2013-03-19 81 views
1

由于我的内部连接,Hiberate正在创建重复的对象。查询将父表与子表连接起来(父表与父子关系一对多)。Hibernate创建重复对象

数据:

Category (Parent) 
ID|Name 
1|A 
2|B 

Testcase (Child) 
ID|CategoryID|Name|Run 
1|1|A|500 
2|1|B|500 
3|1|C|500 
4|2|D|600 

从加入的结果,其中运行= 500

CategoryID|TestcaseID|TestCase Name 
1|1|A 
1|2|B 
1|3|C 

从此,我希望Hibernate来创建一个类的对象包含3个测试用例对象的列表。相反,它创建了3个类别对象,每个对象都带有3个测试用例的正确列表

类别[ID = 1,name = FOO,testCases = [TestCase [ID = 1,name = A,runId = 500],TestCase [ 2,name = B,runId = 500],TestCase [ID = 3,name = C,runId = 500]]]

类别[ID = 1,name = FOO,testCases = [TestCase [名称= C,runId = 500]]]

类别[ID = 1,名称= A,runId = 500],TestCase [ID = 2,名称= B,runId = 500] name = FOO,testCases = [TestCase [ID = 1,name = A,runId = 500],TestCase [ID = 2,name = B,runId = 500],TestCase [ID = 3,name = C,runId = 500 ]]]

型号:

@Entity 
@Table(name = "CATEGORY") 
public class Category implements Serializable 
{ 
    private static final long serialVersionUID = 1L; 

    @Id 
    @Column(name = "CATEGORYID") 
    private int ID; 

    @Column(name = "CATEGORYNAME") 
    private String name; 

    @OneToMany(fetch = FetchType.EAGER) 
    @JoinColumn(name = "CATEGORYID") 
    @Filter(name = "TEST_RUN_ID_FILTER") 
    private Collection<TestCase> testCases; 
} 

@Entity 
@Table(name = "TESTCASE_NEW") 
@FilterDef(name = "TEST_RUN_ID_FILTER", defaultCondition = "TESTRUNID in (:IDS)", parameters = { @ParamDef(name = "IDS", type = "int") }) 
public class TestCase implements Serializable 
{ 
    private static final long serialVersionUID = 1L; 

    @Id 
    @Column(name = "TESTCASEID") 
    private int ID; 

    @Column(name = "TESTCASENAME") 
    private String name; 

    @Column(name = "STATUS") 
    private String status; 

    @Column(name = "TESTRUNID") 
    private int testRunId; 
} 

DAO:

public List<Category> getAllCategoriesForTestRuns(List<Integer> testRunIDs) 
    { 
     Session session = getSession(); 
     session.enableFilter("TEST_RUN_ID_FILTER") 
       .setParameterList("IDS", testRunIDs); 
     Query query = session.createQuery("select c from Category c inner join c.testCases tc"); 
     List<Category> result = query.list(); 
     return result; 
    } 

我可以,如果我恰克的HQL查询来选择不同的,以获得正确的结果,但我想知道是否有更正确的方式。从我的谷歌搜索中,我尝试添加@Fetch(FetchMode.SELECT)到类别中的testCase列表,但没有任何效果。

谢谢!

回答

2

用途使用ResultTransformer:

Query query = session.createQuery("hql") 
        .setResultTransformer(Criteria.DISTINCT_ROOT_ENTITY); 

的问题是,你的内部联接将返回的测试用例集合中的每个实体对象(作为DB查询结果集将为内部连接)。

+0

非常有帮助谢谢。当你将它看作结果集时,它确实有意义!我一直将每个结果对象添加到一个集合中,以滤除总是感觉错误的重复。 – JLove 2014-02-05 16:46:06