ajax+java问题
ajax部分代码如下：function proce() {
if (XMLHttpReq.readyState == 4) {
if (XMLHttpReq.status == 200) {
//var root = XMLHttpReq.responseXML;
var xml = XMLHttpReq.responseXML;
var root = xml.documentElement;
var res = root.getElementsByTagName("content")[0].firstChild.data;
var test = root.getElementsByTagName("test")[0].firstChild.data;
if(res=="success")
{window.location.href("success.jsp");}
else
{window.alert(res);window.location.href("regist.jsp");}
} else {
window.alert("exception");
}
}
}
问题是：
res=success时，怎么还是进入到
{window.alert(res);window.location.href("regist.jsp");}这个里面来呢？？？
LoginAction代码如下：
public class LoginAction extends Action
{
public ActionForward execute(ActionMapping mapping, ActionForm actionform, HttpServletRequest request, HttpServletResponse response)
{
String name = request.getParameter("name");
String ps = request.getParameter("ps");
DB db=new DB();
try {
request.setCharacterEncoding("gb2312");
}
catch (UnsupportedEncodingException e)
{ e.printStackTrace(); }
try
{
String msgStr="";
response.setContentType("text/xml;charset=GB2312");
response.setHeader("Cache-Control","no-cache");
System.out.println("name = "+name+" ps ="+ps);
db.getCon();
db.getStmt();
boolean hasname=db.ifhas(name);
boolean match= db.Match(name, ps);
if(!hasname)
{ msgStr ="无此用户！"; }
else if(!match)
{ msgStr ="密码错误！"; }
else
{
msgStr ="success";
//return (mapping.findForward("success"));
}
System.out.println("msgStr = "+msgStr);
response.getWriter().println("<?xml version='1.0' encoding='GB2312' ?>");
response.getWriter().println("<root>");
response.getWriter().println("<content>");
response.getWriter().print(msgStr);
response.getWriter().println("</content>");
response.getWriter().println("<test>");
response.getWriter().println("正在测试。。。");
response.getWriter().println("</test>");
response.getWriter().println("</root>");
response.getWriter().close();
db.close();
System.out.println("关闭连接！");
return null;
}
catch(Exception ex)
{ ex.printStackTrace();
db.close();
System.out.println("异常，关闭连接！");
return null;
}
}
}
