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Най-ефективният (използвайки over()).
select Grade, count(*) * 100.0 / sum(count(*)) over() from MyTable group by Grade
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Универсален (всяка версия на SQL).
select Grade, count(*) * 100.0 / (select count(*) from MyTable) from MyTable group by Grade;
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С CTE, най-малко ефективни.
with t(Grade, GradeCount) as ( select Grade, count(*) from MyTable group by Grade ) select Grade, GradeCount * 100.0/(select sum(GradeCount) from t) from t;