san*_*ode 3 mysql sql-server oracle
我有一张如下表:
ID Name Department Gender
1 Crib MA MALE
2 Lucy Bsc FEMALE
3 Phil Bcom MALE
4 Ane MA FEMALE
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我有1000行这样的记录.我想找到所有学生的性别(男性和女性)列的比率.
我需要一个查询来执行此操作.
Fab*_*ion 10
MySQL 5.5.32架构设置:
CREATE TABLE table1
(`ID` int, `Name` varchar(4), `Department` varchar(4), `Gender` varchar(6))
;
INSERT INTO table1
(`ID`, `Name`, `Department`, `Gender`)
VALUES
(1, 'Crib', 'MA', 'MALE'),
(2, 'Lucy', 'Bsc', 'FEMALE'),
(3, 'Phil', 'Bcom', 'MALE'),
(4, 'Ane', 'MA', 'FEMALE')
;
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查询1:
SELECT sum(case when `Gender` = 'MALE' then 1 else 0 end)/count(*) as male_ratio,
sum(case when `Gender` = 'FEMALE' then 1 else 0 end)/count(*) as female_ratio
FROM table1
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结果:
| MALE_RATIO | FEMALE_RATIO |
|------------|--------------|
| 0.5 | 0.5 |
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尝试这样的事情
select sum(case when gender = 'MALE' then 1 else 0 end) / count(*) * 100 as perc_male,
sum(case when gender = 'FEMALE' then 1 else 0 end) / count(*) * 100 as perc_female
from students
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这应该给你实际的比率,并且应该在 MySQL 和 SQL Server 中很少或没有修改的情况下工作。您可能需要稍微修改一下强制转换语句 - 我的 MySQL 很生锈,我认为它的处理方式可能略有不同。
SELECT
(CAST((SELECT COUNT(*) FROM tblName WHERE Gender='MALE') AS FLOAT) /
CAST((SELECT COUNT(*) FROM tblName WHERE Gender='FEMALE') AS FLOAT))
AS ratioMaleFemale;
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