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我只想使用 JPA 编写一个简单的搜索引擎。以下是适用于 PHP 的本机 sql 请求示例：

"SELECT mobiles.brand, argus_waitings_mobiles_prices.mobile, argus_waitings_mobiles_prices.argus_buyer_id, "
                . "MATCH(mobile) AGAINST('$mobile_tags' IN BOOLEAN MODE) AS pertinence "
                . "FROM argus_waitings_mobiles_prices "
                . "WHERE argus_waitings_mobiles_prices.argus_buyer_id = $buyerId "
                . "ORDER BY pertinence DESC "
                . "LIMIT 10");

我想这样做，但它不起作用：

  @Override
    @SuppressWarnings("unchecked")
    @Transactional(readOnly=true)
    public List<Thread> findAllWithReferer(String referer) {
        EntityManager entityManager = EntityManagerUtil.createEntityManagerAndOpenTransaction();
        Query query = entityManager.createQuery("SELECT t "
            + "MATCH(t.title) AGAINST(:referer IN BOOLEAN MODE) AS pertinence "
            + "FROM Thread t MATCH(t.title) AGAINST(:referer IN BOOLEAN MODE)"
            + "ORDER BY pertinence DESC "
            + "LIMIT 10");

        return (query.getResultList());
    }

这是错误消息的开头：

Caused by: org.hibernate.hql.internal.ast.QuerySyntaxException: unexpected token: MATCH near line 1, column 40 [SELECT t FROM fr.dorian.model.Thread t MATCH(t.title) AGAINST(:referer IN BOOLEAN MODE)]

有人可以帮助我吗？
在此先感谢您的帮助。

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1 回答 1

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MATCH ... AGAINSTHibernate 不支持。您将不得不修改您的查询或使用 SQLentityManager.createSqlQuery(...)

于 2013-02-20T09:01:40.667 回答