如何使用SQL语句计算百分比

Question

我有一个包含用户及其成绩的SQL Server表.为简单起见,我们只想说有2列 - name&grade.所以一个典型的行是Name:"John Doe",等级:"A".

我正在寻找一个SQL语句,它将找到所有可能答案的百分比.(A,B,C等......)另外,有没有办法在没有定义所有可能的答案的情况下这样做(开放文本字段 - 用户可以输入'通过/失败','无'等...)

我要找的最终输出是A:5%,B:15%,C:40%等等......

Answer 1

我测试了以下内容,这确实有效.gordyii的答案很接近但在错误的地方乘以100并且有一些缺失的括号.

Select Grade, (Count(Grade)* 100 / (Select Count(*) From MyTable)) as Score
From MyTable
Group By Grade

Answer 2

1. 效率最高(使用over()).

   select Grade, count(*) * 100.0 / sum(count(*)) over()
   from MyTable
   group by Grade
   

2. 通用(任何SQL版本).

   select Grade, count(*) * 100.0 / (select count(*) from MyTable)
   from MyTable
   group by Grade;
   

3. 使用CTE,效率最低.

   with t(Grade, GradeCount) 
   as 
   ( 
       select Grade, count(*) 
       from MyTable
       group by Grade
   )
   select Grade, GradeCount * 100.0/(select sum(GradeCount) from t)
   from t;
   

Answer 3

您可以使用不带"partition by"子句的窗口函数,而不是使用单独的CTE来获取总数.

如果您正在使用:

count(*)

要获得组的计数,您可以使用:

sum(count(*)) over ()

得到总数.

例如:

select Grade, 100. * count(*) / sum(count(*)) over ()
from table
group by Grade;

根据我的经验,它往往更快,但我认为在某些情况下它可能会在内部使用临时表(我在运行"set statistics io on"时看到了"工作表").

编辑: 我不确定我的示例查询是否是您正在寻找的,我只是说明了窗口函数的工作原理.

Answer 4

您必须计算成绩总数如果是SQL 2005,您可以使用CTE

    WITH Tot(Total) (
    SELECT COUNT(*) FROM table
    )
    SELECT Grade, COUNT(*) / Total * 100
--, CONVERT(VARCHAR, COUNT(*) / Total * 100) + '%'  -- With percentage sign
--, CONVERT(VARCHAR, ROUND(COUNT(*) / Total * 100, -2)) + '%'  -- With Round
    FROM table
    GROUP BY Grade

Answer 5

您需要在成绩字段上进行分组.这个查询应该可以在几乎任何数据库中为您提供所需的内容.

    Select Grade, CountofGrade / sum(CountofGrade) *100 
    from
    (
    Select Grade, Count(*) as CountofGrade
    From Grades
    Group By Grade) as sub
    Group by Grade

您应该指定您正在使用的系统.

Answer 6

我只需要使用它,我需要计算一个百分比..

ROUND(CAST((Numerator * 100.0 / Denominator) AS FLOAT), 2) AS Percentage

需要注意的是100.0返回小数,而100在它自己将结果四舍五入到最接近的整数,甚至与ROUND()函数!

Answer 7

以下应该有效

ID - Key
Grade - A,B,C,D...

编辑:移动* 100并添加1.0以确保它不进行整数除法

Select 
   Grade, Count(ID) * 100.0 / ((Select Count(ID) From MyTable) * 1.0)
From MyTable
Group By Grade

Answer 8

我相信这是一个通用解决方案,尽管我使用IBM Informix Dynamic Server 11.50.FC3对其进行了测试.以下查询:

SELECT grade,
       ROUND(100.0 * grade_sum / (SELECT COUNT(*) FROM grades), 2) AS pct_of_grades
    FROM (SELECT grade, COUNT(*) AS grade_sum
            FROM grades
            GROUP BY grade
         )
    ORDER BY grade;

在水平线下面显示的测试数据上给出以下输出.该ROUND函数可能是DBMS特定的,但其余(可能)不是.(注意,我将100更改为100.0以确保使用非整数进行计算 - DECIMAL,NUMERIC - 算术;请参阅注释,并感谢迅雷.)

grade  pct_of_grades
CHAR(1) DECIMAL(32,2)
A       32.26
B       16.13
C       12.90
D       12.90
E       9.68
F       16.13

CREATE TABLE grades
(
    id VARCHAR(10) NOT NULL,
    grade CHAR(1) NOT NULL CHECK (grade MATCHES '[ABCDEF]')
);

INSERT INTO grades VALUES('1001', 'A');
INSERT INTO grades VALUES('1002', 'B');
INSERT INTO grades VALUES('1003', 'F');
INSERT INTO grades VALUES('1004', 'C');
INSERT INTO grades VALUES('1005', 'D');
INSERT INTO grades VALUES('1006', 'A');
INSERT INTO grades VALUES('1007', 'F');
INSERT INTO grades VALUES('1008', 'C');
INSERT INTO grades VALUES('1009', 'A');
INSERT INTO grades VALUES('1010', 'E');
INSERT INTO grades VALUES('1001', 'A');
INSERT INTO grades VALUES('1012', 'F');
INSERT INTO grades VALUES('1013', 'D');
INSERT INTO grades VALUES('1014', 'B');
INSERT INTO grades VALUES('1015', 'E');
INSERT INTO grades VALUES('1016', 'A');
INSERT INTO grades VALUES('1017', 'F');
INSERT INTO grades VALUES('1018', 'B');
INSERT INTO grades VALUES('1019', 'C');
INSERT INTO grades VALUES('1020', 'A');
INSERT INTO grades VALUES('1021', 'A');
INSERT INTO grades VALUES('1022', 'E');
INSERT INTO grades VALUES('1023', 'D');
INSERT INTO grades VALUES('1024', 'B');
INSERT INTO grades VALUES('1025', 'A');
INSERT INTO grades VALUES('1026', 'A');
INSERT INTO grades VALUES('1027', 'D');
INSERT INTO grades VALUES('1028', 'B');
INSERT INTO grades VALUES('1029', 'A');
INSERT INTO grades VALUES('1030', 'C');
INSERT INTO grades VALUES('1031', 'F');