如何返回地图< Key，Value>与Hibernate \HQL [英] How to return Map<Key, Value> with Hibernate\HQL

问题描述

  @Entity 
 @Table（name =rank）
 class Rank实现Serializable {
 private int id; 
 private int rank_id; 
私人日期creationDate; 
 ... 
 // getters\setters ... 
} 
  

我想查询这个表并将结果放入一个HashMap！
如下：

  Map< Integer，Rank> = session.createSQL（select new map ...）。list（）; 
  

是否可以将实体作为地图的值？

我可以得到一个代码示例吗？

解决方案

是的，例如：

选择新地图（r.id，r）FROM Rank r

但是，这将返回一个地图列表。你应该看看这里来更好地理解。

编辑：为了更好地解释，返回将如下所示：

 清单<地图< Long，Rank>> （列表< Map< Long，Rank>>）session.createQuery（select new Map< r.rankId，r> FROM Rank r）。 
  

每个记录都是一个Map。不幸的是，它必须是手动的：

  List< Rank>等级=（List< Rank>）session.createQuery（select new r FROM Rank r）.list（）; 
地图< Long，Rank> mapRanks = new HashMap< Long，Rank>（）; 
 for（Rank rank：ranks）{
 if（！map.contains（rank.getId（））{
 map.put（rank.getId（），rank）; 
} 
} 
  

I have an entry as follow:

@Entity
@Table(name="rank")
class Rank implements Serializable {
 private int id;
 private int rank_id;
 private date creationDate;
 ...
 // getters\setters...
}

I would like to query this table and put the results into a HashMap! as follow:

Map<Integer, Rank> = session.createSQL("select new map...").list();

Is this possible to put the entity as the value of the map?

Can I get a code example?

解决方案

Yes, for example:

Select new Map(r.id,r) FROM Rank r

But that will return a list of Maps. You should take a look here to understand better.

Edit: To explain better, the return will be something like this:

List<Map<Long, Rank>> ranks = (List<Map<Long, Rank>>) session.createQuery("select new Map<r.rankId,r> FROM Rank r").list();

It is one Map for each record. To put all in only one map, unfortunately, it has to be manually:

List<Rank> ranks = (List<Rank>) session.createQuery("select new r FROM Rank r").list();
Map<Long, Rank> mapRanks  = new HashMap<Long, Rank>();
for (Rank rank : ranks) {
  if (!map.contains(rank.getId()) {
    map.put(rank.getId(), rank);
  }
}

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