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作者 | 正文 |
发表时间:2008-07-18
本人在使用Spring + hiberante中来根据用户名和用户密码来取得用户的基本信息,可是当用户名存在时,不是按预期所返回null,而在调用此方法时直接出现异常。 //------根据用户名和用户密码来取得用户的基本信息,如有多条,只取第一条 public Ctlm06 getUserInfo(String usrName,String Password) { try { Ctlm06 ctlm06 =null; String queryString = "select ctlm06 from Ctlm06 as ctlm06 where ctlm06.usrName="+usrName+" and ctlm06.usrPwd="+Password; List lstResutl=getHibernateTemplate().find(queryString); if(lstResutl.size()>0) //表示没有找到此用户信息 { ctlm06 = (Ctlm06) lstResutl.get(0); } return ctlm06; }catch (RuntimeException re) { log.error("save failed", re); throw re; } } //---------------------------------------在另外一个类中调用上方法 Ctlm06 ctlm06 = ctlm06Service.getUserInfo(strUserName, strPassword); if(ctlm06!=null){ //表示成功登陆 logger.log("用户名为:"+ctlm06.getUsrName()+"成功登陆本系统"); }但是直接报错;错误如下: javax.servlet.ServletException: Hibernate operation: could not execute query; bad SQL grammar [select ctlm06x0_.com_id as com1_3_, ctlm06x0_.usr_id as usr2_3_, ctlm06x0_.usr_pwd as usr3_3_, ctlm06x0_.usr_name as usr4_3_, ctlm06x0_.dept_id as dept5_3_, ctlm06x0_.wrkgrp_id as wrkgrp6_3_, ctlm06x0_.shgrp_flag as shgrp7_3_, ctlm06x0_.clerk_id as clerk8_3_ from rsgl.ctlm06 ctlm06x0_ where ctlm06x0_.usr_name=ggg and ctlm06x0_.usr_pwd=ggg]; nested exception is com.mysql.jdbc.exceptions.MySQLSyntaxErrorException: Unknown column 'ggg' in 'where clause' org.apache.struts.action.RequestProcessor.processException(RequestProcessor.java:523) org.apache.struts.action.RequestProcessor.processActionPerform(RequestProcessor.java:421) org.apache.struts.action.RequestProcessor.process(RequestProcessor.java:224) org.apache.struts.action.ActionServlet.process(ActionServlet.java:1194) org.apache.struts.action.ActionServlet.doPost(ActionServlet.java:432) javax.servlet.http.HttpServlet.service(HttpServlet.java:709) javax.servlet.http.HttpServlet.service(HttpServlet.java:802) haodong.filter.CharacterEncodingFilter.doFilter(CharacterEncodingFilter.java:23) 请各位高手看下,多谢!!!!!!!!!! 声明:ITeye文章版权属于作者,受法律保护。没有作者书面许可不得转载。
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