你的问题很老了，但我只是想为你的问题添加一个真正的解决方案。这里有一个函数，它将返回您发布帖子的年份数组。您可以将此函数放入函数中。php或插件或任何你想要的东西。

function get_posts_years_array() {
    global $wpdb;
    $result = array();
    $years = $wpdb->get_results(
        $wpdb->prepare(
            "SELECT YEAR(post_date) FROM {$wpdb->posts} WHERE post_status = \'publish\' GROUP BY YEAR(post_date) DESC"
        ),
        ARRAY_N
    );
    if ( is_array( $years ) && count( $years ) > 0 ) {
        foreach ( $years as $year ) {
            $result[] = $year[0];
        }
    }
    return $result;
}

您可以使用此函数获取数组中的所有post years，您可以使用它进行循环。

$terms_year = array(
    \'post_type\'         => array(\'YOUR_CPT\'),
);

$years = array();
$query_year = new WP_Query( $terms_year );

if ( $query_year->have_posts() ) :
    while ( $query_year->have_posts() ) : $query_year->the_post();
        $year = get_the_date(\'Y\');
        if(!in_array($year, $years)){
            $years[] = $year;
        }
    endwhile;
    wp_reset_postdata();
endif;

// Echo the years out wherever you want

echo \'<pre>\';
print_r($years);
echo \'</pre>\';

您可以通过以下方式使用WP函数和短代码：

add_shortcode( \'archives\', \'_get_archived_posts\' );
function _get_archived_posts( $atts ) {

 $a = shortcode_atts( array(
    \'type\'            => \'yearly\',
    \'limit\'           => \'\',
    \'format\'          => \'\', 
    \'before\'          => \'\',
    \'after\'           => \'\',
    \'show_post_count\' => false,
    \'echo\'            => 0,
    \'order\'           => \'DESC\'
), $atts, \'filter_archives_sc\' );

    return wp_get_archives($a);
}

Source

// Get all years posts publish;
function get_posts_years_array($post_type = \'post\') {

    global $wpdb;
    $result = array();

    $query_prepare = $wpdb->prepare("SELECT YEAR(post_date) FROM ($wpdb->posts) WHERE post_status = \'publish\' AND post_type = %s GROUP BY YEAR(post_date) DESC", $post_type);

    $years = $wpdb->get_results($query_prepare);

    if ( is_array( $years ) && count( $years ) > 0 ) {
        foreach ( $years as $year ) {
            $result[] = json_decode(json_encode($year), true);
        }
    }

    return $result;
}