Question

我需要使用大量条件连接和where子句进行条件查询，在这种情况下，代码会变得复杂并且可能会产生重复的连接。

例如，我有以下表格和JPA实体的结构：

ACCOUNT
      ACCOUNT_ID
      ACCOUNT_TYPE


PERSON
      NAME
      AGE
      ACCOUNT_ID ( FK TO ACCOUNT ) 
      ADDRESS_ID ( FK TO ADDRESS ) 

ADDRESS
      ADDRESS_ID
      LOCATION
      COUNTRY

假设我使用静态元模型实现来应用条件查询。

这是可以生成重复联接的错误代码的示例：

CriteriaBuilder cb = entityManager.getCriteriaBuilder();
    CriteriaQuery<Account> cq = cb.createQuery(Account.class);

    cq.select(accountRoot).where(
     cb.and(
      cb.equal(accountRoot.join(Account_.person).get(Person_.name),"Roger"),
      cb.greaterThan(accountRoot.join(Account_.person).get(Person_.age),18),
      cb.equal(accountRoot.join(Account_.person)                                   
              .join(Person_.address).get(Address_.country),"United States")
      )
     )

     TypedQuery<Account> query = entityManager.createQuery(cq);
     List<Account> result = query.getResultList();

上面的代码将生成一个带有同一个表的多个连接的SQL：

 Select
        account0_.account_id as account1_2_,
        account0_.account_type as account2_2_
    from
        account account0_
    inner join
        person person1_
            on account0_.account_id=person1_.account_id
    inner join
        address address2_
            on person1_.address_id=address2_.address_id
    inner join
        person person3_
            on account0_.account_id=person3_.account_id
    inner join
        person person4_
            on account0_.account_id=person4_.account_id
    inner join
        person person5_
            on account0_.account_id=person5_.account_id
    inner join
        address address6_
            on person5_.address_id=address6_.address_id
    where
        person3_.name=?
        and person4_.age>18
        and address6_.country=?

一个简单的解决方案是保持连接的实例在多个谓词中重用：

   Root<Account> accountRoot = cq.from(Account.class);
   Join<Account,Person> personJoin= accountRoot.join(Account_.person);
   Join<Person,Address> personAddressJoin = accountRoot.join(Person_.address);

   cq.select(accountRoot).where(
     cb.and(
      cb.equal(personJoin.get(Person_.name),"Roger"),
      cb.greaterThan(personJoin.get(Person_.age),18),
      cb.equal(personAddressJoin.get(Address_.country),"United States")
      )
     )

好的，它可以工作，但是有一个真正复杂的代码，有几个表和代码的条件连接往往会变成Spaghetti代码！相信我！

避免它的更好方法是什么？

Answer 1

我们使用以下实用程序方法来避免重复的连接

public class CriteriaApiUtils {
  public static <X, Y> ListJoin<X, Y> join(Root<X> criteriaRoot,
                                             ListAttribute<? super X, Y> attribute,
                                             JoinType joinType
  ) {
    return (ListJoin<X, Y>) criteriaRoot.getJoins().stream()
        .filter(j -> j.getAttribute().getName().equals(attribute.getName()) && j.getJoinType().equals(joinType))
        .findFirst()
        .orElseGet(() -> criteriaRoot.join(attribute, joinType));
  }
}

Answer 2

避免它的建议是使用构建器类来封装连接，请参阅下文。

public class AccountCriteriaBuilder {

        CriteriaBuilder cb;
        CriteriaQuery<Account> cq;

        // JOINS INSTANCE
        Root<Account> accountRoot;
        Join<Account,Person> personJoin;
        Join<Person,Address> personAddressJoin;

        public AccountCriteriaBuilder(CriteriaBuilder criteriaBuilder) {
            this.cb =  criteriaBuilder;
            this.cq = cb.createQuery(Account.class);
            this.accountRoot = cq.from(Account.class);
        }

        public CriteriaQuery buildQuery() {
            Predicate[] predicates = getPredicates();
            cq.select(accountRoot).where(predicates);
            return cq;
        }

        public Predicate[] getPredicates() {

           List<Predicate> predicates = new ArrayList<Predicate>();

           predicates.add(cb.equal(getPersonJoin().get(Person_.name), "Roger"));
           predicates.add(cb.greaterThan(getPersonJoin().get(Person_.age), 18));
           predicates.add(cb.equal(getPersonAddressJoin().get(Address_.country),"United States"));

           return predicates.toArray(new Predicate[predicates.size()]);
        }

        public Root<Account> getAccountRoot() {
            return accountRoot;
        }

        public Join<Account, Person> getPersonJoin() {
            if(personJoin == null){
                personJoin = getAccountRoot().join(Account_.person);
            }
            return personJoin;
        }

        public Join<Person, Address> getPersonAddressJoin() {
            if(personAddressJoin == null){
                personAddressJoin = getPersonJoin().join(Person_.address);
            }
            return personAddressJoin;
        }


}

“洞中的王牌”是每个所需连接实例的延迟加载，它将避免重复连接，并简化导航过程。

最后，只需按以下方式调用构建器：

AccountCriteriaBuilder criteriaBuilder = new AccountCriteriaBuilder(em.getCriteriaBuilder());
TypedQuery<Account> query = em.createQuery(criteriaBuilder.buildQuery());
List<Account> result = query.getResultList();

享受:)

JPA标准查询 - 如何避免重复联接

2 个答案: