使用 xmlhttprequest 插入数据库

Question

我想使用 ajax xmlhttprequest 将具有相同 ID 的名字、姓氏、电子邮件和密码插入数据库，然后使用 getElementById.value 获取所有值，但我不知道出了什么问题

function loadDoc() {
    var firstname, lastname, email, password;
    firstname = document.getElementById("firstname").value;
    lastname = document.getElementById("lastname").value;
    email = document.getElementById("email").value;
    password = document.getElementById("password").value;

    var xhttp = new XMLHttpRequest();
    xhttp.onreadystatechange = function() {
        if (this.readyState == 4 && this.status == 200) {
            document.getElementById("hero").innerHTML = this.responseText;
        }
    };
    xhttp.open("POST", "action.php", true);
    xhttp.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
    xhttp.send("firstname=" + firstname "&lastname=" + lastname "&email=" + email "&password=" + password);
}          

<form>
    firstname:<input type="text" id="firstname" name="firstname"><br>
    Lastname:<input type="text" id="lastname" name="lastname"><br>
    Email:<input id="email" type="email" name="email"><br>
    Password:<input id="password" type="password" name="password"><br>
    <input onclickt="loadDoc()" type="button" name="submit" value="Submit">
</form>

<div id="hero"></div>

这里是 $_POST[""] 中的 php 全部赋值

    $firstname = $_POST["firstname"];
    $lastname = $_POST["lastname"];
    $email = $_POST["email"];
    $password = $_POST["password"];

    if(empty($firstname) || empty($lastname) || empty($email) || empty($password)){
        echo '<h2>Empty Firstname</h2>';
        header("Location index.php");
    } else {
        $stmt = mysqli_stmt_init($connection);
        $query = "INSERT INTO user (firstname, lastname, email, password) VALUES (?, ?, ?, ?)";

        mysqli_stmt_prepare($stmt, $query);
        mysqli_stmt_bind_param($stmt, "ssss", $firstname, $lastname, $email, $password);
        mysqli_stmt_execute($stmt);
        mysqli_stmt_close($stmt);
    }
    mysqli_close($connection);

Answer 1

你有一个语法错误

  xhttp.send("firstname=" + firstname "&lastname=" + lastname "&email=" + email "&password=" + password);

改成

  xhttp.send("firstname=" + firstname + "&lastname=" + lastname + "&email=" + email + "&password=" + password);

您缺少“+”。

另外，“onclickt”上的另一个

<input onclickt="loadDoc()" type="button" name="submit" value="Submit">

应该是

<input onclick="loadDoc()" type="button" name="submit" value="Submit">