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Data Science LAB

[MySQL] ν”„λ‘œκ·Έλž˜λ¨ΈμŠ€ μ½”λ”©ν…ŒμŠ€νŠΈ μ—°μŠ΅ 24(μš°μœ μ™€ μš”κ±°νŠΈκ°€ λ‹΄κΈ΄ μž₯λ°”κ΅¬λ‹ˆ) λ³Έλ¬Έ

πŸ“ Coding Test/MySQL

[MySQL] ν”„λ‘œκ·Έλž˜λ¨ΈμŠ€ μ½”λ”©ν…ŒμŠ€νŠΈ μ—°μŠ΅ 24(μš°μœ μ™€ μš”κ±°νŠΈκ°€ λ‹΄κΈ΄ μž₯λ°”κ΅¬λ‹ˆ)

γ…… γ…œ γ…” γ…‡ 2022. 3. 4. 14:27
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λ“œλ””μ–΄ ν”„λ‘œκ·Έλž˜λ¨ΈμŠ€ λ§ˆμ§€λ§‰ 단계인 LEVEL4의 λ¬Έμ œμ΄λ‹€!!

 

문제
CART_PRODUCTS ν…Œμ΄λΈ”μ€ μž₯λ°”κ΅¬λ‹ˆμ— λ‹΄κΈ΄ μƒν’ˆ 정보λ₯Ό 담은 ν…Œμ΄λΈ”μž…λ‹ˆλ‹€. 
CART_PRODUCTS ν…Œμ΄λΈ”μ˜ κ΅¬μ‘°λŠ” λ‹€μŒκ³Ό κ°™μœΌλ©°, 
ID, CART_ID, NAME, PRICE
λŠ” 각각 ν…Œμ΄λΈ”μ˜ 아이디, μž₯λ°”κ΅¬λ‹ˆμ˜ 아이디, μƒν’ˆ μ’…λ₯˜, 가격을 λ‚˜νƒ€λƒ…λ‹ˆλ‹€.

데이터 뢄석 νŒ€μ—μ„œλŠ” 우유(Milk)와 μš”κ±°νŠΈ(Yogurt)λ₯Ό λ™μ‹œμ— κ΅¬μž…ν•œ μž₯λ°”κ΅¬λ‹ˆκ°€ μžˆλŠ”μ§€ μ•Œμ•„λ³΄λ € ν•©λ‹ˆλ‹€. μš°μœ μ™€ μš”κ±°νŠΈλ₯Ό λ™μ‹œμ— κ΅¬μž…ν•œ μž₯λ°”κ΅¬λ‹ˆμ˜ 아이디λ₯Ό μ‘°νšŒν•˜λŠ” SQL 문을 μž‘μ„±ν•΄μ£Όμ„Έμš”. μ΄λ•Œ κ²°κ³ΌλŠ” μž₯λ°”κ΅¬λ‹ˆμ˜ 아이디 순으둜 λ‚˜μ™€μ•Ό ν•©λ‹ˆλ‹€.

 

 

 

 

 

μ˜ˆμ‹œ

 

 

 

 

 

 

문제 풀이
SELECT A.CART_ID
FROM (SELECT DISTINCT CART_ID,NAME
     FROM CART_PRODUCTS
     WHERE NAME = 'MILK' OR NAME = 'YOGURT') A
GROUP BY A.CART_ID
HAVING COUNT (A.CART_ID) >= 2

쀑볡을 μ œκ±°ν•˜κ³  MILKλ‚˜ YOGURTκ°€ λ‹΄κΈ΄ μž₯λ°”κ΅¬λ‹ˆμ˜ IDλ₯Ό κ΅¬ν•œλ’€

HAVING COUNTλ₯Ό μ΄μš©ν•˜μ—¬ 2 이상인 μž₯λ°”κ΅¬λ‹ˆμ˜ IDλ₯Ό κ΅¬ν•œλ‹€. 

 

 

 

SELECT DISTINCT A.CART_ID
FROM CART_PRODUCTS AS A
JOIN CART_PRODUCTS AS B
ON A.CART_ID = B.CART_ID 
AND A.NAME = 'MILK' AND B.NAME = 'YOGURT'

JOIN을 μ΄μš©ν•˜μ—¬ MILK와 YOGURTλ₯Ό 가진 ν…Œμ΄λΈ”μ„ λ”°λ‘œ 뽑아 문제λ₯Ό ν•΄κ²°!

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