Question

我有一张表来记录登录活动。在我的应用仪表板中，我想在图表中显示最后一次登录。每小时间隔记录的用户数。查询将返回如下结果：

time                         count
-------                      --------
9.00.00 - 9.59.59            1
10.00.00 - 10.59.59          0
11.00.00 - 11.59.59          13
12.00.00 - 12.59.59          7
13.00.00 - 13.59.59          4

实体(LoginActivity)有dateLastUpdated字段，查询应在该字段上运行。获得预期结果的查询是什么？还将欣赏正确的本机解决方法查询。

Answer 1

试试这个

Query query = session.createQuery("select la from LoginActivity la where la.dateLastUpdated > :date");
Calendar cal = Calendar.getInstance();
cal.add(Calendar.HOUR, -12);
Date date = cal.getTime();
query.setParameter("date", date);
List result = query.list();

这是一个较短的例子

Query query = session.createQuery("select la from LoginActivity la where la.dateLastUpdated > :date");
query.setParameter("date", new Date(System.currentTimeMillis() - 12*60*60*1000);
List result = query.list();

修改

遵循Guillaume Polet的示例（upvoted），事实证明有一种JPA方法可以做到这一点（比使用数据库供应商特定的SQL函数要好得多）。这段代码对我有用

String query = "select hour(la.dateLastUpdated) as hour, count (la) from LoginActivity la " + "where la.dateLastUpdated > :date " + "group by hour(la.dateLastUpdated) " + "order by hour(la.dateLastUpdated)"; List<Object[]> objects = getSession().createQuery(query).setParameter("date", date).list(); for (Object[] o : objects) { Object hour = o[0]; Object count = o[1]; System.out.println(hour + ".00.00 - " + hour + ".59.59\t" + count); }

Answer 2

您的查询需要依赖一些hibernate细节来处理分组和时间间隔操作。

Coul没有测试它，但它应该是那些行

 String query = "Select day (la.dateLastUpdated) as day, hour(la.dateLastUpdated) as hour, count (la)
 From LoginActivity la
 Where la.dateLastUpdated > :date
 Group by day, hour
 Order by day, hour";

 List <Object []> objects = em.createQuery (query, Object [].class).setParameter ("date", new Date (System.currentTimeInMillis ()-TimeUnit.HOURS.toMillis (12))).getResultList ();
 for(Object [] o:objects) {
      Object hour = o[1];
      Object count = o [2];
      System.out.println (hour + ".00.00 - " + hour + "59.59\t" + count);
 }

用于获取最近12小时记录的hql查询

2 个答案: