如何避免Hibernate的 fails to correctly determine parameter type错误

public List findByNamedQuery(String queryName, String[] paramNames, Object[] paramValues) {
		if (paramNames.length != paramValues.length) {
			throw new IllegalArgumentException();
		}
		Map<String, Object> map = new java.util.HashMap<String, Object>(paramNames.length);
		for (int i = 0; i < paramNames.length; ++i) {
			map.put(paramNames[i], paramValues[i]);
		}
		return getJpaTemplate().findByNamedQueryAndNamedParams(queryName, map);
	}



List findByNamedQuery = findByNamedQuery("findXXXByNameandDateRange", 
								new String[] { "aName","startDate","endDate" },
								new Object[] { 
									aName, 
									(startDate==null ? "" : startDate) ,
									(endDate==null ? "" : endDate) 
								});