JPA标准：草案第

我是新来的JPA加入表自定义，我想使用API标准，以创建这个查询：JPA标准：草案第

SELECT company.* 
FROM company 
LEFT OUTER JOIN address ON company.id = address.company_id AND address.postal_code = '54000' 
ORDER BY company.id;

搜索和反思之后，我尝试这个代码，这似乎是合乎逻辑我说：

... 
CriteriaQuery<Company> query = criteriaBuilder.createQuery(Company.class) 
Root<Company> defCompany = query.from(Company.class); 
Join joinAddresses = defCompany.join("addresses", JoinType.LEFT); 
joinAddresses.on(criteriaBuilder.equal(joinAddresses.get("postalCode"), "54000")); 
query.orderBy(criteriaBuilder.asc(defCompany.get("id"))); 
List<Company> companies = entityManager.createQuer(query) 
    .getResultList(); 
...

但OpenJPA中抛出这个错误：

ERROR OpenEJB- EjbTransactionUtil.handleSystemException: org.apache.openjpa.persistence.criteria.Joins$List.on(Ljavax/persistence/criteria/Expression;)Ljavax/persistence/criteria/Join; 
    java.lang.AbstractMethodError: org.apache.openjpa.persistence.criteria.Joins$List.on(Ljavax/persistence/criteria/Expression;)Ljavax/persistence/criteria/Join;

我希望有人尝试过这样做，我觉得p可以，但我找不到关于此的线索或问题。

来源

2016-12-30 Charles Jourdan