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我正在制作一个网站,我有一个signup.php页面,用户可以在其中注册并将其信息输入到mysqli数据库中。当我这样做,我几乎没有,我只是不停地在这一行获得了一个问题:用户注册注册错误
ajax.send("&u="+u+"&e="+e+"&p="+p1+"&g="+g);
它基本上是发送变量的AJAX/JavaScript的检查,准备运输到服务器。但是我在这条线上遇到了内部服务器500错误。有任何想法吗?如果你想要,我会发布更多的代码。
function ajaxReturn(x){
if(x.readyState == 4 && x.status == 200){
return true;
}
}
function signup(){
var u = _("username").value;
var e = _("email").value;
var p1 = _("pass1").value;
var p2 = _("pass2").value;
var g = _("gender").value;
var status = _("status");
if(u == "" || e == "" || p1 == "" || p2 == "" || g == ""){
status.innerHTML = "Fill out all of the form data";
} else if(p1 != p2){
status.innerHTML = "Your password fields do not match";
} else {
_("signupbtn").style.display = "none";
status.innerHTML = 'please wait ...';
var ajax = ajaxObj("POST", "signup.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.";
}
}
}
type:post;
ajax.send("&u="+u+"&e="+e+"&p="+p1+"&g="+g);
}
}
我得到的东西看起来像这样的阵列(1){[“PHPSESSID”] =>字符串(32)“be48f6f5071fe91a3d5f6c654652a3da”} – elliotanderson
所以,它听起来像变量没有正确发送到服务器。查看使用jQuery的AJAX请求示例的更新答案。 – Travesty3