如何利用Ajax调用成功返回数据到php代码，在同一文件中运行mysql查询。

$.ajax({
  url: 'your_php_file.php',
  type: 'POST',
  data: {param1: value1, param2: value2}, // 根据需要传递的参数进行设置
  success: function(response) {
    // 在这里处理返回的数据
    console.log(response);
  },
  error: function(xhr, status, error) {
    // 处理请求错误
    console.log(error);
  }
});

<?php
// 连接MySQL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);
}

// 获取前端发送的数据
$param1 = $_POST['param1'];
$param2 = $_POST['param2'];

// 执行MySQL查询
$sql = "SELECT * FROM your_table WHERE column1 = '$param1' AND column2 = '$param2'";
$result = $conn->query($sql);

// 处理查询结果
if ($result->num_rows > 0) {
  while($row = $result->fetch_assoc()) {
    // 输出查询结果
    echo "字段1: " . $row["column1"]. " - 字段2: " . $row["column2"]. "<br>";
  }
} else {
  echo "0 结果";
}

// 关闭数据库连接
$conn->close();
?>