分页休眠JPA：ORDERBY + setFirstResult + setMaxResult

我有工作了以下方法：分页休眠JPA：ORDERBY + setFirstResult + setMaxResult

public List<Course> filterOn(String course, String university, List<String> providers, String sortOn, int page) { 
    CriteriaBuilder criteriaBuilder = entityManager.getCriteriaBuilder(); 

    List<Predicate> predicates = new ArrayList<Predicate>(); 
    CriteriaQuery criteriaQuery = criteriaBuilder.createQuery(); 
    Root r = criteriaQuery.from(Course.class); 
    criteriaQuery.select(r); 

    predicates.add(criteriaBuilder.like(r.get("name"), "%" + course + "%")); 
    predicates.add(criteriaBuilder.like(r.get("university").get("name"), "%" + university + "%")); 

    predicates.add(criteriaBuilder.or(criteriaBuilder.like(r.get("type"), providers.get(0)), 
             criteriaBuilder.like(r.get("type"), providers.get(1)), 
             criteriaBuilder.like(r.get("type"), providers.get(2)), 
             criteriaBuilder.like(r.get("type"), providers.get(3)))); 

    if (predicates != null && !predicates.isEmpty()) { 
     criteriaQuery.where(criteriaBuilder.and(predicates.toArray(new Predicate[0]))); 
    } 

    if (sortOn.equals("ranks")) { 
     criteriaQuery.orderBy(criteriaBuilder.asc(r.get("averageScoreLastSemester"))); 
    } else if (sortOn.equals("name")) { 
     criteriaQuery.orderBy(criteriaBuilder.desc(r.get("name"))); 
    } 

    final TypedQuery query = entityManager.createQuery(criteriaQuery); 

    query.setFirstResult(page * COURSES_PER_PAGE); 
    query.setMaxResults(COURSES_PER_PAGE); 

    return query.getResultList(); 
}

的问题是，它只是订单当前页面。我希望它根据数据库中所有匹配的记录进行排序，然后返回当前页面。我怎样才能做到这一点？

从日志：

select 
     course0_.id as id1_14_, 
     course0_.created as created2_14_, 
     course0_.updated as updated3_14_, 
     course0_.averageScoreLastSemester as averageS4_14_, 
     course0_.avgAverageArray as avgAvera5_14_, 
     course0_.avgStructureArray as avgStruc6_14_, 
     course0_.avgTeachersArray as avgTeach7_14_, 
     course0_.avgWorkloadArray as avgWorkl8_14_, 
     course0_.code as code9_14_, 
     course0_.courseLink as courseL10_14_, 
     course0_.credits as credits11_14_, 
     course0_.level as level12_14_, 
     course0_.name as name13_14_, 
     course0_.numberOfAssignments as numberO14_14_, 
     course0_.numberOfExams as numberO15_14_, 
     course0_.numberOfProjects as numberO16_14_, 
     course0_.pace as pace17_14_, 
     course0_.rankCountArray as rankCou18_14_, 
     course0_.requirementsLink as require19_14_, 
     course0_.semester as semeste20_14_, 
     course0_.teacherName as teacher21_14_, 
     course0_.type as type22_14_, 
     course0_.university_id as univers23_14_ 
    from 
     courses course0_ cross 
    join 
     universities university1_ 
    where 
     course0_.university_id=university1_.id 
     and (
      course0_.name like ? 
     ) 
     and (
      university1_.name like ? 
     ) 
     and (
      course0_.type like ? 
      or course0_.type like ? 
      or course0_.type like ? 
      or course0_.type like ? 
     ) 
    order by 
     course0_.name desc limit ? 
Hibernate: 
    select 
     course0_.id as id1_14_, 
     course0_.created as created2_14_, 
     course0_.updated as updated3_14_, 
     course0_.averageScoreLastSemester as averageS4_14_, 
     course0_.avgAverageArray as avgAvera5_14_, 
     course0_.avgStructureArray as avgStruc6_14_, 
     course0_.avgTeachersArray as avgTeach7_14_, 
     course0_.avgWorkloadArray as avgWorkl8_14_, 
     course0_.code as code9_14_, 
     course0_.courseLink as courseL10_14_, 
     course0_.credits as credits11_14_, 
     course0_.level as level12_14_, 
     course0_.name as name13_14_, 
     course0_.numberOfAssignments as numberO14_14_, 
     course0_.numberOfExams as numberO15_14_, 
     course0_.numberOfProjects as numberO16_14_, 
     course0_.pace as pace17_14_, 
     course0_.rankCountArray as rankCou18_14_, 
     course0_.requirementsLink as require19_14_, 
     course0_.semester as semeste20_14_, 
     course0_.teacherName as teacher21_14_, 
     course0_.type as type22_14_, 
     course0_.university_id as univers23_14_ 
    from 
     courses course0_ cross 
    join 
     universities university1_ 
    where 
     course0_.university_id=university1_.id 
     and (
      course0_.name like ? 
     ) 
     and (
      university1_.name like ? 
     ) 
     and (
      course0_.type like ? 
      or course0_.type like ? 
      or course0_.type like ? 
      or course0_.type like ? 
     ) 
    order by 
     course0_.name desc limit ?

我猜的最后一行是很重要的，我读的地方，它应该首先排序的所有记录，然后限制他们。所以这似乎是正确的？

我使用hibernate方言org.hibernate.dialect.MySQL5Dialect和使用com.mysql.jdbc.Driver的数据源。

谢谢。

来源

2015-01-10 nilsi

数据库中的所有记录？但你正在使用限制（与喜欢）。我没有看到任何错误。 –

但是查询很奇怪（如结构），不是通过diferrents实体（课程*和*大学）的常见搜索。 –

我的意思是我想获得与名称，大学，提供者等属性相匹配的所有课程。这工作。但是，我希望通过课程成绩或名称来订购。然后返回一页。这不起作用。它返回一个页面，但它只在该页面内排序，而不是与我想要的所有匹配进行比较。 – nilsi

也许是视图技术的问题？您是否运行Sysout resultList以确认排序顺序问题来自JPA？当你改变到第二页时，过滤器仍然在那里？

来源

2015-06-11 00:35:14

分页休眠JPA：ORDERBY + setFirstResult + setMaxResult

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