"HR-SQL Average Population of Each Continent"의 두 판 사이의 차이

2021년 7월 31일 (토) 10:48 기준 최신판

해커랭크 SQL
문제	DB2	MS SQL	MySQL	Oracle
HR-SQL Basic Join e
HR-SQL Asian Population
HR-SQL African Cities
HR-SQL Average Population of Each Continent
HR-SQL The Report
HR-SQL Top Competitors
HR-SQL Ollivander's Inventory
HR-SQL Challenges
HR-SQL Contest Leaderboard

Africa 274439 
Asia 693038 
Europe 175138 
Oceania 109189 
South America 147435

SELECT CO.CONTINENT, FLOOR(AVG(CI.POPULATION))
FROM CITY AS CI
JOIN COUNTRY AS CO ON CO.CODE = CI.COUNTRYCODE
GROUP BY CO.CONTINENT;

SELECT CO.CONTINENT, FLOOR(AVG(CI.POPULATION))
FROM CITY AS CI, COUNTRY AS CO
WHERE CO.CODE = CI.COUNTRYCODE
GROUP BY CO.CONTINENT;

@@ 1번째 줄: / 1번째 줄: @@
 ==개요==
 ;HR-SQL Average Population of Each Continent
-*
+* https://www.hackerrank.com/challenges/average-population-of-each-continent/problem
 {{HR-SQL 헤더}}
@@ 7번째 줄: / 7번째 줄: @@
 |}
-<source lang='sql'>
+<syntaxhighlight lang='text'>
-</source>
+Africa 274439
+Asia 693038
+Europe 175138
+Oceania 109189
+South America 147435
+</syntaxhighlight>
+==MySQL==
+<syntaxhighlight lang='mysql'>
+SELECT CO.CONTINENT, FLOOR(AVG(CI.POPULATION))
+FROM CITY AS CI
+JOIN COUNTRY AS CO ON CO.CODE = CI.COUNTRYCODE
+GROUP BY CO.CONTINENT;
+</syntaxhighlight>
+<syntaxhighlight lang='mysql'>
+SELECT CO.CONTINENT, FLOOR(AVG(CI.POPULATION))
+FROM CITY AS CI, COUNTRY AS CO
+WHERE CO.CODE = CI.COUNTRYCODE
+GROUP BY CO.CONTINENT;
+</syntaxhighlight>