Hibernate HQL order by Map Key

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Hibernate HQL order by Map Key5

有一需求Hibernate 一实体有一个组件是Map的,Map的Key是一个对象，值是另一个对象，页面表格需要排序，要按照Map的Key的一个属性排序，本人无力应对，特此QiuZhu。


@Entity
@Table(name="_chengJiDengJis")
@Cache(usage= CacheConcurrencyStrategy.READ_WRITE)
@JSONType(orders = {"name","beginDate","endDate","lastEditTime","operater","teachers"}) //FastJson输出顺序
public class ChengJiDengJi extends AbstractIdentityVersionedObject<Long>  implements Serializable
{


private Map<ChengJiDengJiFenShuName,ChengJiDengJiFenShu> chengJiDengJiFenShus = new TreeMap<ChengJiDengJiFenShuName,ChengJiDengJiFenShu>();//成绩等级的分数包含ABCD等级的最大分、最小分


	@ElementCollection(fetch = FetchType.EAGER)
	@MapKeyClass(ChengJiDengJiFenShuName.class)
	@MapKeyEnumerated(EnumType.STRING)
	@MapKeyColumn(name = "dengJiName")
	@CollectionTable(name = "_chengJiDengJis_fenShus", joinColumns = @JoinColumn(name = "chengJiDengJi_ID"))
	public Map<ChengJiDengJiFenShuName, ChengJiDengJiFenShu> getChengJiDengJiFenShus()
	{
		return chengJiDengJiFenShus;
	}
}

public enum ChengJiDengJiFenShuName
{
	A("A等"),B("B等"),C("C等"),D("D等");
	
}

@Embeddable
public class ChengJiDengJiFenShu
{
	private BigDecimal minFenShu;//最低分数
	private BigDecimal maxFenShu;//最高分数
}

想要的结果类似：
from ChengJiDengJi c ordery by c.chengJiDengJiFenShus.ChengJiDengJiFenShuName.A.minFenShu

indices,elements,index等hql关键字不知道是否能用上，如何使用。。

谢谢！

问题补充：类似这样的：

from ChengJiDengJi c order by c.chengJiDengJiFenShus[ChengJiDengJiFenShuName.A].minFenShu

问题补充：问题已经解决一半了，后续暂时没想到办法。
未解决的问题具体如下：maxFenShu 如何换成minFenShu,这两个是同等级的，不知道为何Hibernate选择了maxFenShu，而不选择minFenShu.
order by
chengjiden1_.maxFenShu desc limit ?

控制台输出的Hql,sql语句如下：


Hibernate: 
    /* 
from
    ChengJiDengJi c 
where
    c.school=:operaterSchool  
order by
    c.chengJiDengJiFenShus['A'] desc */ select
        chengjiden0_.id as id0_,
        chengjiden0_.Lock_Version as Lock2_0_,
        chengjiden0_.fenZhi as fenZhi0_,
        chengjiden0_.fuJiaFen as fuJiaFen0_,
        chengjiden0_.guiZeLeiXing as guiZeLei5_0_,
        chengjiden0_.lastEditTime as lastEdit6_0_,
        chengjiden0_.name as name0_,
        chengjiden0_.operater_id as operater8_0_,
        chengjiden0_.school_id as school9_0_ 
    from
        _chengJiDengJis chengjiden0_ cross 
    join
        _chengJiDengJis_fenShus chengjiden1_ 
    where
        chengjiden0_.id=chengjiden1_.chengJiDengJi_ID 
        and chengjiden1_.dengJiName = 'A' 
        and chengjiden0_.school_id=? 
    order by
        chengjiden1_.maxFenShu desc limit ?

order by hibernate hql Map Key

2013年2月16日 15:34

Mr.TianShu
17
0 0 3

2个答案按时间排序按投票排序

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map这个集合本身就是无序的，即使在map中时有序的输出也未必一定是有序的，如果必须要按照某个对象的某个属性进行排序，你不妨把所有的key保存到一个list中，然后对list中的对象进行排序，排序完成以后，你在遍历有序的list中的对象，根据list中的对象（此时list中的对象时有序的key,你根据有序的key去获得相应的值）再去获得map中的对象。

2013年2月16日 21:12

zhangwenbojava
30
0 0 0

添加评论

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    public TreeMap(Comparator<? super K> comparator) {
        this.comparator = comparator;
    }

treemap的构造方法中需要加入key的 comparator

2013年2月16日 15:54

kidding87
1566
1 1 19

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