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;

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

你必须计算总分 如果是SQL 2005，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

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

你们需要在成绩领域分组。这个查询应该为您提供在几乎任何数据库中要查找的内容。

    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。

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;