T1:(多的一方)
public class T1 implements java.io.Serializable {

	// Fields

	private Long id;

	private T t;

	private String name;

	private Date year;

	// Constructors

	/** default constructor */
	public T1() {
	}

	/** full constructor */
	public T1(T t, String name, Date year) {
		this.t = t;
		this.name = name;
		this.year = year;
	}

	// Property accessors

	public Long getId() {
		return this.id;
	}

	public void setId(Long id) {
		this.id = id;
	}

	public T getT() {
		return this.t;
	}

	public void setT(T t) {
		this.t = t;
	}

	public String getName() {
		return this.name;
	}

	public void setName(String name) {
		this.name = name;
	}

	public Date getYear() {
		return this.year;
	}

	public void setYear(Date year) {
		this.year = year;
	}

}

 one的一方：

package com.model;

import java.util.HashSet;
import java.util.Set;

/**
 * T generated by MyEclipse Persistence Tools
 */

public class T implements java.io.Serializable {

	// Fields

	private Long id;

	private String name;

	private Set t1s = new HashSet(0);

	// Constructors

	/** default constructor */
	public T() {
	}

	/** full constructor */
	public T(String name, Set t1s) {
		this.name = name;
		this.t1s = t1s;
	}

	// Property accessors

	public Long getId() {
		return this.id;
	}

	public void setId(Long id) {
		this.id = id;
	}

	public String getName() {
		return this.name;
	}

	public void setName(String name) {
		this.name = name;
	}

	public Set getT1s() {
		return this.t1s;
	}

	public void setT1s(Set t1s) {
		this.t1s = t1s;
	}

}

 查询语句:

 Session session = HibernateSessionFactory.getSession();
  String hql = "from T t where t.id =1 order by t.t1s.year asc";
  Query q = session.createQuery(hql);
  List<T> list = q.list();
  for(T t : list){
   System.out.println("t\t" + t.getId());
   Set<T1> set = t.getT1s();
   for(T1 t1 : set){
    System.out.println("t1\t" + t1.getId());
   }
  }

 输出结果:

Hibernate: select t0_.ID as ID0_, t0_.NAME as NAME0_ from SCOTT.T t0_, SCOTT.T1 t1s1_ where t0_.ID=t1s1_.T_ID and t0_.ID=1 order by t1s1_.YEAR asc
t 1
Hibernate: select t1s0_.T_ID as T2_1_, t1s0_.ID as ID1_, t1s0_.ID as ID1_0_, t1s0_.T_ID as T2_1_0_, t1s0_.NAME as NAME1_0_, t1s0_.YEAR as YEAR1_0_ from SCOTT.T1 t1s0_ where t1s0_.T_ID=?
t1 4
t1 5
t1 3
t 1
t1 4
t1 5
t1 3
t 1
t1 4
t1 5
t1 3

  查看输出结果，发现，每一条数据都输出了三次。

仔细看了下hql语句：

Hibernate: select t0_.ID as ID0_, t0_.NAME as NAME0_ from SCOTT.T t0_, SCOTT.T1 t1s1_ where t0_.ID=t1s1_.T_ID and t0_.ID=1 order by t1s1_.YEAR asc

 发现，竟然是交叉查询。

看来，想在一的一端（T）对多（T1)的一端里面的数据进行排序是不行的。阿门，只能用其他的方式了!