带有Hibernate参数的createSQLQuery

时间:2014-12-05 18:15:12

标签: java hibernate

我试图运行这个:

..
String modelTableName = (model == TableModelType.RING) ? "RING_PLAYERS" : "TOURNY_PLAYERS";

List<Object[]> results = dbs.createSQLQuery("SELECT group_level, COUNT(group_level) FROM :modelTableName WHERE game_id=:gameId GROUP BY group_level")
                .addScalar("group_level", Hibernate.INTEGER)
                .addScalar("COUNT(group_level)", Hibernate.LONG)
                .setString("modelTableName", getModelTableName())
                .setInteger("gameId", getGameId())
                .list();

它给出了一个例外:

Caused by: java.sql.SQLException: ORA-00903: invalid table name

如果我写&#34; RING_PLAYERS&#34;而不是&#34;:tableModelName&#34;然后它工作!

我对这个参数做错了什么?

谢谢。

1 个答案:

答案 0 :(得分:1)

根据这个答案,你不能注入表名:Table name as parameter in HQL

您可能需要将其构建为:

List<Object[]> results = dbs.createSQLQuery("SELECT group_level, COUNT(group_level) FROM " + getModelTableName() + " WHERE game_id=:gameId GROUP BY group_level")
                .addScalar("group_level", Hibernate.INTEGER)
                .addScalar("COUNT(group_level)", Hibernate.LONG)
                .setInteger("gameId", getGameId())
                .list();