Question

我正在使用Hibernate和MySQL并面临有关查询sql的问题

休眠服务：

List<RenterDto> list = new ArrayList<RenterDto>(0);
    try {
        Query query = session.createSQLQuery(sqlString);
        query.setParameter(0, name);
        list = query.setResultTransformer(new AliasToBeanResultTransformer(RenterDto.class)).list();
    } catch (HibernateException e) {
        logger.error("error at RenterDaoImpl.searchByNamePersonalId: " + e.getMessage());
    } finally {
        session.close();
    }
    return list;

SQL：

SELECT 
R.ID, 
R.NAME, 
R.PERSONAL_ID, 
R.PHONENUMBER, 
R.EMAIL
FROM 
  RENTER R 
WHERE 
  LOWER(R.NAME) 
LIKE 
  CONCAT('%', '?')

调试时记录消息：

RenterDaoImpl - error at RenterDaoImpl.searchByNamePersonalId: Position beyond number of declared ordinal parameters. 
Remember that ordinal parameters are 1-based! Position: 1

我该如何解决这个问题，非常感谢！

Answer 1

只需执行错误消息

$(document).on('submit', 'form', function(e) {
  e.preventDefault();
  var query = $(this).serialize()
  console.log(query)
})

请记住，序数参数是基于1的！位置：1在Mysql中

1 个答案: