How to select an employee in SQL, compared to another employee - mysql

Question

I'm new here on the site, I'm learning databases independently, and there's a question I've been trying for a long time, and I decided to ask it here
The question has such a database:
Person (p-id, name, title, hourly-rate, title)
Topics (t-name, description)
Workshops (p-id, t-name, hourly-credit)
Employees (e-id, name, address, mobile)
EmpTraining (e-id, p-id, t-name)
Persona - The lecturer has a unique ID, name, degree, cost per hour
Topics - the topics in which workshops are delivered. Each topic has a unique name (time management, budget management, etc.) and description.
Workshop - A workshop delivered by a specific lecturer on a specific topic. (Several lecturers can deliver a workshop on the same topic and a lecturer can deliver workshops on different topics) Each workshop is defined as the number of hours that entitle the employee to the training allowance.
Employee - The employee has a unique ID, address and telephone name.
EmpTraining - Each employee is documented in all the workshops he has taken
Example of the database I built, I want to point out that the example here is from a test I failed, the database itself I added is just an example, it does not belong to the exercise itself, but that way it is easier for me to explain, and that way easier to understand.
https://www.db-fiddle.com/f/wrDLMHhsquiFQmZT7uS3Z9/0
I need to write it in code in SQL, two things:
Return the names of the employees who have undergone at least the same series of workshops as the outstanding employee whose employee number is 100.
In this question I have to return the names of all the workers, who did all the workshop that a worker with number 100 did
i try to write, i dont thing that is right, because i am not sure how to select same series of workshops:
select name
from Persona as p1, Workshops as w1
where exist {
select *
from Persona as p2, Workshops as w2
where p2.p-id = 100 AND w1.p-is != w2.p-id and w1.t-name = w2.t-name
}
Return the names of all the employees who have accumulated the number of hours of entitlement to training higher than that of the outstanding employee whose employee number is 100.
In this question I have to return the names of all the employees, that the number of hours they did for training is greater than the hours that employee number 100 did
I did not make it 2, was hard to me.
I would be very happy to help, I have been trying to do this for two weeks, thank you very much.
I want to point out that I do not have the tables with the data, this is a question from a test I had, I do not know how to make tables for it, I need an answer that is similar to the solution, like a pseudo-code

Answer 1

As per your example, I'm expecting by Employees you mean Employees table and by no of hours they did you mean hourly_credit in Workshop table, then below are the queries -
1st Query Can be like -
SELECT e.e_id FROM Employees e inner join EmpTraining et on e.e_id = et.e_id where et.t_name in (
SELECT t_name FROM EmpTraining where e_id = 100
)
GROUP BY et.e_id
having count(DISTINCT et.t_name) >= (SELECT count(DISTINCT t_name)
FROM EmpTraining where e_id = 100);
Here first I'm filtering rows on the basis of workshop name of employee id 100, using below sub query
SELECT t_name FROM EmpTraining where e_id = 100
Then if we group the result set and then compare distinct count with the distinct count of workshops of employee hundred we can get the employees set we want
Note: Result will contain employee with id - 100 as well to exclude it from the result we can add another condition in where like -
AND e.e_id != 100
2 nd Query
SELECT any_value(e.name) as name
FROM Employees e inner join EmpTraining et on e.e_id = et.e_id inner join Workshops w on et.t_name = w.t_name
GROUP BY et.e_id
having sum(w.hourly_credit) > (
SELECT sum(w.hourly_credit)
FROM Workshops w inner join EmpTraining etg on etg.t_name = w.t_name
where etg.e_id = 100
);
Here I'm grouping rows of EmpTraining on the basis of e_id and comparing sum of hourly_credit with the sum of hourly_credit of employee id 100 EmpTrainings