hibernate 一对多配置排序

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我使用的是注解的方法

package com.security.model.menu;// Generated 2014-4-4 16:51:48 by Hibernate Tools 3.4.0.CR1import static javax.persistence.GenerationType.IDENTITY;import java.util.ArrayList;import java.util.List;import java.util.Set;import javax.persistence.Column;import javax.persistence.Entity;import javax.persistence.FetchType;import javax.persistence.GeneratedValue;import javax.persistence.Id;import javax.persistence.JoinColumn;import javax.persistence.ManyToOne;import javax.persistence.OneToMany;import javax.persistence.Table;import org.hibernate.annotations.OrderBy;/** * Menu generated by hbm2java */@Entity@Table(name = "menu", catalog = "security")public class Menu implements java.io.Serializable {private Integer id;private Menu menu;private String name;private String url;private Integer orderNum;private List<Menu> menus = new ArrayList<Menu>();public Menu() {}public Menu(Menu menu, String name, String url, List<Menu> menus) {this.menu = menu;this.name = name;this.url = url;this.menus = menus;}public Integer getOrderNum() {return orderNum;}public void setOrderNum(Integer orderNum) {this.orderNum = orderNum;}@Id@GeneratedValue(strategy = IDENTITY)@Column(name = "id", unique = true, nullable = false)public Integer getId() {return this.id;}public void setId(Integer id) {this.id = id;}@ManyToOne(fetch = FetchType.EAGER)@JoinColumn(name = "parentId")public Menu getMenu() {return this.menu;}public void setMenu(Menu menu) {this.menu = menu;}@Column(name = "name")public String getName() {return this.name;}public void setName(String name) {this.name = name;}@Column(name = "url")public String getUrl() {return this.url;}public void setUrl(String url) {this.url = url;}@OneToMany(fetch = FetchType.EAGER, mappedBy = "menu")@OrderBy(clause = "orderNum ASC")public List<Menu> getMenus() {return this.menus;}public void setMenus(List<Menu> menus) {this.menus = menus;}}

注意：

1、要把hibernate生成的原有的Set改成List

2、

@OrderBy(clause = "orderNum ASC")

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