如何在CriteriaBuilder中编写此JPA选择语句

例如，这是基于电子邮件从数据库检索成员：

    CriteriaBuilder cb = em.getCriteriaBuilder();
    CriteriaQuery<Member> criteria = cb.createQuery(Member.class);
    Root<Member> member = criteria.from(Member.class);        

    criteria.select(member).where(cb.equal(member.get("email"), email));

    return em.createQuery(criteria).getSingleResult();

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将where子句更改为如下所示

where(cb.equal(member.get("email"), email), cb.equal(member.get("password", password));

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如果您计划支持具有可选值的多个条件，请将它们收集到一组谓词中，并使用

准则查询where（谓词…限制）方法。
这样，您不仅可以在where子句中包含多个必需的条件过滤器，还可以忽略没有值的条件（例如，因为用户没有填写这些条件）。

范例

where(cb.equal(member.get("email"), email), cb.equal(member.get("password", password));

CriteriaBuilder cb = em.getCriteriaBuilder();
CriteriaQuery<Member> criteria = cb.createQuery(Member.class);
Root<Member> member = criteria.from(Member.class);        

//create a holder for all conditions in `where` clause
List<Predicate> predicates = new ArrayList<>();
criteria.select(member).where(predicates.toArray(new Predicate[predicates.length()]));

//email is required
predicates.add(cb.equal(member.get("email"), email));
//password is required
predicates.add(cb.equal(member.get("password"), password));

//'status' is optional, only add it if it was not empty
if (status != null) {
  predicates.add(cb.equal(member.get("status"), status));
}

return em.createQuery(criteria).getSingleResult();

    <dependency>
        <groupId>org.hibernate</groupId>
        <artifactId>hibernate-jpamodelgen</artifactId>
        <scope>provided</scope>
    </dependency>