我们可以子查询，然后使用COUNT()作为分析函数来检测排名中的平局：

WITH cte AS (
    SELECT u.id,
           CASE WHEN a.result IS NULL THEN 0 ELSE a.result END AS result,
           RANK() OVER(ORDER BY result DESC) AS `rank`
    FROM wp_users u, wp_usermeta um,
         (SELECT pm.meta_value, COUNT(*) result
          FROM wp_postmeta pm
          INNER JOIN wp_posts p ON p.id = pm.post_id
          WHERE p.post_status = 'publish' AND
                pm.meta_key = 'wpcf-userid' AND
                pm.post_id IN (SELECT pm1.post_id FROM wp_postmeta pm1
                               WHERE pm1.meta_key = 'wpcf-badge-name' AND
                                     pm1.meta_value = 'LD' AND
                                     pm1.post_id = pm.post_id) AND
                pm.meta_value IN (SELECT um.user_id FROM wp_usermeta um
                                  WHERE um.meta_value = 'VIC' AND
                                        um.meta_key = 'wpcf-address-state')
          GROUP BY pm.meta_value) a
    WHERE u.id = um.user_id AND a.meta_value = u.id
    GROUP BY u.id
)

SELECT id, result,
       CASE WHEN COUNT(*) OVER (PARTITION BY `rank`) = 1
            THEN `rank` ELSE CONCAT('T', `rank`) END AS `rank`
FROM cte t
ORDER BY t.`rank`;