带有谓词的条件生成器答案

【问题标题】：Criteria Builder with predicate带有谓词的条件生成器
【发布时间】：2019-12-13 16:19:43
【问题描述】：

我有我将从属性文件中读取的对象列表，然后我将使用管道符号拆分，然后我需要使用 or 子句应用。下面是代码。当我尝试应用 builder.or 时，正在生成以下查询，而不是 or 子句。

    String coverageNames = env.getRequiredProperty(AmhiConstants.COVERAGE_NAME);
    String[] coverageName = coverageNames.split(Pattern.quote("|"));
    CriteriaBuilder builder = manager.getCriteriaBuilder();
    TypedQuery typedQuery;
    CriteriaQuery<Integer> criteriaQuery = builder.createQuery(Integer.class);
    Root<BenefitCounter> root = criteriaQuery.from(BenefitCounter.class);
    Join<BenefitCounter, BenefitCounterCoverageDetails> benefitCoverageJoin = root
            .join(BenefitCounter_.benefitCoverageDetails);
    Join<BenefitCounterCoverageDetails, CoverageMaster> coveJoin = benefitCoverageJoin
            .join(BenefitCounterCoverageDetails_.coverageType);
    List<Predicate> conditions = new ArrayList<>();

    conditions.add(builder.and(builder.equal(root.get(BenefitCounter_.isCompleted), 'N')));
    for (String coverage : coverageName) {
        Predicate predicate = builder.like(builder.lower(coveJoin.get(CoverageMaster_.coverageName)),
                AmhiConstants.PERCENTAGE_SYMBOL + coverage + AmhiConstants.PERCENTAGE_SYMBOL);
        conditions.add(conditions.add(builder.or(predicate));

    }

    criteriaQuery.multiselect(root.get(BenefitCounter_.benefitCounterId))
            .where(conditions.toArray(new Predicate[] {}));


    criteriaQuery.orderBy(builder.asc(root.get(BenefitCounter_.benefitCounterId)));
    criteriaQuery.distinct(true);
    typedQuery = manager.createQuery(criteriaQuery);
    List<Integer> list = typedQuery.getResultList();
    return list;

生成的查询：

select distinct benefitcou0_.utilization_dtl_id as col_0_0_ from claims.utilization_dtls benefitcou0_ inner join claims.utilization_coverage_dtls benefitcov1_ on benefitcou0_.utilization_dtl_id=benefitcov1_.utilization_dtl_id inner join claims.coverage_master coveragema2_ on benefitcov1_.coverage_id=coveragema2_.coverage_id where (lower(coveragema2_.coverage_name) like ?) and (lower(coveragema2_.coverage_name) like ?) and (lower(coveragema2_.coverage_name) like ?) and (lower(coveragema2_.coverage_name) like ?) and benefitcou0_.is_completed=? order by benefitcou0_.utilization_dtl_id asc

需要实际查询：

select distinct benefitcou0_.utilization_dtl_id as col_0_0_ from claims.utilization_dtls benefitcou0_ inner join claims.utilization_coverage_dtls benefitcov1_ on benefitcou0_.utilization_dtl_id=benefitcov1_.utilization_dtl_id inner join claims.coverage_master coveragema2_ on benefitcov1_.coverage_id=coveragema2_.coverage_id where (lower(coveragema2_.coverage_name) like ?) or (lower(coveragema2_.coverage_name) like ?) or(lower(coveragema2_.coverage_name) like ?) or (lower(coveragema2_.coverage_name) like ?) and benefitcou0_.is_completed=? order by benefitcou0_.utilization_dtl_id asc

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标签： hibernate jpa criteria-api

【解决方案1】：

这是因为CriteriaQuery<T>.where(Predicate... restrictions)，根据JavaDoc，这是您使用的方法，

...修改查询，根据指定的限制谓词的连词来限制查询结果

连词表示一组“和”谓词。为了得到“或”，你必须使用CriteriaBuilder.or(Predicate... restrictions)：

Predicate andPredicate = builder.equal(root.get(BenefitCounter_.isCompleted), 'N');
for (String coverage : coverageName) {
    Predicate predicate = builder.like(builder.lower(coveJoin.get(CoverageMaster_.coverageName)),
            AmhiConstants.PERCENTAGE_SYMBOL + coverage + AmhiConstants.PERCENTAGE_SYMBOL);
    conditions.add(predicate);

}
Predicate orPredicate = builder.or(conditions.toArray(new Predicate[] {}));
Predicate andPredicate = builder.and(andPredicate, orPredicate);

criteriaQuery.multiselect(root.get(BenefitCounter_.benefitCounterId))
        .where(andPredicate);

【讨论】：