在'表達式中制定JPA標準'

Question

我在請求幫助以瞭解如何使用javax.persistence.criteria程序包制定'in'條件。

我正在創建基於聯繫人類的搜索條件CriteriaQuery。聯繫人可以屬於0到許多聯繫人類型。搜索標準可以包括姓氏值，聯繫人類型或兩者。

當我試試這個：

Expression<ContactType> param = criteriaBuilder.parameter(ContactType.class);   
Expression<List<ContactType>> contactTypes = fromContact.get("contactTypes"); 
Predicate newPredicate = param.in(this.getContactType(), contactTypes); 

我得到：

org.apache.openjpa.persistence.ArgumentException: Cannot execute query; declared parameters "ParameterExpression<ContactType>" were not given values. You must supply a value for each of the following parameters, in the given order: [ParameterExpression<ContactType>] 

我一直沒能找到如何做到這一點的好例子。任何幫助和指導都非常重要。完整的代碼如下。

public CriteriaQuery<Contact> getSearchCriteriaQuery(EntityManager entityManager) { 
    CriteriaBuilder criteriaBuilder = entityManager.getCriteriaBuilder(); 
    CriteriaQuery<Contact> criteriaQuery = criteriaBuilder.createQuery(Contact.class); 
    Root<Contact> fromContact = criteriaQuery.from(Contact.class); 
    Predicate whereClause = criteriaBuilder.equal(fromContact.get("domain"), this.getDomain()); 

    if (!StringUtils.isEmpty(this.getLastName())) { 
     Predicate newPredicate = criteriaBuilder.equal(fromContact.get("lastName"), this.getLastName()); 
     whereClause = criteriaBuilder.and(whereClause, newPredicate); 
    } 

    if (this.getContactType() != null) { 
     Expression<ContactType> param = criteriaBuilder.parameter(ContactType.class); 
     Expression<List<ContactType>> contactTypes = fromContact.get("contactTypes"); 
     Predicate newPredicate = param.in(this.getContactType(), contactTypes); 
     whereClause = criteriaBuilder.and(whereClause, newPredicate); 
    } 

    return criteriaQuery.where(whereClause); 
} 

@Entity 
@Table(name = "contact") 
public class Contact implements Serializable { 

    private static final long serialVersionUID = -2139645102271977237L; 
    private Long id; 
    private String firstName; 
    private String lastName; 
    private Domain domain; 
    private List<ContactType> contactTypes; 

    public Contact() { 
    } 

    @Id 
    @GeneratedValue(strategy = GenerationType.AUTO) 
    @Column(unique = true, nullable = false) 
    public Long getId() { 
     return this.id; 
    } 

    public void setId(Long id) { 
     this.id = id; 
    } 

    @Column(name = "FIRST_NAME", length = 20) 
    public String getFirstName() { 
     return this.firstName; 
    } 

    public void setFirstName(String firstName) { 
     this.firstName = firstName; 
    } 

    @Column(name = "LAST_NAME", length = 50) 
    public String getLastName() { 
     return this.lastName; 
    } 

    public void setLastName(String lastName) { 
     this.lastName = lastName; 
    } 

    //bi-directional many-to-one association to Domain 
    @ManyToOne(fetch = FetchType.LAZY) 
    @JoinColumn(name = "DOMAIN") 
    public Domain getDomain() { 
     return this.domain; 
    } 

    public void setDomain(Domain domain) { 
     this.domain = domain; 
    } 

    @ManyToMany 
    @JoinTable(name = "CONTACT_CNTTYPE", 
    joinColumns = { 
     @JoinColumn(name = "CONTACT", referencedColumnName = "ID")}, 
    inverseJoinColumns = { 
     @JoinColumn(name = "CONTACT_TYPE", referencedColumnName = "ID")}) 
    public List<ContactType> getContactTypes() { 
     return this.contactTypes; 
    } 

    public void setContactTypes(List<ContactType> contactTypes) { 
     this.contactTypes = contactTypes; 
    } 
} 

Answer 1

你必須參數值設置爲查詢時，您列出的結果：

TypedQuery<Entity> q = this.entityManager.createQuery(criteriaQuery); 
q.setParameter(ContactType.class, yourContactTypeValueToFilter); 
q.getResultList(); 

criteriaBuilder.parameter(ContactType.class); 

做什麼，是建立在查詢一個參數，即你需要稍後綁定。