I need to send a request through JS to a php file via API [closed]-CodePudding

Help me please. There are two php files. When you enter data in the zip field, you need to send a request to the php file, and if zip is available, send the data to another php file and parse the response in the field I will be grateful for the help)

CodePudding user response：

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

CodePudding user response：

**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);
});