Hibernate性能问题

Question

我对Hibernate很新.在这里我想比较两个选项.

第一种选择

我的hibernate pojo类如下.

Stock {
  @Id
  @GeneratedValue(strategy = GenerationType.AUTO)
  @Column(name = "stock_id")
  private Long stockId;

  @Column(name = "stock_name")
  private String stockName;

  @ManyToMany(fetch = FetchType.EAGER)
  @JoinTable(name = "stock_characteristics", joinColumns =  {@JoinColumn(name = "stock_id")}, inverseJoinColumns = {@JoinColumn(name = "ct_id")})
  private List<Characteristic> characteristics = new ArrayList<>();

  //constructor, getters and setters

}

Characteristics {
  @Id
  @GeneratedValue(strategy = GenerationType.AUTO)
  @Column(name = "ct_id", nullable = false)
  private Long id;

  @Column(name = "name", nullable = false, length = 32)
  private String name;

  //constructor, getters and setters
}

每个库存包含一系列特征.每当我获取股票时,特征条目列表将关联并且结果正在获得.

我的库存表包含超过100万条记录,每条库存与10个特征相关联(因此stock_characteristics包含超过1000万行).当我们获取整个结果时,库存和特征之间的关联可能会变慢.

第二种选择.

我按如下方式重写了我的pojo类.

Stock {
  @Id
  @GeneratedValue(strategy = GenerationType.AUTO)
  @Column(name = "stock_id")
  private Long stockId;

  @Column(name = "stock_name")
  private String stockName;

  //constructor, getters and setters

}

特点-同上和

StockCharacteristics {
      @Id
      @GeneratedValue(strategy = GenerationType.AUTO)
      @Column(name = "id")
      private Long id;
      @Column(name = "stock_id", nullable = false)
      private Long stockId;
      @Column(name = "ct_id", nullable = false)
      private Long ctId;
    }

为了获得我的结果集,我只传递了一组特征.例如,如果将特征传递为2,那么首先我找到具有两个特征的股票ID.然后我将从Stock类投射股票详细信息.这是我的第一个选项的示例代码.

    criteria.createAlias("stock.characteristics", "stockCharacteristics",       CriteriaSpecification.INNER_JOIN).add(Restrictions.in("stockCharacteristics.id", listOfSelectedCharacteristics));
    List<Object[]> projectedList = criteria.setProjection(Projections.projectionList().add(Projections.count("id")).add(Projections.groupProperty("id"))).list();
    List<Long> stockIdList = new ArrayList<>();
    for(Object[] entry: projectedList){
      if(((Long) entry[0]).intValue() == listOfSelectedCharacteristics.size()){
           stockIdList.add((Long)entry[1]);
      }
    }

    if(!stockIdList.isEmpty()){
       Criteria criteriaWithCharacteristics =   getDb(true).createCriteria(Stock.class, "stock").add(Restrictions.in("id", stockIdList));
       selectedStocks = criteriaWithCharacteristics.setResultTransformer(Criteria.DISTINCT_ROOT_ENTITY).list();
    }

在这里,您可以看到在库存和特征之间执行连接查询,这可能会减慢速度和速度

这是我的第二个选项的示例代码

List<Object[]> stockIdList = //gets the stock id list from StockCharacteristics
    if(!stockIdList.isEmpty()){ 
       Criteria criteriaWithCharacteristics =   getDb(true).createCriteria(Stock.class, "stock").add(Restrictions.in("id", stockIdList));
       selectedStocks = criteriaWithCharacteristics.setResultTransformer(Criteria.DISTINCT_ROOT_ENTITY).list();
    }

作为一个程序观点,哪个是最佳选择？或者我应该使用哪一个以获得更好的性能？

Answer 1

我认为你应该在休眠中映射实体之间的所有关系。如果您想使用 hql 和条件，它将有所帮助，否则您将无法在实体之间进行联接。

为了提高性能，请将所有映射标记为 LAZY 并阅读以下内容：

https://zeroturnaround.com/rebellabs/how-to-use-jpa- Correctly-to-avoid-complaints-of-a-slow-application/

我经常使用 MS SQL Server 并检查慢速查询的执行计划，以确保我的查询有良好的索引。对于 mysql，您可以使用“显示索引”