Question

假设我有两个相关的实体，例如帐户类型和帐户看起来像这样

//Accounts.java
@Entity
@Table(name = "accounts")
@XmlRootElement
public class Accounts implements Serializable {

 @Id
 @GeneratedValue(strategy = GenerationType.IDENTITY)
 @Basic(optional = false)
 @Column(name = "account_id")
 private Long accountId;

 @JoinColumn(name = "account_type_id", referencedColumnName = "account_type_id")
 @ManyToOne(optional = false)
 private AccountTypes accountTypeId;

 @JoinColumn(name = "customer_id", referencedColumnName = "customer_id")
 @ManyToOne(optional = false)
 private CustomerInformation customerId;


//AccountTypes.java
@Entity
@Table(name = "account_types")
@XmlRootElement
public class AccountTypes implements Serializable {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
@Basic(optional = false)
@Column(name = "account_type_id")
private Integer accountTypeId;

@Basic(optional = false)
@NotNull
@Size(min = 1, max = 45)
@Column(name = "account_type_name")
private String accountTypeName;

@OneToMany(cascade = CascadeType.ALL, mappedBy = "accountTypeId")
private Collection<CustomerAccounts> customerAccountsCollection;

我希望能够

select from accounts where accounts.customerId=? and accountTypes.accountTypeName=?.

任何符合JPA标准的解决方案都适用于我，我只是不会运行两个不同的查询。

Answer 1

我假设CustomerInformation在customerId类型中有Long字段：

String hqlString = "select account from Accounts account where account.customerId.customerId=:customerId and account.accountTypeId.accountTypeName=:accountTypeName";

Query query = getSession().createQuery(hqlString);
query.setLong("customerId", customerId);
query.setString("accountTypeName", accountTypeName);

Accounts account = (Accounts) query.uniqueResult(); //if it's not unique then query.list()

我建议您将班级重命名为Account和AccountType。也不要像你一样用private CustomerInformation customerId;来命名一个具有id的对象。你将看到如何编写搞砸的hql。

Answer 2

您可以根据您创建的实体类使用以下查询。

select from Account acnt
where acnt.customerId=:customerId
and acnt.accountTypeId.accountTypeName = :accountTypeName

Answer 3

感谢您的回答，但这对我有用 -

String hql_string = "from Accounts acnt where acnt.customerId.id=:custmerID and acnt.accountTypeId.accountTypeName=:typeName";
    Query query = entityManager.createQuery(hql_string);
    query.setParameter("custmerID", new Long(customer_id));
    query.setParameter("typeName", "Account Type Name");

感谢#Mustafa，会更改变量名称。

JPA / Hibernate从两个不同列的实体中选择

3 个答案: