Question

我使用带有openJPA的蓝图DSL的jBoss Fuse 6.1.0。我使用容器管理事务（JTA）和Aspects管理的事务处理提交和回滚截至目前

我有以下作为JPA实体的类。

@Entity
@Table(name="CLIENT")
@NamedQuery(name="Client.findAll", query="SELECT c FROM Client c")
public class Client implements Serializable {
    private static final long serialVersionUID = 1L;

    //Had to add this for avoiding exception. And it works as expected
    //Dummy constructor for JPA - Workaround
    public Client(String s1, String s2){}

    @Column(name="requestid", unique=true,nullable=false)
    private String requestId;

    @Id
    @Column(name="clientid", unique=true, nullable=false, length=128)
    private String clientId;


    @OneToOne(fetch=FetchType.LAZY)
    @JoinColumn(name="REQUESTID", nullable=false)
    private RoccoRequest roccoRequest;

    //bi-directional One-To-Many association to ClientGroup
    @OneToMany(mappedBy="client",fetch=FetchType.LAZY)
    private List<ClientGroup> clientGroups;
....
,...
...
}

@Entity
@Embeddable
@Table(name="CLIENTGROUP")
@NamedQuery(name="ClientGroup.findAll", query="SELECT c FROM ClientGroup c")
public class ClientGroup implements Serializable {
    private static final long serialVersionUID = 1L;

    @EmbeddedId
    private ClientGroupPK id;

    @Column(length=32)
    private String type;

    @Column(name="clientid", length=128)
    private String clientId;

    //bi-directional many-to-one association to Client
    @ManyToOne(fetch=FetchType.EAGER)
    @MapsId("clientid")
    @JoinColumn(name="CLIENTID", nullable=true, insertable=false,  updatable=false)
    private Client client;
..
.
.
.
}

@Entity
@Table(name="ROCCOREQUEST")
@NamedQuery(name="RoccoRequest.CHECK_EXISISTING_CLIENT_DETAILS",
            query="SELECT r FROM RoccoRequest r JOIN r.client c WHERE c.crmId = :crmId")
public class RoccoRequest implements Serializable {
    private static final long serialVersionUID = 1L;

    @Id
    @Column(name="requestid", unique=true, nullable=false, length=128)
    private String requestId;

@OneToOne(mappedBy="roccoRequest", fetch=FetchType.LAZY, cascade={CascadeType.PERSIST, CascadeType.REMOVE})
    private Client client;
..
..
..

CriteriaQuery<Client> criteriaQuery =  criteriaBuilder.createQuery(Client.class);
            Root<Client> clientRoot =  criteriaQuery.from(Client.class);
            //Join the Client table with the RoccoRequest table
            final Join<Client, RoccoRequest> clientRoccoJoin = clientRoot.join(Client_.roccoRequest,JoinType.INNER); 

            final Path<String> _requestStatus = clientRoccoJoin.get(RoccoRequest_.statusCode);
            final Path<String> _requestId = clientRoccoJoin.get(RoccoRequest_.requestId);

            final Predicate _crmIdPredicate = criteriaBuilder.equal(clientRoot.get(Client_.crmId), CRMId);

            criteriaQuery.multiselect(_requestId,_requestStatus);
            criteriaQuery.where(_crmIdPredicate);

            //Get list of details of existing requests for the client with the request type as ACO
            clientDetails = entityManager.createQuery(criteriaQuery).getResultList();

            if(null != clientDetails) for(Client clientDetail : clientDetails){
                StatusBO statusDetails = new StatusBO();
                    statusDetails.setCode((clientDetail.getRoccoRequest().getStatusCode()));
                PreInitiationBO preinitiateDetails = new PreInitiationBO();
                    preinitiateDetails.getCaseHeader().setRequestId(requestId);
                    preinitiateDetails.setStatus(statusDetails);
                exisitngRequestInfo.add(preinitiateDetails);    
            }

我做了一些Criteria获取实体。但我得到的例外情况如下：

无法找到＆＃34;类com.xxx.xxx.model.Client＆＃34;的构造函数。同参数类型＆＃34; [类java.lang.String，类java.lang.String]＆＃34;至填写数据。

为什么JPA期望一个参数Constructor？它与协会有什么关系？我尝试删除OneToMany关系，但我仍然得到错误。

请注意我添加了一个对我来说毫无意义的2参数构造函数。但如果给出它，它就有效。日志根级别已启用调试。关于例外的信息非常少。

请帮忙。

Answer 1

正如JBNizet指出的那样，我通过添加带有两个字符串的多选来犯了一个愚蠢的错误，但是它有一个类型为Client.class的CrtieriaQuery。

这可以通过删除多选（不在我的情况下）或通过使用Tuples.class而不是Client.class制作CriteriaQuery和其他类型来解决，并通过元组循环并获取为tuple.get（0）等

问题已解决。谢谢@Neil和@JBNizet

JPA异常 - 无法使用参数类型＆＃34; [类java.lang.String，类java.lang.String]＆＃34;找到<class>的构造函数。填写数据

1 个答案: