문제코드SELECT YEAR(ONLINE_SALE.SALES_DATE) AS YEAR, MONTH(ONLINE_SALE.SALES_DATE) AS MONTH, USER_INFO.GENDER, COUNT(DISTINCT ONLINE_SALE.USER_ID) AS USERSFROM USER_INFOJOIN ONLINE_SALE ON USER_INFO.USER_ID = ONLINE_SALE.USER_IDWHERE USER_INFO.GENDER IS NOT NULLGROUP BY YEAR(ONLINE_SALE.SALES_DATE), MONTH(ONLINE_SALE.SALES_DATE), USER_INFO.GENDERORDER BY Y..

문제코드SELECT APPOINTMENT.APNT_NO, PATIENT.PT_NAME, PATIENT.PT_NO, DOCTOR.MCDP_CD, DOCTOR.DR_NAME, APPOINTMENT.APNT_YMDFROM DOCTOR JOIN APPOINTMENT ON DOCTOR.DR_ID = APPOINTMENT.MDDR_ID JOIN PATIENT ON PATIENT.PT_NO = APPOINTMENT.PT_NOWHERE APPOINTMENT.APNT_CNCL_YN = 'N' AND DOCTOR.MCDP_CD = 'CS' AND DATE_FORMAT(APPOINTMENT.APNT..

문제코드SELECT FP.PRODUCT_ID, FP.PRODUCT_NAME, SUM(FP.PRICE * FO.AMOUNT) AS TOTAL_SALESFROM FOOD_PRODUCT AS FPJOIN FOOD_ORDER AS FO ON FP.PRODUCT_ID = FO.PRODUCT_IDWHERE DATE_FORMAT(FO.PRODUCE_DATE, '%Y-%m') = '2022-05'GROUP BY FP.PRODUCT_ID, FP.PRODUCT_NAMEORDER BY TOTAL_SALES DESC, FP.PRODUCT_ID; 출처: 프로그래머스 코딩 테스트 연습, https://school.programmers.co.kr/learn/courses/30/lessons/131117

문제코드SELECT CATEGORY, PRICE, PRODUCT_NAMEFROM FOOD_PRODUCTWHERE (CATEGORY, PRICE) IN ( SELECT CATEGORY, MAX(PRICE) FROM FOOD_PRODUCT WHERE CATEGORY IN ('과자', '국', '김치', '식용유') GROUP BY CATEGORY)ORDER BY PRICE DESC; 출처: 프로그래머스 코딩 테스트 연습, https://school.programmers.co.kr/learn/courses/30/lessons/131116

문제코드SELECT AI.ANIMAL_ID, AI.ANIMAL_TYPE, AI.NAMEFROM ANIMAL_INS AS AI JOIN ANIMAL_OUTS AS AO ON AI.ANIMAL_ID = AO.ANIMAL_IDWHERE SEX_UPON_INTAKE = 'Intact Male' AND AO.SEX_UPON_OUTCOME = 'Neutered Male' OR SEX_UPON_INTAKE = 'Intact Female' AND AO.SEX_UPON_OUTCOME = 'Spayed Female' 출처: 프로그래머스 코딩 테스트 연습, https://school.programmers.co.kr/learn/courses/30/lessons/59045

문제코드SELECT id, (CASE WHEN ntile_rank = 1 THEN 'CRITICAL' WHEN ntile_rank = 2 THEN 'HIGH' WHEN ntile_rank = 3 THEN 'MEDIUM' ELSE 'LOW' END) AS colony_nameFROM ( SELECT id, NTILE(4) OVER (ORDER BY size_of_colony DESC) AS ntile_rank FROM ecoli_data) ranked_dataORDER BY id; 출처: 프로그래머스 코딩 테스트 연습, https://school.pro..