Question

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());
    }

以下是错误消息的开头：

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。

Answer 1

Hibernate不支持

MATCH ... AGAINST。您必须修改查询或使用SQL entityManager.createSqlQuery(...)

使用createQuery的搜索引擎不起作用

1 个答案: