     
     
     this
     
     
     
     .sex) .toHashCode(); }