SQLD 기출문제 예상문제 265번

다음 문제를 풀어보세요.

다음 중 아래 SQL과 동일한 결과를 반환하며, 일반적으로 더 효율적이라고 알려진 쿼리를 고르시오.

[테이블 구조]
CREATE TABLE CUSTOMERS (
    CUSTOMER_ID NUMBER PRIMARY KEY,
    CUSTOMER_NAME VARCHAR2(100),
    CITY VARCHAR2(50)
);

CREATE TABLE ORDERS (
    ORDER_ID NUMBER PRIMARY KEY,
    CUSTOMER_ID NUMBER,
    ORDER_DATE DATE,
    AMOUNT NUMBER,
    FOREIGN KEY (CUSTOMER_ID) REFERENCES CUSTOMERS(CUSTOMER_ID)
);

[제공된 SQL]
SELECT CUSTOMER_ID, CUSTOMER_NAME
FROM CUSTOMERS
WHERE CUSTOMER_ID IN (SELECT CUSTOMER_ID FROM ORDERS);

SELECT C.CUSTOMER_ID, C.CUSTOMER_NAME FROM CUSTOMERS C WHERE EXISTS (SELECT 1 FROM ORDERS O WHERE O.CUSTOMER_ID = C.CUSTOMER_ID);

SELECT C.CUSTOMER_ID, C.CUSTOMER_NAME FROM CUSTOMERS C JOIN ORDERS O ON C.CUSTOMER_ID = O.CUSTOMER_ID;

SELECT C.CUSTOMER_ID, C.CUSTOMER_NAME FROM CUSTOMERS C WHERE C.CUSTOMER_ID NOT IN (SELECT CUSTOMER_ID FROM ORDERS);

SELECT C.CUSTOMER_ID, C.CUSTOMER_NAME FROM CUSTOMERS C JOIN ORDERS O ON C.CUSTOMER_ID = O.CUSTOMER_ID GROUP BY C.CUSTOMER_ID, C.CUSTOMER_NAME HAVING COUNT(O.ORDER_ID) > 0;