你可以在你的from查询中使用子选择(未经测试，不确定哪个更快):

SELECT Grade, COUNT(*) / TotalRows
FROM (SELECT Grade, COUNT(*) As TotalRows
      FROM myTable) Grades
GROUP BY Grade, TotalRows

Or

SELECT Grade, SUM(PartialCount)
FROM (SELECT Grade, 1/COUNT(*) AS PartialCount
      FROM myTable) Grades
GROUP BY Grade

Or

SELECT Grade, GradeCount / SUM(GradeCount)
FROM (SELECT Grade, COUNT(*) As GradeCount
      FROM myTable
      GROUP BY Grade) Grades

你也可以使用存储过程(Firebird语法不好意思):

SELECT COUNT(*)
FROM myTable
INTO :TotalCount;

FOR SELECT Grade, COUNT(*)
FROM myTable
GROUP BY Grade
INTO :Grade, :GradeCount
DO
BEGIN
    Percent = :GradeCount / :TotalCount;
    SUSPEND;
END

你可以在你的from查询中使用子选择(未经测试，不确定哪个更快):

SELECT Grade, COUNT(*) / TotalRows
FROM (SELECT Grade, COUNT(*) As TotalRows
      FROM myTable) Grades
GROUP BY Grade, TotalRows

Or

SELECT Grade, SUM(PartialCount)
FROM (SELECT Grade, 1/COUNT(*) AS PartialCount
      FROM myTable) Grades
GROUP BY Grade

Or

SELECT Grade, GradeCount / SUM(GradeCount)
FROM (SELECT Grade, COUNT(*) As GradeCount
      FROM myTable
      GROUP BY Grade) Grades

你也可以使用存储过程(Firebird语法不好意思):

SELECT COUNT(*)
FROM myTable
INTO :TotalCount;

FOR SELECT Grade, COUNT(*)
FROM myTable
GROUP BY Grade
INTO :Grade, :GradeCount
DO
BEGIN
    Percent = :GradeCount / :TotalCount;
    SUSPEND;
END

在任何sql server版本中，您都可以使用一个变量来表示所有等级的总和，如下所示:

declare @countOfAll decimal(18, 4)
select @countOfAll = COUNT(*) from Grades

select
Grade,  COUNT(*) / @countOfAll * 100
from Grades
group by Grade

我认为这是一种通用的解决方案，不过我使用IBM Informix Dynamic Server 11.50.FC3对其进行了测试。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;

给出关于水平规则下面显示的测试数据的以下输出。ROUND函数可能是特定于dbms的，但其余部分(可能)不是。(注意，我将100更改为100.0，以确保计算使用非整数- DECIMAL, NUMERIC -算术;请参阅评论，感谢Thunder。)

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');

您可以使用没有“partition by”子句的窗口函数，而不是使用单独的CTE来获得总数。

如果你正在使用:

count(*)

要获得一个组的计数，您可以使用:

sum(count(*)) over ()

来获得总数。

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在内部使用临时表(我在运行“set statistics io on”时看到过“Worktable”)。

编辑: 我不确定我的示例查询是否是你要找的，我只是说明了窗口函数是如何工作的。

以下操作应该可以工作

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

编辑:移动* 100并添加1.0以确保它不做整数除法

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