虽然我不知道您为什么要这样做，但请查看下面的工作代码。请注意行中的注释。我希望这有帮助。

class test{
    public function __construct() {
        add_action( \'wp_ajax_test2\', array( $this, \'test2\' ) );
        /* Front end ajax needs this. */
        add_action( \'wp_ajax_nopriv_test2\', array( $this, \'test2\' ) );
        add_action( \'wp_enqueue_scripts\', array( $this, \'test1\' ) );
    }

    public function test1() {
      /* in JavaScript, object properties are accessed as ajax_object.ajax_url, ajax_object.we_value */
      wp_localize_script(\'test-script-ajax\', \'ajaxobj\', array(\'ajaxurl\' => admin_url(\'admin-ajax.php\')));
      /* Moved your js to a separate js file and enquing it the WordPress way */ 
      wp_enqueue_script( \'test-script-ajax\', get_stylesheet_directory_uri() . \'/js/test-ajax.js\', array( \'jquery\' ));
    }

    public function test2(){
        echo "success";
        exit;
    }
}

$ob_call = new test;
$ob_call->test1();

jQuery(document).ready(function($){
  var data = { \'action\':\'test2\' };
  jQuery.post( ajaxobj.ajaxurl, data, function( response ) {
            alert(response);
    });
});

现在可以了

----------Function.php---------------

add_action( \'wp_ajax_test2\', array( \'test\', \'test2\' ) );

class test{

    public function __construct() {
        
    }                       
    public function test1() { ?>
        <script>
            jQuery(document).ready(function($){
                var ajaxurl = "<?php echo get_site_url();?>/wp-admin/admin-ajax.php";
                var data = {\'action\':\'test2\'};
                 $.ajax({ 
                    url: ajaxurl,
                    type: "POST",
                    data: data,
                    success: function(val) {
                        alert(val);
                    },
                }); 
            });             
        </script><?php 
    }               
    public function test2(){
        echo "success";
        exit;
    } 
          
}

$ob_call = new test;

$ob_call->test1();

添加wp_die(); 在功能结束时。