我有一个ajax函数，可以通过ajax将用户帖子的视图类型从列表模式切换到网格模式，但问题是，当调用wp查询时，它无法获取正在查看的用户配置文件的id。当我手动输入id号时，它会起作用，但当我尝试使用$user->id或$user\\u id或$user[\'id]时，它不会起作用。有人能帮我弄清楚吗？

这是我当前的php函数，它是在我的函数中编写的。php文件：

function profile_view(){

$args = array(
        \'post_type\' => \'listings\',
        \'post_status\'    => \'publish\',
        \'meta_query\' => array(
            array(
                \'key\' => \'stm_car_user\',
                \'value\' => $user->ID
            )
        )
    );

$query = new WP_Query($args);
//$query = stm_user_listings_query(\'1\', \'publish\');

$response = array();

//Grid/list settings
$view_list = \'\';
$view_grid = \'\';
$view_map = \'\';
$current_link_args = array();
if(!empty($_GET)){
    $current_link_args = $_GET;
}

$view_list_link = $view_grid_link = $view_map_link = $current_link_args;
$view_list_link[\'view_type\'] = \'list\';
$view_grid_link[\'view_type\'] = \'grid\';
$view_map_link[\'view_type\'] = \'map\';


if(!empty($_GET[\'view_type\'])) {
    if ( $_GET[\'view_type\'] == \'list\' ) {
        $view_list = \'active\';
    } elseif ( $_GET[\'view_type\'] == \'grid\' ) {
        $view_grid = \'active\';
        $current_link_args[\'view_type\'] = \'grid\';
    }
    elseif ( $_GET[\'view_type\'] == \'map\' ) {
        $view_map = \'active\';
        $current_link_args[\'view_type\'] = \'map\';
    }
} else {
    $view_list = \'active\';
}


    if ( $query->have_posts() ):
    ob_start();

    $template = \'partials/listing-cars/listing-list-directory-loop\';
    if(!empty($_GET[\'view_type\']) and $_GET[\'view_type\'] == \'grid\') {

    $template = \'partials/listing-cars/listing-grid-directory-loop\';

    } elseif(!empty($_GET[\'view_type\']) and $_GET[\'view_type\'] == \'map\') {

    $template = \'partials/listing-cars/listing-map\';

    } else {

    $template = \'partials/listing-cars/listing-list-directory-loop\';

    }


    while ( $query->have_posts() ) {
        $query->the_post();

        get_template_part($template);

    }

    $response[\'html\'] = ob_get_contents();
    ob_end_clean();
else:

endif;
wp_reset_postdata();

$show_pagination = true;
if(!empty($query->found_posts) and !empty($query->query_vars[\'posts_per_page\'])) {
    if($query->found_posts < $query->query_vars[\'posts_per_page\']) {
        $show_pagination = false;
    }
}

$response = json_encode( $response );

echo $response;
exit;
}

 add_action(\'wp_ajax_profile_view\' , \'profile_view\');
 add_action(\'wp_ajax_nopriv_profile_view\',\'profile_view\');

我也不想使用wp\\u get\\u current\\u user（），因为它只显示当前登录用户的帖子，我希望在查看某个用户的个人资料时，即使我没有登录，也能看到该用户的所有帖子。。。

以下是ajax函数：

$(\'.stm-view-by-profile a:not(.view-map)\').on(\'click\', function(e){
        e.preventDefault();
        var viewType = $(this).data(\'view\');
        $(\'#stm_view_type_profile\').val(viewType);
        $(\'.stm-view-by-profile a\').removeClass(\'active\');
        $(this).addClass(\'active\');

        var data_form = $(\'.user_posts\').serialize();

        $.ajax({
            url: ajaxurl,
            dataType: \'json\',
            context: this,
            data: data_form + \'&action=profile_view\',
            beforeSend: function(){
                $(\'.stm-ajax-row\').addClass(\'stm-loading\');
            },
            success: function (data) {
            console.log(data);
           $(\'.stm-isotope-sorting\').html(data.html);
           $(\'.stm-ajax-row\').removeClass(\'stm-loading\');
            }
        });
    });

function profile_view() {
$user_id = $_GET["user_id"]; // name of hidden input

$args = array(
        \'post_type\' => \'listings\',
        \'post_status\'    => \'publish\',
        \'meta_query\' => array(
            array(
                \'key\' => \'stm_car_user\',
                \'value\' => $user_id
            )
        )
    );
...