如何使用php和ajax编辑基于国家的状态

<?php
// 连接数据库
$servername = "数据库服务器地址";
$username = "数据库用户名";
$password = "数据库密码";
$dbname = "数据库名称";

$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error) {
    die("连接失败: " . $conn->connect_error);
}

// 处理Ajax请求
if ($_SERVER["REQUEST_METHOD"] == "POST") {
    $country = $_POST["country"];
    $status = $_POST["status"];

    // 更新状态
    $sql = "UPDATE 表名 SET status='$status' WHERE country='$country'";
    if ($conn->query($sql) === TRUE) {
        echo "状态更新成功";
    } else {
        echo "状态更新失败: " . $conn->error;
    }
}

// 获取国家状态数据
$sql = "SELECT * FROM 表名";
$result = $conn->query($sql);

if ($result->num_rows > 0) {
    // 输出数据到前端
    while($row = $result->fetch_assoc()) {
        echo "国家：" . $row["country"]. " - 状态：" . $row["status"]. "<br>";
    }
} else {
    echo "没有数据";
}

$conn->close();
?>

function updateStatus(country, status) {
    var xhttp = new XMLHttpRequest();
    xhttp.onreadystatechange = function() {
        if (this.readyState == 4 && this.status == 200) {
            console.log(this.responseText);
        }
    };
    xhttp.open("POST", "后端处理代码的URL", true);
    xhttp.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
    xhttp.send("country=" + country + "&status=" + status);
}