如何与JPA查询进行时间戳比较？

Question

我们需要确保在JPQL查询中仅返回过去30天内的结果.一个例子如下:

Date now = new Date();
Timestamp thirtyDaysAgo = new Timestamp(now.getTime() - 86400000*30);

Query query = em.createQuery(
  "SELECT msg FROM Message msg "+
  "WHERE msg.targetTime < CURRENT_TIMESTAMP AND msg.targetTime > {ts, '"+thirtyDaysAgo+"'}");
List result = query.getResultList();

这是我们收到的错误:

<openjpa-1.2.3-SNAPSHOT-r422266:907835 nonfatal user error> org.apache.openjpa.persistence.ArgumentException: An error occurred while parsing the query filter 'SELECT msg FROM BroadcastMessage msg WHERE msg.targetTime < CURRENT_TIMESTAMP AND msg.targetTime > {ts, '2010-04-18 04:15:37.827'}'. Error message: org.apache.openjpa.kernel.jpql.TokenMgrError: Lexical error at line 1, column 217.  Encountered: "{" (123), after : ""

救命!

Answer 1

因此,您输入的查询不是JPQL(您可以通过参考JPA规范看到).如果要将字段与日期进行比较,则将日期作为参数输入到查询中

msg.targetTime < CURRENT_TIMESTAMP AND msg.targetTime > :param

这不是SQL.

Answer 2

您正在使用的OpenJPA版本可能不支持JDBC转义语法.最新1.2.x版本的文档位于:http://openjpa.apache.org/builds/1.2.2/apache-openjpa-1.2.2/docs/manual/manual.html#jpa_langref_lit.

前面提到的文档涉及OpenJPA 2.0.0(最新)的文档:http://openjpa.apache.org/builds/latest/docs/manual/jpa_langref.html#jpa_langref_lit

那说你有什么理由想在你的JPQL中注入一个字符串？如下所示的片段呢？

Date now = new Date();
Date thirtyDaysAgo = new Date(now.getTime() - (30 * MS_IN_DAY));

Query q = em.createQuery("Select m from Message m " 
    + "where m.targetTime < :now and m.targetTime > :thirtyDays");
q.setParameter("now", now); 
q.setParameter("thirtyDays", thirtyDaysAgo);

List<Message> results = (List<Message>) q.getResultList();