MySQL的查詢給予不準確的結果

Question

這是我當前的查詢：

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; 

這會產生不準確的結果。 baseball的可能值爲TRUE和FALSE，這些值出現在我的結果的baseball列中。但是，my_percent的值全部關閉。差遠了。

如果我運行這個查詢（不同的GROUP BY），我在my_percent中得到了正確的值FALSE，但TRUE不在結果中。

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; 

我在這裏錯過了什麼？

Answer 1

如何更簡單的東西：

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

這個查詢給女性是播放器/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' 
    ; 

，最後一個，具有真/在行假，如最終確定的要求：

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

看起來你試圖展示打棒球的人的百分比。

你不想

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 

您可能還需要。

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 

如果你想知道'm'和'f'性別百分比分別具有棒球屬性。

的100.0 *將您的分數成比例

你LIMIT 10使得在這種情況下，沒有任何意義，因爲你總查詢不包含十個不同的類別。