This table, it is necessary with the help of the window function, it is necessary to show what percentage of the total salary in the department (job) is the salary of each employee

name | job | salary -----------+---------+-------- Joe | develop | 2500 Connor | manager | 3500 Callum | sales | 1400 Kyle | develop | 1200 Jacob | sales | 950 John | manager | 1300

It should go like this

  name | job | salary | percent -----------+---------+--------+--------- Joe | develop | 2500 | 67% Kyle | develop | 1200 | 33% Connor | manager | 3500 | 73% John | manager | 1300 | 27% Callum | sales | 1400 | 59% Jacob | sales | 950 | 41%

My option, but nothing works.

  SELECT name, job, salary, percent(sum(salary)*100/salary) OVER (PARTITION BY job) FROM emp;
  • percent function does not exist. you probably meant (sum(salary) OVER (PARTITION BY job)) *100/salary as percent - Mike

1 answer 1

  SELECT name, job, salary, salary / (sum(salary) OVER (partition BY job)) * 100 FROM emp