在找到一些结果时，我发现了以下两种方法，但无法确定哪种查询具有更好的性能。希望有人会建议！
SELECT SQL_CALC_FOUND_ROWS *, MATCH (m.firstname, m.lastname) AGAINST ('*keyword1* *keyword2* *keyword3*' IN BOOLEAN MODE) AS relavance FROM members m WHERE m.status = 1 AND m.membership > 1 AND (SELECT count(media_id) FROM arts AS media WHERE media.active = 1 AND media.owner = m.mem_id) > 0 AND MATCH (m.firstname, m.lastname) AGAINST ('*keyword1* *keyword2* *keyword3*' IN BOOLEAN MODE) ORDER BY relavance DESC, m.firstname ASC, m.lastname ASC
SELECT SQL_CALC_FOUND_ROWS *, (SELECT count(media_id) FROM arts AS media WHERE media.active = 1 AND media.owner = m.mem_id) AS arts_count, MATCH (m.firstname, m.lastname) AGAINST ('*keyword1* *keyword2* *keyword3*' IN BOOLEAN MODE) AS relavance FROM members m WHERE m.status = 1 AND m.membership > 1 AND MATCH (m.firstname, m.lastname) AGAINST ('*keyword1* *keyword2* *keyword3*' IN BOOLEAN MODE) HAVING arts_count > 0 ORDER BY relavance DESC, m.firstname ASC, m.lastname ASC
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我也找到了另一种方法，现在比较是在三个查询之间。以下是我发现的新查询以及从性能角度来看更好的所有三个查询？
SELECT SQL_CALC_FOUND_ROWS *, MATCH (m.firstname, m.lastname) AGAINST ('*keyword1* *keyword2* *keyword3*' IN BOOLEAN MODE) AS relavance FROM members m LEFT JOIN (SELECT count(media_id) AS arts_count, owner AS ar_owner FROM arts GROUP BY media_id) AS ar ON ar.ar_ar_owner = m.owner WHERE m.status = 1 AND m.membership > 1 AND MATCH (m.firstname, m.lastname) AGAINST ('*keyword1* *keyword2* *keyword3*' IN BOOLEAN MODE) AND ar.arts_count > 0 ORDER BY relavance DESC, m.firstname ASC, m.lastname ASC
查询目标：要查找匹配关键字的用户，并且至少有一个上传的图片/艺术品，还有一些检查最终按相关性进行排序。