DATABASE STEP 19 - SubQuery

데이터베이스

DATABASE STEP 19 - SubQuery

IT의 큰손 2023. 3. 21. 22:06

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★ MainQuery

여태까지의 SQL
일반적인 SQL
하나의 문장안에 하나의 select(insert, update, delete)로 구성된 쿼리

★ Sub Query, 서브 쿼리, 부속 질의

하나의 문장(select, insert, update, delete)안에 또 다른 문장(select)이 들어있는 쿼리
하나의 select문 안에 들어있는 또 다른 select 문
삽입 위치 > 거의 대다수의 절(select 절, from 절, where 절 < 자주 사용 :: 자주사용 안함 > group by 절, having 절, order by 절)

■ 사용 ex)

ex1)

-- tblCountry.  인구수가 가장 많은 나라의 이름?
select *
from tblCountry;

update tblCountry set population = 120060 where name = '중국';

select max(population) -- 인구수가 가장 많은 > 120660 
from tblCountry;
select name from tblCountry where population = 120660; --중국

-- 합치면
-- 장점
-- 1. 2개의 SQL > 1개의 SQL
-- 2. 변화에 강하다. (인구수 변동) > 개발자 편의성 향상
select name from tblCountry 
    where population = (select max(population) from tblCountry);

ex2)

-- tblComedian. 체중이 가장 많이 나가는 사람의 이름?
select max(weight) from tblComedian;

select * from tblComedian
    where weight = (select max(weight) from tblComedian);

ex3)

-- tblInsa. 급여 1등?
select max(basicpay) from tblInsa; -- 2650000

select * from tblInsa
    where basicpay = (select max(basicpay) from tblInsa);

ex4)

-- tblInsa. 막내 직원? 이정석 05/09/26
select * from tblInsa where ibsadate = (select max(ibsadate) from tblInsa);

-- tblIsa. 왕고참? 김인수 95/02/23
select * from tblInsa where ibsadate = (select min(ibsadate) from tblInsa);

-- tblInsa. 평균 급여보다 더 많이 받는 직원?
select avg(basicpay) from tblInsa;

select * from tblInsa where basicpay >= (select avg(basicpay) from tblInsa);

-- tblInsa. '홍길동'보다 급여가 많은 직원?
select basicpay from tblInsa where name = '홍길동';

select * from tblInsa 
    where basicpay >= (select basicpay from tblInsa where name = '홍길동') and name <> '홍길동';

★ 서브 쿼리 삽입 위치

1. 조건절
2. 컬럼리스트
3. from 절

■ 1. 조건절

비교 대상(비교값) 역할 > 값
where 절, having 절, case문
a. 반환값이 1행 1열 > 단일 값 반환 > 값 1개로 취급

select * from tblInsa
    where basicpay >= (select avg(basicpay) from tblInsa);

b. 반환값이 N행 1열 > 다중값(같은 성질의 여러개의 데이터) 반환 > 값 N개로 취급

-- 급여가 260만원 이상 받는 직원이 근무하는 부서 직원 명단을 가져오시오. > 기획부 + 총무부
-- 01427. 00000 -  "single-row subquery returns more than one row"

select * from tblInsa
    -- where buseo(비교대상 1개) = 기획부, 총무부(비교대상 2개)
    -- where buseo = '기획부' or buseo = '총무부';
    -- where buseo in ('기획부', '총무부');  같은 성질의 N개의 데이터 > 열거형
    where buseo in (select buseo from tblInsa where basicpay >= 2600000);
    
    
-- b. 반환값이 N행 1열 > 다중 값 (같은 성질의 여러개의 데이터) 반환 > 값 N개로 취급  
-- '홍길동'과 같은 지역 + 같은 직위 > 소속 부서 직원 명단
select *from tblInsa where name = '홍길동';
select * from tblInsa where city='서울' and jikwi = '부장' ;

select city from tblInsa where name = '홍길동'; -- 서울
select jikwi from tblInsa where name = '홍길동'; --부장

select buseo from tblInsa
    where city = (select city from tblInsa where name = '홍길동')
        and jikwi = (select jikwi from tblInsa where name = '홍길동')
            and name <> '홍길동';

select * from tblInsa
    where buseo in (select buseo from tblInsa
    where city = (select city from tblInsa where name = '홍길동')
        and jikwi = (select jikwi from tblInsa where name = '홍길동')
            and name <> '홍길동');

c. 반환값이 1행 N열 > 다중 값(서로 다른 성질의 여러개의 데이터) 반환

select city from tblInsa where name ='홍길동';
select jikwi from tblInsa where name = '홍길동';

select * from tblInsa where city = '서울' and jikwi = '부장';

select * from tblInsa
    where city = (select city from tblInsa where name ='홍길동')
        and jikwi = (select jikwi from tblInsa where name = '홍길동');
        
select * from tblInsa
    where (city, jikwi) = (select city, jikwi from tblInsa where name = '홍길동');
    
    
select * from tblAddressBook; 

select * from tblAddressBook
    where (gender, age, job) = (select gender, age, job
                                                                    from tblAddressBook where name = '기예주');

d. 반환값이 N행 N열 > 다중값 반환

-- 급여가 260만원 이상 받는 직원과 같은 부서 + 같은 지역에 있는 모든 직원?
select * from tblInsa where basicpay >= 2600000;

-- 서울 + 기획부 && 경남 + 총무부
select city, buseo from tblInsa where basicpay >= 2600000;

select * from tblInsa
    where (city, buseo) in (select city, buseo from tblInsa where basicpay >= 2600000);
    
select
    buseo,
    avg(basicpay)
from tblInsa
    group by buseo
        having avg(basicpay) >= (select avg(basicpay) from tblInsa where buseo = '개발부');

■ 2. 컬럼 리스트

단일 데이터(값) > 원자 값 > 관계형 데이터베이스 > 표 > 속성(컬럼)의 값은 원자값이어야한다.
a. 컬럼명
b. 상수
c. 연산
d. 함수

select 
    name as "컬럼",
    100 as "상수",
    basicpay + 100 as "연산",
    length(name) as "함수"
from tblInsa;

select
    name, buseo, basicpay,
    (select round(avg(basicpay)) from tblInsa) as "평균 급여"
from tblInsa;

select
    name, buseo, basicpay,
    (select round(avg(basicpay)) from tblInsa where buseo = i.buseo) as "소속 부서 평균 급여"
from tblInsa i;

■ 3. from 절

서브쿼리의 결과셋을 하나의 테이블이라고 생각하고, 또 다른 select를 실행
구문을 단순화하기 위해서 사용

-- employees. 'Munich'에 위치한 부서에 소속된 직원 명단?
select * from employees; --department_id 소속 부서
select * from departments; -- location_id 위치 정보
select * from locations;

select location_id from locations
    where city = 'Munich';
select department_id from departments
    where location_id = (select location_id from locations
    where city = 'Munich');

select * from employees
    where department_id in (select department_id from departments
        where location_id = (select location_id from locations
            where city = 'Seattle'));

■ 종합 요구사항 예제 모음

1. tblMen. tblWomen. 서로 짝이 있는 사람 중 남자와 여자의 정보를 모두 가져오시오. > select절

select * from tblMen;
select * from tblWomen;

select
    name, height, weight,
    (select height from tblWomen where name = m.couple) as "여자친구 키",
    (select name from tblWomen where name = m.couple) as "여자친구 이름",
    (select weight from tblWomen where name = m.couple) as "여자 몸무게"
from tblMen m;

2. tblAddressBook. 가장 많은 사람들이 가지고 있는 직업은 주로 어느 지역 태생(hometown)인가? > where절

select * from tblAddressBook;

select max(count(*)) from tblAddressBook group by job;

select job from tblAddressBook
    group by job
        having count(*) = (select max(count(*)) from tblAddressBook group by job);
        
select distinct hometown from tblAddressBook
    where job = (select job from tblAddressBook
    group by job
        having count(*) = (select max(count(*)) from tblAddressBook group by job));

3. tblAddressBook. 이메일 도메인들 중 평균 아이디 길이가 가장 긴 이메일 사이트의 도메인은 무엇인가? > group by + having

select 
    substr(email, instr(email, '@') +1) as "도메인",
    avg(length(substr(email, 1, instr(email, '@') -1))) as "평균 아이디 길이"
from tblAddressBook
    group by substr(email, instr(email, '@') +1)
        having avg(length(substr(email, 1, instr(email, '@') -1))) = 
        (select 
            max(avg(length(substr(email, 1, instr(email, '@') -1))))
        from tblAddressBook
            group by substr(email, instr(email, '@') +1));
    
select 
    max(avg(length(substr(email, 1, instr(email, '@') -1))))
from tblAddressBook
    group by substr(email, instr(email, '@') +1);

4. tblAddressBook. 평균 나이가 가장 많은 출신(hometown)들이 가지고 있는 직업 중 가장 많은 직업은? > where + group by + having

select * from tblAddressBook;

select 
    max(avg(age))
from tblAddressBook
    group by hometown;
    
select
    hometown
from tblAddressBook
    group by hometown
        having avg(age) = 
        (select 
            max(avg(age))
        from tblAddressBook
        group by hometown);
    

select 
    hometown,
    round(avg(age))
from tblAddressBook
    group by hometown
        having max(avg(age));

select 
        max(count(*))
from tblAddressBook
    where hometown = '광주'
        group by job;

select
    max(avg(age))
from tblAddressBook
    group by hometown; -- 38.6875

select
    hometown
from tblAddressBook
    group by hometown
        having avg(age) = (select
                                max(avg(age))
                            from tblAddressBook
                                group by hometown);

select
    max(count(*))
from tblAddressBook
    where hometown = '광주'
        group by job;

select
    job
from tblAddressBook
    where hometown = (select
                            hometown
                        from tblAddressBook
                            group by hometown
                                having avg(age) = (select
                                                        max(avg(age))
                                                    from tblAddressBook
                                                        group by hometown))
        group by job
            having count(*) = (select
                                    max(count(*))
                                from tblAddressBook
                                    where hometown = (select
                                                            hometown
                                                        from tblAddressBook
                                                            group by hometown
                                                                having avg(age) = (select
                                                                                        max(avg(age))
                                                                                    from tblAddressBook
                                                                                        group by hometown))
                                        group by job);

5. tblAddressBook. 남자 평균 나이보다 나이가 많은 서울 태생 + 직업을 가지고 있는 사람들을 가져오시오. > where절

select * from tblAddressBook;
select 
    avg(age)
from tblAddressBook
    where gender = 'm';
    
select *
from tblAddressBook
    where job is not null;
    
select 
    name, job, hometown
from tblAddressBook
    where age > (select 
                            avg(age)
                        from tblAddressBook
                            where gender = 'm') and hometown = '서울' and job <> '백수';

6. tblAddressBook. gmail.com을 사용하는 사람들의 성별 > 세대별(10,20,30,40대) 인원수를 가져오시오. > where절

select * from tblAddressBook;

select 
    gender, age
from tblAddressBook
    where substr(email, instr(email, '@') +1) = 'gmail.com';
    
select 
    count(case
        when gender = 'm' and age between 10 and 19 then 1 
    end) as "남자 10대",
    count(case
        when gender = 'f' and age between 10 and 19 then 1 
    end) as "여자 10대",
    count(case
        when gender = 'm' and age between 20 and 29 then 1 
    end) as "남자 20대",
    count(case
        when gender = 'f' and age between 20 and 29 then 1 
    end) as "여자 20대",
    count(case
        when gender = 'm' and age between 30 and 39 then 1 
    end) as "남자 30대",
    count(case
        when gender = 'f' and age between 30 and 39 then 1 
    end) as "여자 30대",
    count(case
        when gender = 'm' and age between 40 and 49 then 1 
    end) as "남자 40대",
    count(case
        when gender = 'f' and age between 40 and 49 then 1 
    end) as "여자 40대"
from (select 
            gender, age
        from tblAddressBook
            where substr(email, instr(email, '@') +1) = 'gmail.com');

7. -- tblAddressBook. 가장 나이가 많으면서 가장 몸무게가 많이 나가는 사람과 같은 직업을 가지는 사람들을 가져오시오. > where절

select * from tblAddressBook;

select 
    max(age)
from tblAddressBook;

select
    max(weight)
from tblAddressBook;

select
    job
from tblAddressBook
    where age = (select 
                            max(age)
                        from tblAddressBook) 
        and weight = (select
                            max(weight)
                        from tblAddressBook);



select *
from tblAddressBook
    where job = (select
                            job
                        from tblAddressBook
                            where age = (select 
                                                    max(age)
                                                from tblAddressBook) 
                                and weight = (select
                                                    max(weight)
                                                from tblAddressBook));

8. tblAddressBook. 동명이인이 여러명 있습니다. 이 중 가장 인원수가 많은 동명이인(모든 이도윤)의 명단을 가져오시오. > where절

select * from tblAddressBook;

select 
    max(count(*))
from tblAddressBook
    group by name;


    
select 
    name
from tblAddressBook
    group by name
        having count(*) = (select 
                                    max(count(*))
                                from tblAddressBook
                                    group by name);
    
select *
from tblAddressBook
    where name = (select 
                            name
                        from tblAddressBook
                            group by name
                                having count(*) = (select 
                                                            max(count(*))
                                                        from tblAddressBook
                                                            group by name));

9. tblAddressBook. 가장 사람이 많은 직업의(332명) 세대별 비율을 구하시오.> where + group by + having

SELECT 
    job,
    round(count(CASE
        when age between 10 and 19 then 1
    end)/count(*)*100,2)||'%' as "[10대]",
    round(count(CASE
        when age between 20 and 29 then 1
    end)/count(*)*100,2)||'%' as "[20대]",
    round(count(CASE
        when age between 30 and 39 then 1
    end)/count(*)*100,2)||'%' as "[30대]",
    round(count(CASE
        when age between 40 and 49 then 1
    end)/count(*)*100,2)||'%' as "[40대]"
FROM tbladdressbook group by job having count(job) = (SELECT max(count(job)) FROM tbladdressbook group by job);

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