如何使用 AJAX 和 PHP 有效地將多個表單輸入提交到資料庫而無需刷新頁面？

<?php

   echo "<form method='post'>";

   $i=1;

    while($i <= $num_to_enter){

    $form_output .= "First Name:

    <input>

function MyFunction(){

  var i = 1;
  var x = $('#num_to_enter').val(); // Get the number of form sets

  var dataString = ''; // Initialize an empty data string

  // Iterate through the number of form sets
  while (i <= x){

    // Concatenate the data string using JavaScript's += operator
    dataString += "fname[" + i + "]=" + $('#fname[' + i + ']').val() + "&";
    dataString += "lname[" + i + "]=" + $('#lname[' + i + ']').val() + "&";
    dataString += "email[" + i + "]=" + $('#email[' + i + ']').val() + "&";

    i++;
  }

  // Remove the trailing '&' character from the data string
  dataString = dataString.substr(0, dataString.length - 1);

  // Send the data via AJAX to the 'process.php' script
  $.ajax({
    url: 'process.php', 
    type: 'POST', 
    data: dataString, // The concatenated data string is passed here
    success: function(data){

      // Handle the response from 'process.php'
      // Your code goes here
    },
    complete: function(){
      // Perform tasks after the AJAX request is complete
      // Your code goes here
    }
  });
}

<?php
  if($_SERVER['REQUEST_METHOD'] === 'POST') {

    // Parse the data from the POST request
    parse_str($_POST['data'], $formData);

    // Extract the individual data sets
    $fnames = $formData['fname'];
    $lnames = $formData['lname'];
    $emails = $formData['email'];

    // Initialize the success flag
    $success = true;

    // Perform the database insertions
    for ($i = 0; $i < count($fnames); $i++) {

      // Insert into database
      // ...

      // Check if any insertion failed
      if(!...){
        $success = false;
        break;
      }
    }

    // Return the success flag as JSON
    echo json_encode(array('success' => $success));
  }
?>