Question

以下是对问题的更多更好的描述：

我做一个简单的选择查询。返回列表包含确切的记录/对象数，就像我在数据库中执行相同的查询一样，但问题是所有对象都相同/相同。

例如，这是DB中的结果（我删除了空值）：
26801 01-JAN-00 7 31-DEC-99 7 Obere Kirchstr。 26 CH 8304 Walliselln
26801 01-JAN-00 2 31-DEC-99 2 Obere Kirchstr。 26 CH 8304 Walliselln

这是变量地址的内容，因为我可以在执行查询后的调试期间在Eclipse中看到它：

地址ArrayList（id = 81）
  elementData Object [10]（id = 86）
    [0] DLDBAddress（id = 82）
    [1] DLDBAddress（id = 82）
    [2] null
    ...
  modCount 1
  2号

DLDBAddress [persid = 26801，valPeriodStart = 1900-01-01，valPeriodEnd = 9999-12-31，addressNr = 7，addressType = 7，addressRow1 = null，addressRow2 = Obere Kirchstr。 26，addressRow3 = null，country = CH，postalCode = 8304，city = Walliselln，phoneNr = null，faxNr = null，sekretaryPhoneNr = null，alternatPhoneNr = null，pagerNr = null]

DLDBAddress [persid = 26801，valPeriodStart = 1900-01-01，valPeriodEnd = 9999-12-31，addressNr = 7，addressType = 7，addressRow1 = null，addressRow2 = Obere Kirchstr。 26，addressRow3 = null，country = CH，postalCode = 8304，city = Walliselln，phoneNr = null，faxNr = null，sekretaryPhoneNr = null，alternatPhoneNr = null，pagerNr = null]]

如您所见，这两个对象是相同的。它们应该与addressNr和addressType ...

不同

这是构建查询的代码段：

public static <T> List<T> findBy(EntityManager eM, Class<T> boClass, String whereClause, String whereValue)  
 {
    EntityManager entityManager = eM;
    Query query = entityManager.createQuery("from " + boClass.getName() + " s where s." + whereClause + " = " + whereValue);
    ...
    return (List<T>) query.getResultList();
}

这是（非常简单的）结果查询：

来自ch.ethz.id.wai.pdb.bo.DLDBAddress，其中s.persid = 26801

这是生成的查询：

Hibernate: 
select
dldbaddres0_.PERSID as PERSID0_,
dldbaddres0_.ADRNUM as ADRNUM0_,
dldbaddres0_.ADRZEIL1 as ADRZEIL3_0_,
dldbaddres0_.ADRZEIL2 as ADRZEIL4_0_,
dldbaddres0_.ADRZEIL3 as ADRZEIL5_0_,
dldbaddres0_.ADRTYP as ADRTYP0_,
dldbaddres0_.ADRAUSWTEL as ADRAUSWTEL0_,
dldbaddres0_.ADRORT as ADRORT0_,
dldbaddres0_.ADRLAND as ADRLAND0_,
dldbaddres0_.ADRFAX as ADRFAX0_,
dldbaddres0_.ADRPSA as ADRPSA0_,
dldbaddres0_.ADRTEL as ADRTEL0_,
dldbaddres0_.ADRPLZ as ADRPLZ0_,
dldbaddres0_.ADRSEKTEL as ADRSEKTEL0_,
dldbaddres0_.BISDAT as BISDAT0_,
dldbaddres0_.VONDAT as VONDAT0_ 
from
NETHZ.V_DLDB_ADRESSE dldbaddres0_ 
where
dldbaddres0_.PERSID=26801

这里是实体：

@Entity
@Table(name = "V_DLDB_ADRESSE", schema="NETHZ")
public class DLDBAddress
{
  @Id
  @Column(name = "PERSID", insertable = false, updatable = false)
  private Integer   persid;
  @Temporal(TemporalType.DATE)
  @Column(name = "VONDAT", insertable = false, updatable = false)
  private Date   valPeriodStart;
  @Temporal(TemporalType.DATE)
  @Column(name = "BISDAT", insertable = false, updatable = false)
  private Date   valPeriodEnd;
  @Column(name = "ADRNUM", insertable = false, updatable = false)
  private Integer addressNr;
  @Column(name = "ADRTYP", insertable = false, updatable = false)
  private Integer addressType;
  @Column(name = "ADRZEIL1", insertable = false, updatable = false)
  private String addressRow1;
  @Column(name = "ADRZEIL2", insertable = false, updatable = false)
  private String addressRow2;
  @Column(name = "ADRZEIL3", insertable = false, updatable = false)
  private String addressRow3;
  @Column(name = "ADRLAND", insertable = false, updatable = false)
  private String country;
  @Column(name = "ADRPLZ", insertable = false, updatable = false)
  private String postalCode;
  @Column(name = "ADRORT", insertable = false, updatable = false)
  private String city;
  @Column(name = "ADRTEL", insertable = false, updatable = false)
  private String phoneNr;
  @Column(name = "ADRFAX", insertable = false, updatable = false)
  private String faxNr;
  @Column(name = "ADRSEKTEL", insertable = false, updatable = false)
  private String secretaryPhoneNr;
  @Column(name = "ADRAUSWTEL", insertable = false, updatable = false)
  private String alternatPhoneNr;
  @Column(name = "ADRPSA", insertable = false, updatable = false)
  private String pagerNr;

...

我错过了什么吗？

啊，我正在连接到Oracle DB。

提前致谢弗朗西斯

Answer 1

where dldbaddres0_.PERSID=26801

@Id
  @Column(name = "PERSID", insertable = false, updatable = false)

您将PERSID定义为@Id，这是一个主键。它对您的应用来说真的很独特吗？从行为来看并非如此。但对于Hib来说，它必须是。

那会发生什么：

数据库中有两条+记录，PERSID = 26801
您查询WHERE PERSID = 26801
SQL Query返回两个+行
Hib加载第一个，并以PERSID作为键进入会话（因为它标记为@Id）。对象放在结果列表中。
Hib加载第二个，注意到具有相同@Id的对象已经在会话中，并且只是将引用放入结果列表中。行数据被忽略。
因此你得到两份+副本。

Hibernate - TypedQuery.getResultList（）返回同一对象的列表

1 个答案: