好。我正在尝试登录Twitter。此代码未打开该窗口。收到警报的响应不是null，而是指向登录屏幕的链接。有任何想法吗？

var url = "./twitter_login.php";
var con = createPHPRequest();

con.open("POST",url,true);
con.setRequestHeader("Content-type","application/x-www-form-urlencoded");
con.send("");

var response = "";

con.onreadystatechange = function() {

    if(con.readyState==4 && con.status==200) {

        response = con.responseText;    
        alert(response);
        window.open(response,"twitter","menubar=1,resizable=1,width=350,height=500");

    }

}