일 | 월 | 화 | 수 | 목 | 금 | 토 |
---|---|---|---|---|---|---|
1 | 2 | 3 | 4 | |||
5 | 6 | 7 | 8 | 9 | 10 | 11 |
12 | 13 | 14 | 15 | 16 | 17 | 18 |
19 | 20 | 21 | 22 | 23 | 24 | 25 |
26 | 27 | 28 | 29 | 30 | 31 |
Tags
- 개발자_조이킴
- 정규표현식
- 코딩테스트
- for문
- 코플릿
- 프로그래머스
- 자바스크립트
- JavaScript
- MySQL
- Algorithms
- Hackerrank
- Where
- 배열
- Programmers
- node.js
- join
- array.push()
- 코딩공부
- 코드스테이츠
- 최강의 인생
- 알고리즘
- 블록체인
- array.slice()
- 개발자의 책장
- Developer_JoyKim
- select
- array
- 재귀함수
- SQL
- 역행자
Archives
- Today
- Total
CodingSpace
[HackerRank/SQL] Basic SELECT - Employee Salaries 본문
Problem. Basic SELECT - Employee Salaries
Link.
https://www.hackerrank.com/challenges/salary-of-employees/problem?isFullScreen=true
Description.
Write a query that prints a list of employee names (i.e.: the name attribute) for employees in Employee having a salary greater than per month who have been employees for less than months.
Sort your result by ascending employee_id.
Key Point.
SELECT,
WHERE,
My Answer.
SELECT E.name
FROM Employee E
WHERE E.salary > 2000 AND E.months < 10;
References.
'HackerRank > SQL' 카테고리의 다른 글
[HackerRank/SQL] Revising Aggregations - The Sum Function (0) | 2022.07.10 |
---|---|
[HackerRank/SQL] Revising Aggregations - The Count Function (0) | 2022.07.05 |
[HackerRank/SQL] Basic Join - The Report (0) | 2022.06.27 |
[HackerRank/SQL] Basic Join - Average Population of Each Continent (0) | 2022.06.08 |
[HackerRank/SQL] Basic Join - Population Census (0) | 2022.05.26 |
Comments