使用SQL,如何计算具有特定值的行的百分比？

Question

我正在使用PHP/MySQL创建一个显示比赛结果的网站.我想显示10个最难的比赛的统计数据,即大多数人DNF的比赛.我是这样做的:

select raceid,race.name,race.location,race.date,count(result.raceid) as dnfs 
from result 
inner join race on result.raceid=race.id 
where result.place=0 
group by result.raceid 
order by dnfs desc limit 10

但这并没有考虑到比赛中的人数.如何修改查询以返回percentDNF(dnfs/totalracers)并按该列排序？地点= 0表示DNF.

Answer 1

删除WHERE子句并使用CASE语句来计算did-not-finish值.然后由总赛手划分.像这样的东西:

  SELECT result.raceid, race.name, race.location, race.date, 
         COUNT(result.raceid) AS TOTAL, 
         SUM(CASE WHEN result.place = 0 THEN 1 ELSE 0 END) AS DNF, 
         SUM(CASE WHEN result.place = 0 THEN 1 ELSE 0 END) / 
             COUNT(result.raceid) AS PCT_DNF
    FROM result 
         JOIN race ON result.raceid=race.id 
GROUP BY result.raceid, race.name, race.location, race.date
ORDER BY SUM(CASE WHEN result.place = 0 THEN 1 ELSE 0 END) / 
             COUNT(result.raceid) DESC LIMIT 10