Java 如何通过将列表的参数添加为对象来发送列表

我有一个带有对象参数的模型

如果有3个不同的候选者，它应该显示3个候选者，但我的输出将重复最后一个候选者的详细信息3次。我没有得到前两个候选人的详细信息

public class CandidateFeedbackDisplay implements Serializable{  
    private Candidate candidate;
    private List<Integer> feedbackIds;
//setter and getters
}




public List<CandidateFeedbackDisplay> list(Integer cID, Integer jID, String accepted) throws Exception {
        Session session = this.sessionFactory.getCurrentSession();
        List<Candidate> candidateList = null;
        CandidateFeedbackDisplay feedbackDisplay = new CandidateFeedbackDisplay();
        List<CandidateFeedbackDisplay> feedbackDisplayList = new ArrayList();
//      List<CandidateFeedbackDisplay> feedbackDisplayListTest = null;
        try {   
          Query query = session.createQuery("from Candidate WHERE phoneNumber IN (select DISTINCT mobileNo from InviteCandidates WHERE c_id= :cID AND j_id= :jID AND status= :accepted)");               
          query.setInteger("cID", cID);       
          query.setInteger("jID", jID);
          query.setString("accepted", accepted);          
          candidateList = query.list();
          Iterator itr = candidateList.iterator();
          while(itr.hasNext()){
              Candidate candidate = (Candidate) itr.next();
              System.out.println("candidate.getCandidateID() : " + candidate.getCandidateID());
              List<CandidateFeedback> candidateFeedback = this.getFeedback(cID, jID, candidate.getCandidateID());
              Iterator itr1 = candidateFeedback.iterator();
              List<Integer> feedbackid = new ArrayList<Integer>();
              while(itr1.hasNext()){                  
                  CandidateFeedback Feedback =  (CandidateFeedback) itr1.next();
                  feedbackid.add(Feedback.getFeedbackID());               
              }
              feedbackDisplay.setFeedbackIds(feedbackid);
              feedbackDisplay.setCandidate(candidate);            
              feedbackDisplayList.add(feedbackDisplay);
//            feedbackDisplayListTest.add(feedbackDisplay); // null pointer access error
          }
          }catch (Exception e) {
              e.printStackTrace();
              this.logger.error("Error while fetching List :" + e);
              return null;
          }
        return feedbackDisplayList;
    }

---

将同一对象添加到列表中三次。每次都必须通过移动此行来创建新对象：

CandidateFeedbackDisplay feedbackDisplay = new CandidateFeedbackDisplay();

进入while循环，否则将继续更改先前放置的对象的属性。事实上，您正在更改同一对象，并将其添加到列表中三次

public List<CandidateFeedbackDisplay> list(Integer cID, Integer jID, String accepted) throws Exception {
    ...
    // DELETE HERE
    // CandidateFeedbackDisplay feedbackDisplay = new CandidateFeedbackDisplay();
    List<CandidateFeedbackDisplay> feedbackDisplayList = new ArrayList();
    try {   
        ...
        while(itr.hasNext()) {
            ...
            // INSERT HERE
            CandidateFeedbackDisplay feedbackDisplay = new CandidateFeedbackDisplay();
            feedbackDisplay.setFeedbackIds(feedbackid);
            feedbackDisplay.setCandidate(candidate);            
            feedbackDisplayList.add(feedbackDisplay);
        }
    } catch (Exception e) {
        ...
    }
    return feedbackDisplayList;
}

附加说明：为了防止下次出现此类错误，您可以将参数对象CandidateFeedbackDisplay从setters更改为基于构造函数的实现：

public class CandidateFeedbackDisplay {
    private final Candidate candidate;
    private final List<Integer> feedbackIds;

    public CandidateFeedbackDisplay(Candidate candidate, List<Integer> feedbackIds) {
        this.candidate = candidate;
        this.feedbackIds = feedbackIds;
    }

    // no setters
    // add getters or make fields public, but keep final
}

这样，您就可以真正证明这个对象只是一个不可变的值持有者。您不能再犯同样的错误，构造函数可能会将代码缩短一点。当然，优点和缺点取决于您的具体情况。

放置候选反馈显示feedbackDisplay feedbackDisplay=new CandidateFeedbackDisplay；在while循环内，因为您正在尝试准备候选反馈显示列表

请查找下面的代码，希望对您有所帮助

while(itr.hasNext()){
        CandidateFeedbackDisplay feedbackDisplay = new CandidateFeedbackDisplay();
        Candidate candidate = (Candidate) itr.next();
        System.out.println("candidate.getCandidateID() : " + candidate.getCandidateID());
        List<CandidateFeedback> candidateFeedback = this.getFeedback(cID, jID, candidate.getCandidateID());
        Iterator itr1 = candidateFeedback.iterator();
        List<Integer> feedbackid = new ArrayList<Integer>();
        while(itr1.hasNext()){
            CandidateFeedback Feedback =  (CandidateFeedback) itr1.next();
            feedbackid.add(Feedback.getFeedbackID());
        }
        feedbackDisplay.setFeedbackIds(feedbackid);
        feedbackDisplay.setCandidate(candidate);
        feedbackDisplayList.add(feedbackDisplay);
    }

---

首先，你需要解释你的代码，放上注释，指出你有问题的确切代码，并删除不必要的代码谢谢你的快速回复谢谢你。“你救了我的命。”普里亚当然，听到这个消息很高兴。我在结尾加了一个小纸条，这有助于防止将来发生这样的错误。谢谢你的回复