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Question

我在javascript函数中有ajax响应和内部html标记的奇怪错误。我试图在函数中更改一些但没有改变，它应该删除表单中的所有内部html并在PHP signup_success的ajax响应时添加一个“谢谢”语句，它只是echo ( signup_success )而没有别的发生。这是我的网站：web

这是函数的代码：

    function signup(){
    var u = _("username").value;
    var e = _("email").value;
    var p1 = _("pass1").value;
    var p2 = _("pass2").value;
    var c = _("country").value;
    var g = _("gender").value;
    var status = _("status");
    if(u == "" || e == "" || p1 == "" || p2 == "" || c == "" || g == ""){
        status.innerHTML = "Fill out all of the form data";
    } else if(p1 != p2){
        status.innerHTML = "Your password fields do not match";
    } else if( _("terms").style.display == "none"){
        status.innerHTML = "Please view the terms of use";
    } else {
        _("signupbtn").style.display = "none";
        status.innerHTML = 'please wait ...';
        var ajax = ajaxObj("POST", "core/register.php");
        ajax.onreadystatechange = function() {
            if(ajaxReturn(ajax) == true) {
                if(ajax.responseText != "signup_success"){
                    status.innerHTML = ajax.responseText;
                    _("signupbtn").style.display = "block";
                } else {
                    window.scrollTo(0,0);
                    _("signupform").innerHTML = "OK "+u+", check your email inbox and junk mail box at <u>"+e+"</u> in a moment to complete the sign up process by activating your account. You will not be able to do anything on the site until you successfully activate your account.";
                }
            }
        }
        ajax.send("u="+u+"&e="+e+"&p="+p1+"&c="+c+"&g="+g);
    }}