문제
MySQL
: 대소문자 구분하지 않아도 됨
내가 작성한 정답1 - like
select city
from station
where city like 'i%' or city like 'e%' or city like 'a%' or city like 'o%' or city like 'u%';내가 작성한 정답2 - 정규표현식
select city
from station
where city regexp '^[aeiou]'; // ^: 문자열의 시작내가 작성한 정답3 - substring
select city
from station
where substring(city,1,1) in ('a','e','i','o','u');내가 작성한 정답4 - left
select city
from station
where left(city,1) in ('a','e','i','o','u');Oracle
: 대소문자 구분 필요
내가 작성한 정답1 - like
: 도시의 첫글자는 대문자
select city
from station
where city like 'I%' or city like 'E%' or city like 'A%' or city like 'O%' or city like 'U%';내가 작성한 정답2 - 정규표현식 regexp_like
select city
from station
where regexp_like(lower(city),'^[aeiou]');내가 작성한 정답3 - substr
select city
from station
where lower(substr(city,1,1)) in ('a','e','i','o','u');MS SQL Server - MySQL과 유사
: 대소문자 구분 x
내가 작성한 정답1 - like
select city
from station
where city like 'i%' or city like 'e%' or city like 'a%' or city like 'o%' or city like 'u내가 작성한 정답2 - like2
select city
from station
where city like '[aeiou]%';내가 작성한 정답3 - substring
select city
from station
where substring(city,1,1) in ('a','e','i','o','u');내가 작성한 정답4 - left
select city
from station
where left(city,1) in ('a','e','i','o','u');DB2 - Oracle과 유사
내가 작성한 정답1 - like
select city
from station
where city like 'I%' or city like 'E%' or city like 'A%' or city like 'O%' or city like 'U%';내가 작성한 정답2 - substr
select city
from station
where lower(substr(city,1,1)) in ('a','e','i','o','u');정리!
DBMS | 문자추출 | left 유무 | 정규표현식 |
MS SQL Server | substring | o | x → like로 표현가능 |
MySQL | substring | o | o → regexp |
Oracle | substr | x | o → regexp_like(colunm,정규식) |
DB2 | substr | x | x → substr로 표현가능 |
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