배수빈: [SQL] 언어별 개발자 분류하기_241122

Author: baexxbin

SQL/11주차/SB_언어별 개발자 분류하기.sql

@@ -0,0 +1,78 @@
+-- 코드를 작성해주세요
+-- GRADE가 존재하는 개발자의 GRADE별 개발자의 정보를 조회
+
+-- CTE 이용
+WITH FE AS(
+    SELECT SUM(CODE) AS CODE
+    FROM SKILLCODES
+    WHERE CATEGORY = "Front End"
+)
+, Python AS(
+    SELECT CODE
+    FROM SKILLCODES
+    WHERE NAME = "Python"
+)
+, C AS(
+    SELECT CODE
+    FROM SKILLCODES
+    WHERE NAME = "C#"
+)
+, GRADES AS(
+    SELECT
+        CASE 
+            WHEN d.SKILL_CODE & f.CODE !=0 AND d.SKILL_CODE & p.CODE !=0 THEN 'A'
+            WHEN d.SKILL_CODE & c.CODE != 0 THEN 'B'
+            WHEN d.SKILL_CODE & f.CODE != 0 THEN 'C'
+        END AS GRADE
+        , d.ID, d.EMAIL
+    FROM DEVELOPERS d, FE f, Python p, C c
+)
+SELECT *
+FROM GRADES g
+WHERE GRADE IS NOT NULL
+ORDER BY g.GRADE, g.ID
+
+-- 각 CTE를 테이블 처럼 암시적 조인으로 사용
+-- 각 개발자마다 CTE반복 참조하며 CASE WHEN조건 매번 계산
+
+
+
+-- 최적화
+WITH FE AS (
+    SELECT SUM(CODE) AS CODE
+    FROM SKILLCODES
+    WHERE CATEGORY = "Front End"
+),
+SKILLS AS (
+    SELECT 
+        MAX(CASE WHEN NAME = "C#" THEN CODE END) AS C_CODE,
+        MAX(CASE WHEN NAME = "Python" THEN CODE END) AS P_CODE,
+        (SELECT CODE FROM FE) AS FE_CODE
+    FROM SKILLCODES
+),
+DEVELOPER_FLAGS AS (
+    SELECT
+        d.ID,
+        d.EMAIL,
+        d.SKILL_CODE,
+        (d.SKILL_CODE & s.FE_CODE) AS IS_FE,
+        (d.SKILL_CODE & s.P_CODE) AS IS_PYTHON,
+        (d.SKILL_CODE & s.C_CODE) AS IS_C
+    FROM DEVELOPERS d
+    CROSS JOIN SKILLS s
+),
+GRADES AS (
+    SELECT
+        CASE 
+            WHEN IS_FE != 0 AND IS_PYTHON != 0 THEN 'A'
+            WHEN IS_C != 0 THEN 'B'
+            WHEN IS_FE != 0 THEN 'C'
+        END AS GRADE,
+        ID,
+        EMAIL
+    FROM DEVELOPER_FLAGS
+)
+SELECT *
+FROM GRADES
+WHERE GRADE IS NOT NULL
+ORDER BY GRADE, ID;