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我必須閱讀hibernate中的一些特定列。我已經寫了這樣的方法在hibernate中讀取對象格式的特定列
Branch getBranch(long branchId)
{
String query = "select distinct bran.branch.branchId,bran.branch.branchName from Location bran where bran.branch.branchId =:branchId";
Session session = getSession();
session.beginTransaction();
Query buildQuery = getSession().createQuery(query).setParameter("branchId", branchId);
List<Object[]> countryList = buildQuery.list();
Branch branch = null;
for(Object[] obj : countryList)
{
branch = new Branch((Long)obj[0] , (String)obj[1]);
}
session.close();
return branch;
}
但是在這裏,我必須手動構造我不想要的對象。我想讀取對象形式的結果,我不想手動構建對象。我的實體是這樣的
public final class Location implements Comparable<Location>{
@Id
@Column(name = Tables.LOCATION.LOCATION_ID, nullable = false, precision = 15, scale = 0, updatable = false, insertable = false)
private long locationId;
@Column(name = Tables.LOCATION.LOCATION_NAME, length = 60, updatable = false, insertable = false)
private String locationName;
@Embedded
private Country country;
@Embedded
private Branch branch;
分支不映射到實體,而它是可嵌入的。我想以DTO的形式讀取數據,所以我已經通過了ResultTransformer類,但它只能用sqlQuery而不是HQL(如果我錯了,請更正)。我不能改變我的查詢。請幫我
感謝亞歷山大。 。它工作正常,但讓我知道package.className始終是必需的,即使當我從同一個包調用它因爲已經嘗試過這種方式之前,它不工作時,我只使用類名。 – rishi