PHP post请求未识别索引错误

你需要正确使用jquery。
例如 var email = $("email").val(); //IS WRONG 应该是(如果你有输入id="email") var email = $("#email").val(); 如果你只有名字，你可以使用 var email = $("[name='email']").val();

有点跑题了:如果你正在使用表单ajax提交考虑jquery方法序列化https://api.jquery.com/serialize/获取所有表单值(或一些jquery ajax插件)。

,请!不要使用不安全的mysql语句。看在上帝的份上，使用准备好的语句。如果您需要非常基本的东西，只需使用预处理语句或考虑https://phpdelusions.net/pdo/pdo_wrapper

还有一个小提示:在echo json之前制作json header <?php header('Content-type:application/json;charset=utf-8');

我认为你是错误的与你的jquery数据，他们应该有标识符像id表示的'#'和类表示的'。'，这样做，你有id="字段的名称"的输入参数:

$(document).ready(function() {
$("#x").click(function() {
    var email = $("#email").val();
    var pass = $("#password").val();
    var confirmPass = $("#confirmPassword").val();
    var name = $("#name").val();
    var question = $("#question").val();
    var answer = $("#answer").val();
    if(pass != confirmPass) {
        alert("Passwords do not match!");
        return;
    }
    var stuff = {email: email, pass: pass, name: name, question: question, answer: answer};
    $.ajax({method: "POST", url: "addAccount.php", data: stuff, success: function(result) {
        alert(result);
        window.location.href = "../Dashboard";
    }});
});
});

或者像这样，在输入参数中有class="name of the field"

$(document).ready(function() {
$("#x").click(function() {
    var email = $(".email").val();
    var pass = $(".password").val();
    var confirmPass = $(".confirmPassword").val();
    var name = $(".name").val();
    var question = $(".question").val();
    var answer = $(".answer").val();
    if(pass != confirmPass) {
        alert("Passwords do not match!");
        return;
    }
    var stuff = {email: email, pass: pass, name: name, question: question, answer: answer};
    $.ajax({method: "POST", url: "addAccount.php", data: stuff, success: function(result) {
        alert(result);
        window.location.href = "../Dashboard";
    }});
});
});

或者如果您想直接使用名称，请遵循以下命令:

$(document).ready(function() {
$("#x").click(function() {
    var email = $("input[name='email']").val();
    var pass = $("input[name='pasword']").val();
    var confirmPass = $("input[name='confirmPassword']").val();
    var name = $("input[name='name']").val();
    var question = $("input[name='question']").val();
    var answer = $("input[name='answer']").val();
    if(pass != confirmPass) {
        alert("Passwords do not match!");
        return;
    }
    var stuff = {email: email, pass: pass, name: name, question: question, answer: answer};
    $.ajax({method: "POST", url: "addAccount.php", data: stuff, success: function(result) {
        alert(result);
        window.location.href = "../Dashboard";
    }});
});
});

希望这对你有帮助

您的代码不能正常工作的原因有很多。@AucT和@gentle已经解决了Javascript方面的问题，所以我将专注于PHP。您的查询代码是:

$conn = new mysqli($servername, $username, $password, $dbname);
$sql = "...";
$conn->close();

注意:

• 永远不会执行您的查询。$sql只是一个保存在内存中的字符串。
• 你混合mysqli功能与mysql_功能(mysql_affected_rows);这行不通
• 你直接插入POST数据到你的查询，所以你很容易受到SQL注入
• 最后，您回显JSON，但您没有告诉浏览器期望这种格式

这样做:

$conn = new mysqli(...);
//SQL with ? in place of values is safe against SQL injection attacks
$sql = "INSERT INTO accounts (accountEmail, accountPassword,
          accountName, accountQuestion, accountRecover) VALUES (?, ?, ?, ?, ?)";
$error = null;
//prepare query and bind params. save any error
$stmt = $conn->prepare($sql);
$stmt->bind_param('sssss',$email,$pass,$name,$question,$answer)
        or $error = $stmt->error;
//run query. save any error
if(!$error) $stmt->execute() or $error = $stmt->error;
//error details are in $error
if($error) $response = "Error creating new account";
else $response = "Successfully created new account";
//set content-type header to tell the browser to expect JSON
header('Content-type: application/json');
$pre = ['Response' => $response];
echo json_encode($pre);