How to calculate percentage with a SQL statement

Question

I have a SQL Server table that contains users & their grades. For simplicity's sake, lets just say there are 2 columns - name & grade. So a typical row would be Name: "John Doe", Grade:"A".

I'm looking for one SQL statement that will find the percentages of all possible answers. (A, B, C, etc...) Also, is there a way to do this without defining all possible answers (open text field - users could enter 'pass/fail', 'none', etc...)

The final output I'm looking for is A: 5%, B: 15%, C: 40%, etc...

Thanks in advance!

Answer 1

You have to calculate the total of grades If it is SQL 2005 you can use 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 2

You need to group on the grade field. This query should give you what your looking for in pretty much any database.

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

You should specify the system you're using.

Answer 3

In any sql server version you could use a variable for the total of all grades like this:

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

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

Answer 4

You can use a subselect in your from query (untested and not sure which is faster):

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

You can also use a stored procedure (apologies for the Firebird syntax):

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

Answer 5

The following should work

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

EDIT: Moved the * 100 and added the 1.0 to ensure that it doesn't do integer division

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

Answer 6

This is, I believe, a general solution, though I tested it using IBM Informix Dynamic Server 11.50.FC3. The following query:

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;

gives the following output on the test data shown below the horizontal rule. The ROUND function may be DBMS-specific, but the rest (probably) is not. (Note that I changed 100 to 100.0 to ensure that the calculation occurs using non-integer - DECIMAL, NUMERIC - arithmetic; see the comments, and thanks to 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');

Answer 7

I have tested the following and this does work. The answer by gordyii was close but had the multiplication of 100 in the wrong place and had some missing parenthesis.

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

Answer 8

Instead of using a separate CTE to get the total, you can use a window function without the "partition by" clause.

If you are using:

count(*)

to get the count for a group, you can use:

count(*) over ()

to get the total count.

For example:

SELECT Grade, ( COUNT(*) / count(*) over () ) * 100.
FROM table
GROUP BY Grade;

It tends to be faster in my experience, but I think it might internally use a temp table in some cases (I've seen "Worktable" when running with "set statistics io on").

EDIT: I'm not sure if my example query is what you are looking for, I was just illustrating how the windowing functions work.