11

我需要使用大量条件联接和 where 子句进行条件查询，在这种情况下，代码往往会变得复杂并且可能会产生重复的联接。

例如，我有以下表结构和 JPA 实体：

ACCOUNT
      ACCOUNT_ID
      ACCOUNT_TYPE


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

ADDRESS
      ADDRESS_ID
      LOCATION
      COUNTRY

所以假设我使用静态元模型实现来应用标准查询。

这是可以生成重复连接的错误代码示例：

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 SQL 与同一个表的多个连接：

 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留 Joins 的实例以在多个谓词中重用，例如：

   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，但是对于具有多个表和代码条件连接的真正复杂代码往往会变成意大利面条代码！相信我 ！

避免它的更好方法是什么？

4

2 回答 2

6

避免它的建议是使用构建器类来封装连接，见下文。

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，只需像下面这样调用构建器：

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

享受 ：）

于 2015-07-21T03:33:01.800 回答
5

我们使用以下实用方法来避免重复连接

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));
  }
}
于 2020-09-21T15:03:18.577 回答