[๋‹ฌ๋ฆฌ๊ธฐ๋ฐ˜] SQL ๋ฌธ๋ฒ• ์—ฐ์Šต 7

2025. 1. 25. 02:11ยท๋‚ด์ผ๋ฐฐ์›€์บ ํ”„/์‚ฌ์ „์บ ํ”„ ๊ณผ์ œ

LV5. ์˜ˆ์‚ฐ์ด ๊ฐ€์žฅ ํฐ ํ”„๋กœ์ ํŠธ๋Š”?

๋‚ด๊ฐ€ ํ‹€๋ฆฐ/๋†“์นœ ๋ถ€๋ถ„์€ ๋ฏผํŠธ์ƒ‰์œผ๋กœ ํ‘œ์‹œ!

 

1.
๊ฐ ์ง์›์ด ์†ํ•œ ๋ถ€์„œ์—์„œ ๊ฐ€์žฅ ๋†’์€ ์›”๊ธ‰์„ ๋ฐ›๋Š” ์ง์›๋“ค๋งŒ ํฌํ•จ๋œ ๊ฒฐ๊ณผ๋ฅผ ์กฐํšŒํ•˜๋Š” SQL ์ฟผ๋ฆฌ๋ฅผ ์ž‘์„ฑํ•ด์ฃผ์„ธ์š”.
a.
 ์ถœ๋ ฅ ๊ฒฐ๊ณผ์—๋Š” ์ง์›์˜ ์ด๋ฆ„, ๋ถ€์„œ, ๊ทธ๋ฆฌ๊ณ  ์›”๊ธ‰์ด ํฌํ•จ๋˜์–ด์•ผ ํ•ฉ๋‹ˆ๋‹ค. 

๋‚˜์˜ ๋‹ต
select e.Name,
e.Department,
e.Salary
from eemployees e left join
(select Department,
max(salary) max_sal
from eemployees
group by 1) a on e.Department=a.Department
where e.Salary=a.max_sal
์ •๋‹ต
SELECT
e.Name,
e.Department,
e.Salary
FROM
Employees e
WHERE
e.Salary = (
SELECT
MAX(Salary)
FROM
Employees e2
WHERE
e2.Department = e.Department
);
์˜ค๋‹ต ๋…ธํŠธ where์ ˆ์— ์„œ๋ธŒ์ฟผ๋ฆฌ๋ฅผ ์ž‘์„ฑํ•˜๋ฉด ๋๋‹ค!
๊ณ ๋ฏผํ•œ ์  ๋ถ„๋ช…ํžˆ join์—†์ด ์„œ๋ธŒ์ฟผ๋ฆฌ๋งŒ์œผ๋กœ ๋ฐ์ดํ„ฐ๋ฅผ ์ถ”์ถœํ•  ์ˆ˜ ์žˆ์—ˆ๋Š”๋ฐ.. from์—๋‹ค๊ฐ€ eemployees ํ…Œ์ด๋ธ”์„ ์•ˆ๋ถ€๋ฅด๊ณ  ๋ฐ”๋กœ ์„œ๋ธŒ์ฟผ๋ฆฌ๋ฅผ ์ž‘์„ฑํ•˜๋‹ˆ๊นŒ name์ด ์•ˆ๋– ์„œ join์œผ๋กœ eemployees ํ…Œ์ด๋ธ”์„ ์—ฐ๊ฒฐํ–ˆ๋‹ค.

 

 

2.
์ง์›์ด ์ฐธ์—ฌํ•œ ํ”„๋กœ์ ํŠธ ์ค‘ ์˜ˆ์‚ฐ์ด 10,000 ์ด์ƒ์ธ ํ”„๋กœ์ ํŠธ๋งŒ์„ ์กฐํšŒํ•˜๋Š” SQL ์ฟผ๋ฆฌ๋ฅผ ์ž‘์„ฑํ•ด์ฃผ์„ธ์š”. 
a.
์ถœ๋ ฅ ๊ฒฐ๊ณผ์—๋Š” ์ง์› ์ด๋ฆ„, ํ”„๋กœ์ ํŠธ ์ด๋ฆ„, ๊ทธ๋ฆฌ๊ณ  ํ”„๋กœ์ ํŠธ ์˜ˆ์‚ฐ์ด ํฌํ•จ๋˜์–ด์•ผ ํ•ฉ๋‹ˆ๋‹ค. 

๋‚˜์˜ ๋‹ต
select e.Name, p.ProjectName, p.Budget
from employeeprojects ep join projects p on ep.projectID=p.ProjectID
join eemployees e on ep.EmployeeID=e.EmployeeID
where p.Budget>=10000
์ •๋‹ต
SELECT
e.Name,
p.ProjectName,
p.Budget
FROM
Employees e
JOIN
EmployeeProjects ep ON e.EmployeeID = ep.EmployeeID
JOIN
Projects p ON ep.ProjectID = p.ProjectID
WHERE
p.Budget >= 10000;
์˜ค๋‹ต ๋…ธํŠธ -
๊ณ ๋ฏผํ•œ ์  JOIN๋งŒ ์‚ฌ์šฉํ•˜๋ฉด ๋ผ์„œ ์˜์™ธ๋กœ ์•„์ฃผ ๊ฐ„๋‹จํ–ˆ๋‹ค.

 

'๋‚ด์ผ๋ฐฐ์›€์บ ํ”„ > ์‚ฌ์ „์บ ํ”„ ๊ณผ์ œ' ์นดํ…Œ๊ณ ๋ฆฌ์˜ ๋‹ค๋ฅธ ๊ธ€

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[๋‹ฌ๋ฆฌ๊ธฐ๋ฐ˜] SQL ๋ฌธ๋ฒ• ์—ฐ์Šต 7
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