問題描述

問題解答

select user_id,a.sc,min(created_at) tm from (select user_id,max(score) sc from active_gamescore group by user_id) a join active_gamescore b on a.user_id=b.user_id and a.sc=b.score group by a.user_id,a.sc order by a.sc desc,tm asc limit 20;回答2：

select t.userid,t.score from (select * from active_gamescore order by score desc,created desc) as t group by t.userid limit 20;

更新：一個(gè)人只能回復(fù)一次，很憂傷感謝幾位抽出時(shí)間去幫我解題測(cè)試了一下(5次平均)

在1w數(shù)據(jù)量的情況下，@clcx_1315 ：0.004s@伊拉克 : 0.009s@邢愛明 :0.006s我自己的 ：0.016s

在20w的數(shù)據(jù)量下：@clcx_1315 ：0.104s@伊拉克 : 0.141s@邢愛明 :0.165s我自己的 ：0.171s

所以@clcx_1315的方法最優(yōu)，十分感謝，學(xué)到了一個(gè)思路。從explain看，@clcx_1315的寫法只做了2次全表遍歷，其他都是3次，或許這就是原因了。

===========================之前的分隔線============================琢磨了一種寫法，但效率有待提高

select ta.user_id,ta.max_score,tb.min_timefrom (select a.user_id, max(a.score) max_score from active_gamescore a where a.active_id=’58’ group by a.user_id) tajoin (select a.user_id,a.score,min(a.created_at) min_time from active_gamescore a where a.active_id=’58’ group by a.user_id,a.score) tb on ta.user_id=tb.user_id and ta.max_score=tb.scoreorder by ta.max_score desc,tb.min_time asc limit 20回答4：

假設(shè)同一用戶下的created_at字段值不重復(fù)，可以試試下面的語句：

SELECT t1.user_id, t1.score, t1.created_atFROM (SELECT @row_num:=IF(@prev_col1=t.user_id, @row_num+1, 1) AS row_number, t.*, @prev_col1:=t.user_idFROM (SELECT * FROM active_gamescore ORDER BY user_id, score DESC, created_at) t, (SELECT @row_num:=1, @prev_col1:=NULL) var) t1 WHERE row_number = 1ORDER BY t1.score DESC, t1.created_atLIMIT 20回答5：

SELECT yws0.user_id, yws0.score, min(create_time) AS create_timeFROM active_gamescore yws0WHERE (user_id, score) IN (SELECT yws.user_id, yws.max_scoreFROM (SELECT user_id, max(score) AS max_scoreFROM active_gamescoreGROUP BY user_id ) yws )GROUP BY yws0.user_id, yws0.scoreORDER BY SCORE DESCLIMIT 3