Question

我有以下java JPA项目的错误。谁能帮助我？谢谢。

类型例外报告

消息您试图为参数cityMakeMakeID设置类型类data.CityMake的值，期望类型为int from query string选择f from Foodmodel f WHERE f.cityMakeMakeID =：cityMakeMakeID order by f .foodModelName。

说明服务器遇到意外情况，无法完成请求。

异常

java.lang.IllegalArgumentException：您试图为参数cityMakeMakeID设置类型类data.CityMake的值，期望类型为int from query string选择f from Foodmodel f WHERE f.cityMakeMakeID =：cityMakeMakeID order by f.foodModelName

org.eclipse.persistence.internal.jpa.QueryImpl.setParameterInternal（QueryImpl.java:933）

org.eclipse.persistence.internal.jpa.EJBQueryImpl.setParameter（EJBQueryImpl.java:593）

dao.FoodModelDAO.getModels（FoodModelDAO.java:69）

servlets.StartServlet.processRequest（StartServlet.java:67）

servlets.StartServlet.doGet（StartServlet.java:86）

javax.servlet.http.HttpServlet.service（HttpServlet.java:635）

javax.servlet.http.HttpServlet.service（HttpServlet.java:742）

org.apache.tomcat.websocket.server.WsFilter.doFilter（WsFilter.java:52）

注意服务器日志中提供了根本原因的完整堆栈跟踪。

-------------------------------- ------------ FoodModelDAO.java ---------

公共类FoodModelDAO {

  public static List<Foodmodel> getModels(int makeid)
  {

    CityMake make = CityMakeDAO.findMakeById(makeid);

    EntityManager em = DBUtil.getEmFactory().createEntityManager();
    String qString = "Select f from Foodmodel f " +
                     "WHERE f.cityMakeMakeID = :cityMakeMakeID  " +
                     "order by f.foodModelName";

    TypedQuery<Foodmodel> query = em.createQuery(qString, Foodmodel.class);

      query.setParameter("cityMakeMakeID", make );

    try
    {
        List<Foodmodel> models = query.getResultList();
        return models;
    }

    catch(NoResultException e) 
    {
        return new ArrayList<Foodmodel>();
    }
    finally 
    {
        em.close();
    }
  }

}

--------------- FoodModel.java post 4/2/2018 ----------------------- 我再次更新了FoodModel的帖子。

enter code here




@Entity
public class FoodModel {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private Long    foodModelId;
private String  foodModelName;


@OneToOne
private CityMake cityMakeMakeID;

public FoodModel() {

    foodModelName = "";

}

public FoodModel(Long m, String v, CityMake makeObj) {
    foodModelId = m;
    foodModelName = v;
    //citymake = makeObj;
    cityMakeMakeID = makeObj;
}

public Long getModelId() {
    return foodModelId;
}

public void setModelId(Long id) {
    foodModelId = id;
}

public String getModelName() {
    return  foodModelName;
}

public void setModelName(String value) {
    this.foodModelName = value;
}

public CityMake getMake() {
    //return citymake;
    return cityMakeMakeID;
}

public void setMake(CityMake make) {
    //this.citymake = make;
    this.cityMakeMakeID = make;

 }

}

Answer 1

    query.setParameter("cityMakeMakeID", makeid);

为什么？您正在为Foodmodel.class创建一个类型化查询，因此参数类型应与模型相对应（查看FoodModel，该属性定义为int）。

    @Column(name = "CityMake_MakeID")
    private int cityMakeMakeID;

如果您的模型是这样定义的，那么您的工作方式应该有效：

    @OneToOne //Or any type of relation that matches your needs
    private CityMake cityMakeMakeID; //See the variable Type

Java JPA错误;您试图从查询字符串中设置类型类数据的期望类型int的值

1 个答案: