
SELECT ANIMAL_ID
FROM ANIMAL_INS
WHERE NAME IS NULL
ORDER BY 1;

SELECT COUNT(USER_ID)
FROM USER_INFO
WHERE JOINED LIKE '2021%' AND AGE BETWEEN 20 AND 29;

SELECT ANIMAL_ID, NAME,
IF(SEX_UPON_INTAKE REGEXP 'Spayed|Neutered', 'O', 'X') AS '중성화'
FROM ANIMAL_INS;

SELECT SUBSTR(PRODUCT_CODE,1,2) CATEGORY, COUNT(PRODUCT_ID) PRODUCTS
FROM PRODUCT
GROUP BY 1
ORDER BY 1;

SELECT ANIMAL_TYPE,COUNT(ANIMAL_ID) count
FROM ANIMAL_INS
GROUP BY 1
ORDER BY 1;

SELECT HOUR(DATETIME) AS HOUR, COUNT(*) COUNT
FROM ANIMAL_OUTS
WHERE HOUR(DATETIME) BETWEEN 9 AND 19
GROUP BY 1
ORDER BY 1;

SELECT MCDP_CD '진료과코드', COUNT(*) '5월예약건수'
FROM APPOINTMENT
WHERE APNT_YMD LIKE '2022-05%'
GROUP BY 1
ORDER BY 2,1;

SELECT PT_NAME, PT_NO,GEND_CD,AGE,COALESCE(TLNO,'NONE')
FROM PATIENT
WHERE GEND_CD='W' AND AGE<=12
ORDER BY AGE DESC,PT_NAME;

SELECT FLAVOR
FROM FIRST_HALF
ORDER BY TOTAL_ORDER DESC, SHIPMENT_ID
30. 자동차 종류 별 특정 옵션이 포함된 자동차 수 구하기

SELECT CAR_TYPE,COUNT(*) CARS
FROM CAR_RENTAL_COMPANY_CAR
WHERE OPTIONS LIKE ( '%통풍시트%') OR
OPTIONS LIKE ( '%열선시트%') OR
OPTIONS LIKE ( '%가죽시트%')
GROUP BY 1
ORDER BY 1;
'데일리루틴(알고리즘) > SQL문제' 카테고리의 다른 글
| 10/28 SQL 문제풀이 (0) | 2024.10.28 |
|---|---|
| 10/22 SQL문제 풀이 (0) | 2024.10.22 |