用INFORMATION_SCHEMA中的statistics找出沒用到的index。
SELECT DISTINCT s.TABLE_SCHEMA, s.TABLE_NAME, s.INDEX_NAME
FROM information_schema.statistics s LEFT JOIN information_schema.index_statistics `is` ON(s.TABLE_SCHEMA=is.TABLE_SCHEMA AND
s.TABLE_NAME=is.TABLE_NAME AND s.INDEX_NAME=is.INDEX_NAME)
WHERE is.TABLE_SCHEMA IS NULL;

