Quiero entregar el valor html a través de JS para el manejo de PHP, pero aún recibo el error 400 (Solicitud incorrecta).

Esto es HTML:

 </div> <form class="row g-3" id="my_form"> <div class="col-md-12"> <label for="url" class="form-label">Input URL</label> <input type="text" class="form-control my_info" id="url" required> <div class="col-12"> <button class="btn btn-primary submit" type="button">Submit form</button> </div> </form> </div>

Esto es JavaScript:

 function getBaseUrl() { var re = new RegExp(/^.*\//); var my_local_url = re.exec(window.location.href)[0]; return my_local_url; } console.log(getBaseUrl()) jQuery(document).ready(function($) { console.log('test'); // We'll pass this variable to the PHP function example_ajax_request var my_data = { 'action':'crawler_info', 'my_info' : document.getElementById('url').value }; $('.submit').click(function (){ // This does the ajax request $.ajax({ url: getBaseUrl() + 'admin-ajax.php', data: my_data,, success:function(data) { // This outputs the result of the ajax request console.log(data); }, error: function(errorThrown){ console.log(errorThrown); } }); }) });

Esto es PHP:

 function get_info() { // The $_REQUEST contains all the data sent via ajax if ( isset($_REQUEST) ) { $infos = $_REQUEST['my_info']; // Now we'll return it to the javascript function // Anything outputted will be returned in the response echo $infos; // If you're debugging, it might be useful to see what was sent in the $_REQUEST print_r($_REQUEST); } // Always die in functions echoing ajax content or it will display 0 or another word die(); } add_action( 'wp_ajax_crawler_info', 'get_info' ); add_action( 'wp_ajax_nopriv_crawler_info', 'get_info' );

Intenté muchas formas de solucionarlo, pero aún muestra este error nuevamente. Alguien sabe como solucionar mi problema, gracias