MySQL查询给出不正确的结果

Question

这是我当前的查询：

select baseball , const.cnt/count(*) as my_percent 
from full_db3 inner join
 (select count(*) as cnt from full_db3 where gender = 'f') const
group by baseball
order by my_percent desc limit 10;

这会导致结果不准确。 baseball的可能值为TRUE和FALSE，它们显示在我的结果的baseball栏中。 但是， my_percent的值全部关闭。 差远了。

如果我运行此查询（不同的GROUP BY ）， my_percent在my_percent获得FALSE的正确值，但结果中不包含TRUE。

select baseball , const.cnt/count(*) as my_percent 
from full_db3 inner join
 (select count(*) as cnt from full_db3 where gender = 'f') const
group by const.cnt
order by my_percent desc limit 10;

我在这里想念什么？

Answer 1

简单的事情如何：

select baseball , 
      (sum(case when gender = 'f' then 1 else 0 end) / count(*)) * 100 as pct
FROM full_db3
group by baseball;

该查询给出了女性玩家（非玩家）的百分比；

select gender, 
   (sum(baseball) / count(baseball)) * 100 as players, 
   (1 - (sum(baseball) / count(baseball))) * 100 as non_players
    from full_db3
    where gender = 'f'
    ;

最后一个在行中具有true / false，最后确定为要求：

 select baseball, 
    (count(*) / 
       (select count(gender) from full_db3 where gender = 'f')) * 100 as pct
 from full_db3 
 where gender = 'f'
 group by baseball;

小提琴： http ://sqlfiddle.com/#! 9/15866/6

Answer 2

您似乎想显示出打棒球的人的百分比。

你不想要

SELECT baseball, gender, (100.0 * COUNT(*) / const.cnt) AS my_percent
  FROM full_db3
  JOIN (SELECT COUNT(*) AS cnt FROM full_db3) CONST
 GROUP BY baseball, gender

您可能还想要。

SELECT baseball, gender, (100.0 * COUNT(*) / const.cnt) AS my_percent
  FROM full_db3
  JOIN (
     SELECT COUNT(*) AS cnt, gender
       FROM full_db3
      GROUP BY gender
       ) CONST ON full_db3.gender = CONST.gender
 GROUP BY baseball, gender

如果您想知道分别有'm'和'f'性别的棒球属性的百分比。

100.0 *将您的分数变成一个百分比

在这种情况下，您的LIMIT 10毫无意义，因为您的汇总查询中没有十个不同的类别。