如何使用php mysql在弹出窗口中显示记录

<?php
// 连接到数据库
$servername = "localhost";
$username = "your_username";
$password = "your_password";
$dbname = "your_database";

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

// 查询数据库
$sql = "SELECT * FROM your_table";
$result = $conn->query($sql);

// 生成HTML和JavaScript代码
echo '<script>
function showRecord(id) {
    // 创建弹出窗口
    var popup = window.open("", "Popup", "width=400,height=300");

    // 查询数据库以获取特定记录
    var xmlhttp = new XMLHttpRequest();
    xmlhttp.onreadystatechange = function() {
        if (this.readyState == 4 && this.status == 200) {
            // 将记录填充到弹出窗口中
            popup.document.body.innerHTML = this.responseText;
        }
    };
    xmlhttp.open("GET", "get_record.php?id=" + id, true);
    xmlhttp.send();
}
</script>';

// 显示记录
if ($result->num_rows > 0) {
    while ($row = $result->fetch_assoc()) {
        $id = $row["id"];
        $name = $row["name"];

        // 在页面上显示记录，并在点击时调用showRecord函数显示弹出窗口
        echo '<a href="javascript:void(0);" onclick="showRecord(' . $id . ');">' . $name . '</a><br>';
    }
} else {
    echo "没有记录";
}

$conn->close();
?>