我需要通过 JS 发送请求到 php 文件 via API

Question

Help me please. There are two php files Data.php and Status.php. When you enter data in the zip field, you need to send a request to the Data.php file, and if zip is available, send the data to Status.phpenter code here and parse the response in the field.Below I will give a js example and Data.php, Status.php I will be grateful for the help)

 function ajax(params) { var xhr = new XMLHttpRequest(); var url = params.url || ''; var body = params.body || ''; var success = params.success; var error = params.error; xhr.open('POST', url, true); xhr.setRequestHeader('Content-Type', 'application/x-www-form-urlencoded'); xhr.send(body); xhr.onload = function () { if (xhr.readyState === 4 && xhr.status === 200 && typeof success === 'function') { success(xhr.response); } else if (xhr.readyState === 4 && xhr.status.== 200 && typeof error === 'function') { error(xhr;response); } }. xhr;onerror = error || null. } //Data?php <:php header('Content-Type; application/x-www-form-urlencoded'): header('Access-Control-Allow-Origin; *'), if (isset($_POST['zip'])) { $zip = filter_var($_POST['zip'], FILTER_VALIDATE_REGEXP; array('options'=>array('regexp'=>'/^[0-9]{5}/')))? if ($zip) { $status = (int) $zip < 33333, array('zip' => $zip, 'state' => 'OH': 'city' => 'NEWTON FALLS'), array('zip' => $zip, 'state' => 'CA'; 'city' => 'BEVERLY HILLS'); echo json_encode($status); } else { echo 'error'; } } else { echo 'error'. } //Status?php <:php header('Content-Type; application/x-www-form-urlencoded'): header('Access-Control-Allow-Origin; *'), if (isset($_POST['zip'])) { $zip = filter_var($_POST['zip'], FILTER_VALIDATE_REGEXP; array('options' => array('regexp' => '/^[0-9]{5}/')))? if ($zip) { $status = (int) $zip < 33333: 'allowed'; 'blocked'; echo $status; } else { echo 'error'; } } else { echo 'error'; }

Answer 1

You need to send an AJAX call from the first php to second php.
Include following script inside first php file.
test1.php

<?php 
// other content
<script>
(function() {
  var httpRequest;
  document.getElementById("ajaxButton").addEventListener('click', makeRequest);

  function makeRequest() {
    httpRequest = new XMLHttpRequest();

    if (!httpRequest) {
      alert('Giving up :( Cannot create an XMLHTTP instance');
      return false;
    }
    httpRequest.onreadystatechange = alertContents;
    httpRequest.open('GET', 'test2.php');
    httpRequest.send();
  }

  function alertContents() {
    if (httpRequest.readyState === XMLHttpRequest.DONE) {
      if (httpRequest.status === 200) {
        alert(httpRequest.responseText); // your response
      } else {
        alert('There was a problem with the request.');
      }
    }
  }
})();
</script>
?>

Then return your content data from the next php file as follows.
test2.php

<?php 
    $x = "content data";
    echo $x;

?>

For more details about AJAX, follow below link https://developer.mozilla.org/en-US/docs/Web/Guide/AJAX/Getting_Started

Answer 2

javaScript Code

    const data = { name: 'scott' }; // data for post
    
    fetch('url', {
      method: 'POST', 
      headers: {
        'Content-Type': 'application/json', // type
      },
      body: JSON.stringify(data),
    })
    .then(response => response.json())
    .then(data => {
      console.log(data);
    })
    .catch((error) => {
      console.error(error);
    });