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MySQL查询给出不正确的结果

[英]mySQL query giving inaccurate results

This is my current query: 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;

This yields inaccurate results. 这会导致结果不准确。 The possible values for baseball are TRUE and FALSE, which appear in my result's baseball column. baseball的可能值为TRUE和FALSE,它们显示在我的结果的baseball栏中。 However, the values for my_percent are all off. 但是, my_percent的值全部关闭。 Not even close. 差远了。

If I run this query (different GROUP BY ), I get the correct value in my_percent for FALSE, but TRUE isn't in the result. 如果我运行此查询(不同的GROUP BY ), my_percentmy_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;

What am I missing here? 我在这里想念什么?

How about something simpler: 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;

This query gives percentage of females that are players / non players; 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'
    ;

And the last one, that has the true / false in the rows, as finally determined as the requirement: 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;

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

It looks like you're trying to show percentages of people who play baseball. 您似乎想显示出打棒球的人的百分比。

Don't you want 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

You might also want. 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

If you want to know what percentage of 'm' and 'f' genders separately have the baseball attribute. 如果您想知道分别有'm'和'f'性别的棒球属性的百分比。

The 100.0 * turns your fraction into a percentage 100.0 *将您的分数变成一个百分比

Your LIMIT 10 makes no sense in this case because your aggregate query doesn't contain ten different categories. 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。

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