Question

我有以下hibernate查询字符串：

 String queryString = "select \r\n" +
                    "cms.my_status_id as 'myStatusId',\r\n" +
                    "cms.status_label as 'statusLabel',\r\n" +
                    "csl.status_id as 'companyStatusLabel'\r\n" +
                    "from "+client+".corresponding_status cms \r\n" +
                    "join "+client+".company_status_label csl on csl.status_id = cms.my_status_id";

我的对应实体是：

@Entity(name = "corresponding_status")
@Table(name = "corresponding_status")
public class CorrespondingStatus implements Serializable {

    @Id
    @JsonProperty
    @Column(name = "my_status_id")
    private Integer myStatusId;

    // varchar(255)
    @JsonProperty
    @Column(name = "status_label")
    private String statusLabel;

    @JsonProperty
    @Transient
    private String companyStatusLabel;

但是，当我运行查询时，我得到：

Column 'my_status_id' not found

即使它确实在数据库中。

这是什么问题？

Answer 1

在HQL中，必须使用属性而不是数据库列名称。将您的HQL更改为

String queryString = "select \r\n" +
                "cms.myStatusId as 'myStatusId',\r\n" +
                "cms.statusLabel as 'statusLabel',\r\n" +
                "csl.status_id as 'companyStatusLabel'\r\n" +
                "from "+client+".corresponding_status cms \r\n" +
                "join "+client+".company_status_label csl with csl.status_id = cms.myStatusId";

编辑：您可能需要相应地更改company_status_label实体

EDIT2：更改为WITH

Answer 2

我建议不要使用criteria API来手动构建JPA查询。您上面的查询将更改为：

String queryString = "select \r\n" +
                    "cms.my_status_id as 'myStatusId',\r\n" +
                    "cms.status_label as 'statusLabel',\r\n" +
                    "csl.status_id as 'companyStatusLabel'\r\n" +
                    "from "+client+".corresponding_status cms \r\n" +
                   "join "+client+".company_status_label csl on csl.status_id = cms.my_status_id";

类似于：

Session session = HibernateUtil.getHibernateSession();
CriteriaBuilder cb = session.getCriteriaBuilder();
CriteriaQuery<Entity> cq = cb.createQuery(Entity.class);
Root<Entity> root = cq.from(Entity.class);
cq.select(root);
Query<Entity> query = session.createQuery(cq);
List<Entity> results = query.getResultList();

SQLGrammarException：无法执行查询：找不到列？

2 个答案: