ibatis in not in备案

xiaoxiao2026-03-16  9

Map<Long, MSUser> ret = null; String params = StringUtils.toStringValue(ids); if(params==null||params.length()==0)return ret; Map<String, String> map = new HashMap<String, String>(); map.put("id", "1,2,3,4"); map.put("tt", "2,3"); List<MSUser> msUsers = getSqlMapClientTemplate().queryForList("getTsByIds",map); System.out.println(msUsers.size());//2条,即1,4 return ret; <select id="getTsByIds" resultClass="MSUser"> select * from ms_user where id in ($id$) and id not in($tt$) </select> 虽然是猜测的,结果中了,果然是通过map的key来区分的,如果你传递的是一个字符串则会把id和tt都替换成同一个值了。
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