各位高手,我写了个ajax程序,用来登录(使用ajax验证用户密码),验证通过,返回到jsp,如何重定向到另外页面 http_request.open("POST",linkurl,false);
http_request.setRequestHeader("Content-Type","application/x-www-form-urlencoded");
http_request.send(null); var returntxt=http_request.responseText;
if (returntxt!="ok")
{
alert(returntxt);
//lbl_verifyCode.innerHTML=returntxt;
window.location=returntxt;
// refreshVerifyCode();
}
else
{
window.open(returntxt);
}
此方法好像不行啊 ,谢谢了
http_request.setRequestHeader("Content-Type","application/x-www-form-urlencoded");
http_request.send(null); var returntxt=http_request.responseText;
if (returntxt!="ok")
{
alert(returntxt);
//lbl_verifyCode.innerHTML=returntxt;
window.location=returntxt;
// refreshVerifyCode();
}
else
{
window.open(returntxt);
}
此方法好像不行啊 ,谢谢了
window.location=".......";
怎么可以能不行呢?
你输出returntxt的值看看到底是不是url呢?
if(http_request.readyState==4)
{
if(http_request.status==200)
{
//请求成功返回
window.open("abc.jsp");
}
}
}
{
alert(returntxt);
//lbl_verifyCode.innerHTML=returntxt;
window.location="XYXYXY.jsp";
// refreshVerifyCode();
}
else
{
window.open("XXXX.jsp"); }
我原先后台这样写, targetPage = "message.jsp?msg=" + e.getExMsg();
out.println(targetPage);
不行,谢谢各位