JPA 2 CriteriaQuery问题

时间:2011-01-20 02:23:23

标签: java jpa jpa-2.0 criteriaquery

我刚刚开始使用JPA 2标准查询API并且发现它很难学习。看了一下网络,但还没有找到好的例子/教程。有人可以建议一个很好的教程和/或帮我处理以下我想编写的简单查询吗?

我有一个名为Transaction的类,它引用了它所属的Account:

public class Transaction {
    private Account account;
    ...
}

public class Account {
    private Long id;
    ...
}

我需要编写一个查询,根据帐户ID获取帐户的所有交易。这是我尝试这样做(显然不起作用):

public List<Transaction> findTransactions(Long accountId) {        
    CriteriaBuilder builder = entityManager.getCriteriaBuilder();
    CriteriaQuery<Transaction> query = builder.createQuery(Transaction.class);
    Root<Transaction> transaction = query.from(Transaction.class);

    // Don't know if I can do "account.id" here
    query.where(builder.equal(transaction.get("account.id"), accountId));
    return entityManager.createQuery(query).getResultList();
}

有人能指出我正确的方向吗?

感谢。 纳雷什

1 个答案:

答案 0 :(得分:7)

解决方案: -

public List<Transaction> findTransactions(Long accountId) { 
        CriteriaBuilder builder = entityManager.getCriteriaBuilder();
        CriteriaQuery<Transaction> query = builder.createQuery(Transaction.class);
        Root<Transaction> _transaction = query.from(Transaction.class);

        Path<Account> _account = _transaction.get(Transaction_.account);
        Path<Long> _accountId = _account.get(Account_.id);

        query.where(builder.equal(_accountId, accountId));
        return entityManager.createQuery(query).getResultList();
    }

要理解上述代码的含义,请阅读: - Dynamic, typesafe queries in JPA 2.0

要了解/生成JPA Metamodel,请阅读: - Hibernate Metamodel Generator Reference Guide