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...

Answer 1

1. The most efficient (using over()).

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

2. Universal (any SQL version).

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

3. With CTE, the least efficient.

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

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

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

Answer 4

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 5

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');