hibernate - 从表中选择全部休眠

Question

所以我有以下代码：

Query query = session.createQuery("from Weather");
        List<WeatherModel> list = query.list();
        WeatherModel w = (WeatherModel) list.get(0);

我不想从表 Weather 中获取所有项目，但我不断收到以下错误：（第 23 行是我创建查询的地方）

java.lang.NullPointerException
    at action.WeatherAction.validate(WeatherAction.java:23)
    at com.opensymphony.xwork2.validator.ValidationInterceptor.doBeforeInvocation(ValidationInterceptor.java:251)
    at com.opensymphony.xwork2.validator.ValidationInterceptor.doIntercept(ValidationInterceptor.java:263)............

有什么问题？

score 11 · Accepted Answer

Query query = session.createQuery("from Weather"); //You will get Weayher object
List<WeatherModel> list = query.list(); //You are accessing  as list<WeatherModel>

他们都是不同的实体

Query query = session.createQuery("from Weather"); 

 List<Weather> list = query.list(); 

Weather w = (Weather) list.get(0);

score 6 · Accepted Answer

在复杂的项目中，我不喜欢在字符串文字中硬编码实体名称：

Session session = sessionFactory.getCurrentSession();

CriteriaBuilder criteriaBuilder = session.getCriteriaBuilder();
CriteriaQuery<WeatherModel> criteriaQuery = 
    criteriaBuilder.createQuery(WeatherModel.class);

// Your underlying table name might change in the future.
// Let hibernate take care of the names.
Root<WeatherModel> root = criteriaQuery.from(WeatherModel.class);
criteriaQuery.select(root);

Query<WeatherModel> query = session.createQuery(criteriaQuery);
List<WeatherModel> weatherModelList = query.getResultList();

score 1 · Accepted Answer

我刚刚遇到了类似的问题，它似乎已经通过提供您尝试查询的对象的完整路径来解决。所以，当我让它看起来像这样时： session.createQuery("from com.mystuff.something.or.other.MyEntity") 它起作用了。

score 0 · Accepted Answer

Weather 将是与 WeatherModel 不同的实体。您的列表将包含 Weather 对象，只有当它是 WeatherModel 的子类型时才能强制转换

hibernate - 从表中选择全部休眠

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