SELECT r.uid, SUM(r.point) as allpoints, (SUM(r.level)/SUM(r.totimes)) as acc, (SUM(r.usetime)/SUM(r.level)) as avgtime, m.username, m.groupid, m.osspath, n.bpath FROM plugin_logicox_fish_rank r LEFT JOIN common_member m ON m.uid=r.uid WHERE r.usetime > 0 AND m.groupid NOT IN('38', '4', '5') GROUP BY r.uid ORDER BY level DESC, acc DESC, avgtime ASC, allpoints DESC, r.uid ASC LIMIT 0, 20
时间: 2023-03-08 12:05:29 浏览: 57
我无法理解你的问题,但我可以尝试解释一下这个查询语句:它从表plugin_logicox_fish_rank中查询uid、point和usetime,并从common_member表中查询username和osspath,最后将查询结果根据level、acc、avgtime、allpoints和uid进行排序,取出前20条记录。