这个在MS SQL中工作得很好。它将varchar转换为两个小数限制浮点数的结果。

Select field1, cast(Try_convert(float,(Count(field2)* 100) / 
Try_convert(float, (Select Count(*) From table1))) as decimal(10,2)) as new_field_name 
From table1 
Group By field1, field2;

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

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。

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');

我也遇到过类似的问题。您应该能够得到乘以1.0而不是100的正确结果。参见附图示例

选择Grade， (Count(Grade)* 1.0 / (Select Count(*) From MyTable))作为来自MyTable Group By Grade的分数

最有效的(使用over())。 count(*) * 100.0 / sum(count(*)) over() 从MyTable 按年级分组 通用(任何SQL版本)。 select count(*) * 100.0 / (select count(*) from MyTable) 从MyTable 按年级分组; CTE的效率最低。 用t(Grade, GradeCount) 作为 （ select Grade, count(*) 从MyTable 按年级分组 ） (select sum(GradeCount) from t) 从t;