迭代列表<Object>來自加入查詢使用JPA/PLay

Question

這是我的查詢使用實體管理器。試圖加入2表與玩框架和jpa。

List<Object> joinQryResult = JPA.em().createNativeQuery(
     "select e.elementname as elementname, " + 
     "c.comparetype as comparetype, " + 
     "jd.matchvalue as matchvalue " + 
     "from details jd " + 
     "join elements e on jd.elementnamerid = e.rid " + 
     "join comparers c on jd.comparetyperid = c.rid " + 
     "where jd.rid = " + temp.rid).getResultList(); 

Not sure how to iterate and get the values from List<Object> 

我想這

List<MyClass> myClass = (List<MyClass>)(Object)joinQryResult; 


for(MyClass myC:jd) 
{ 
System.out.println(myC.ElementName); //intellisense shows the property here 
} 

MyClass的定義：ttrying到列表轉換爲這種類型的

public class MyClass { 

    public String ElementName; 

    public String CompareType; 

    public String MatchValue; 

    public JobDetails(String ElementName, String CompareType, String MatchValue) 
    { 
     this.ElementName = ElementName; 

     this.CompareType = CompareType; 

     this.MatchValue = MatchValue; 
    } 
} 

得到這個錯誤

ClassCastException occured : [Ljava.lang.Object; cannot be cast to models.MyClass 

Answer 1

查詢的結果將成爲一個列表與th的元素e對象是相應的對象類型。

你可以做如下映射到你的對象

List<Object[]> results = query.getResultList(); 
    JobDetails jobDetail = null; 
    for (Object[] objects : results) { 
     jobDetail = new JobDetail((String) objects[0],(String) objects[1],(String) objects[2]) 
    }