2011-04-22 67 views

回答

18
UPDATE a 
INNER JOIN (SELECT AVG(b.score) avg_score, 
        COUNT(b.id) cnt_id, 
        b.app_id 
       FROM b 
      GROUP BY b.app_id) x ON x.app_id = a.app_id 
     SET remark_avg = x.avg_score, 
      remark_count = x.cnt_id; 
+0

什么是“x”等于在这个答案? – alexk 2014-03-08 02:05:06

+0

@alexk:实际上它是一个嵌套的查询别名我错过了,现在修复 – zerkms 2014-03-08 03:50:25

相关问题