hibernate oracle char character 只查出一个

public List queryTradeConfirmBusinFlag() {
		final String sql = "select C_BUSINFLAG, C_BUSINNAME from tbusinflag";
		
		return (List) getHibernateTemplate().execute(new HibernateCallback() {
			public Object doInHibernate(Session session) throws HibernateException {
				Query query = session.createSQLQuery(sql);
				
				List<InitBusinFlag> list = query.list();
				Iterator iter = list.iterator();
				list = new ArrayList<InitBusinFlag>();
				
				while(iter.hasNext()) {
					Object[] obj = (Object[])iter.next();
					InitBusinFlag businFlag = new InitBusinFlag();
					businFlag.setBusinFlag(obj[0].toString());
					businFlag.setBusinName(obj[1].toString());
					list.add(businFlag);
				}
				logger.debug("读取交易确认查询交易类型数据成功...");
				return list;
			}
		});
	}

C_BUSINFLAG 只能查出一个字符来

C_BUSINFLAG 在oracle数据库中是char(2)类型

改成final String sql = "select cast(C_BUSINFLAG as varchar2(2)), C_BUSINNAME from tbusinflag";

OK

评论

2 楼 zeallf 2010-10-21  

数据库设计遗留下来的问题，如果类型不匹配，会导致结果截断，只拿到一个字符

1 楼 genier200 2010-07-29  

谢谢，找了半天，终于在这里得到了正确的答案！