Question

我有这样的表格。

table ppsscore
string name, int playlevel , int scoretotal,

playlevel需要0-1
scoretotal需要0-100

行数超过100万。

我想按playlevel和scoretotal对行进行分类，然后计算行数。

然而，这需要太多时间。

起初，我制作了索引“playlevel”和“scoretotal”，但没有那么多不同。

还有其他优化方法可以优化计数吗？

  hash['level1']['over100'] = Ppscore.where("playlevel = '1' AND scoretotal = '100'").count    
  hash['level1']['over90'] = Ppscore.where("playlevel = '1' AND scoretotal >= '90' AND scoretotal != '100'").count
  hash['level1']['over80'] = Ppscore.where("playlevel = '1' AND scoretotal >= '80' AND scoretotal < '90'").count
  hash['level1']['over70'] = Ppscore.where("playlevel = '1' AND scoretotal >= '70' AND scoretotal < '80'").count
  hash['level1']['over60'] = Ppscore.where("playlevel = '1' AND scoretotal >= '60' AND scoretotal < '70'").count
  hash['level1']['over50'] = Ppscore.where("playlevel = '1' AND scoretotal >= '50' AND scoretotal < '60'").count
  hash['level1']['over40'] = Ppscore.where("playlevel = '1' AND scoretotal >= '40' AND scoretotal < '50'").count
  hash['level1']['over30'] = Ppscore.where("playlevel = '1' AND scoretotal >= '30' AND scoretotal < '40'").count
  hash['level1']['over20'] = Ppscore.where("playlevel = '1' AND scoretotal >= '20' AND scoretotal < '30'").count
  hash['level1']['over10'] = Ppscore.where("playlevel = '1' AND scoretotal >= '10' AND scoretotal < '20'").count
  hash['level1']['over0'] = Ppscore.where("playlevel = '1' AND scoretotal >= '0' AND scoretotal < '10'").count


  hash['level0']['over100'] = Ppscore.where("playlevel = '0' AND scoretotal = '100'").count    
  hash['level0']['over90'] = Ppscore.where("playlevel = '0' AND scoretotal >= '90' AND scoretotal != '100'").count
  hash['level0']['over80'] = Ppscore.where("playlevel = '0' AND scoretotal >= '80' AND scoretotal < '90'").count
  hash['level0']['over70'] = Ppscore.where("playlevel = '0' AND scoretotal >= '70' AND scoretotal < '80'").count
  hash['level0']['over60'] = Ppscore.where("playlevel = '0' AND scoretotal >= '60' AND scoretotal < '70'").count
  hash['level0']['over50'] = Ppscore.where("playlevel = '0' AND scoretotal >= '50' AND scoretotal < '60'").count
  hash['level0']['over40'] = Ppscore.where("playlevel = '0' AND scoretotal >= '40' AND scoretotal < '50'").count
  hash['level0']['over30'] = Ppscore.where("playlevel = '0' AND scoretotal >= '30' AND scoretotal < '40'").count
  hash['level0']['over20'] = Ppscore.where("playlevel = '0' AND scoretotal >= '20' AND scoretotal < '30'").count
  hash['level0']['over10'] = Ppscore.where("playlevel = '0' AND scoretotal >= '10' AND scoretotal < '20'").count
  hash['level0']['over0'] = Ppscore.where("playlevel = '0' AND scoretotal >= '0' AND scoretotal < '10'").count

Answer 1

在一个查询中执行此操作：

select playlevel, concat('over',floor(scoretotal/10)*10) score, count(*) c
from ppsscore
group by playlevel, concat('over',floor(scoretotal/10)*10)

用于分类的SQL优化和用整数值

1 个答案: