Question

代码类TypeCompetence

的代码

@OneToMany(cascade = CascadeType.ALL, fetch = FetchType.LAZY, mappedBy = "typeCompetence")
    public Set<Criteres> getCritereses() {
        return this.critereses;
    }

Criteres类的代码

@ManyToOne(fetch = FetchType.LAZY)
    @JoinColumn(name = "id_type_compete")
    public TypeCompetence getTypeCompetence() {
        return this.typeCompetence;
    }

CriteresDAO类的代码

public List<Criteres> FindidAspetComp(){

        String queryString = "from Criteres c ,TypeCompetence t where c.id_type_compete = t.id_type_compete AND t.id_type_compete = 4 ";
        return getHibernateTemplate().find(queryString);
    }

错误

javax.servlet.ServletException: could not resolve property: id_type_compete of: tn.com.ModuleEVL.domaine.Criteres [from tn.com.ModuleEVL.domaine.Criteres c ,tn.com.ModuleEVL.domaine.TypeCompetence t where c.id_type_compete = t.id_type_compete AND t.id_type_compete = 4 ];

Answer 1

HQL 从不使用表名和列名。它始终使用实体名称和字段/属性名称。你所需要的只是

select c from Criteres c where c.typeCompetence.id = 4

阅读the documentation。请修改你的命名。实体不应具有复数形式，因此Criteres应命名为Critere。 getCritereses()应命名为getCriteres()。我甚至无法理解FindidAspetComp()应该是什么意思。它不遵守Java命名约定：方法以小写字母开头，并且是camelCaded ..

查询加入HibernateTemplate（）

1 个答案: