如何以百分比的形式获得mysql匹配的结果？

SELECT (mthCount / ttlCount) AS mPercent
FROM (
  SELECT COUNT( * ) AS mthCount
  FROM articles WHERE (
     MATCH(body) AGAINST(body_var) 
     )
) AS MCount JOIN (
  SELECT COUNT( * ) AS ttlCount
  FROM articles
) AS TCount;

SELECT FORMAT((mthCount / ttlCount),2) AS mPercent
FROM (
  SELECT COUNT( * ) AS mthCount
  FROM articles WHERE (
     MATCH(body) AGAINST(body_var) 
     )
) AS MCount JOIN (
  SELECT COUNT( * ) AS ttlCount
  FROM articles
) AS TCount;

SELECT
MAX(MATCH (table.col1)  AGAINST ('text 1'  IN NATURAL LANGUAGE MODE)) AS  bscore_col1 ,
MAX(MATCH (table.col2) AGAINST ('text 2'                          
IN NATURAL LANGUAGE MODE)) AS bscore_col2
FROM table
ORDER BY bscore_name col1 DESC, bscore_col2 DESC) AS bests

SELECT *,
MATCH (table.col1) AGAINST ('text 1' IN NATURAL LANGUAGE MODE)/bests.bscore_col1 AS score_col1 ,
MATCH (table.col2) AGAINST ('text 2' IN NATURAL LANGUAGE MODE)/bests.bscore_col2 AS score_col2
FROM (table,
(SELECT
MAX(MATCH (table.col1)  AGAINST ('text 1'  IN NATURAL LANGUAGE MODE)) AS bscore_col1 ,
MAX(MATCH (table.col2) AGAINST ('text 2' IN NATURAL LANGUAGE MODE)) AS bscore_col2
FROM table
ORDER BY bscore_col2 DESC, bscore_col1 DESC) AS bests)
WHERE
MATCH (table.col1)  AGAINST ('text 1'  IN NATURAL LANGUAGE MODE)/bests.bscore_col1 > 0.5 AND
MATCH (table.col2) AGAINST ('text 2'IN NATURAL LANGUAGE MODE)/bests.bscore_col2 > 0.5
ORDER BY score_col2 DESC, score_col1 DESC