SQL 명령어 요약

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* SELECT 칼럼들,... FROM 테이블
    WHERE 칼럼='문자값', 칼럼<=수,  칼럼 LIKE='%문자%' ,... (AND|OR)
    ORDER BY 칼럼 ASC(작은수,오름)|DESC(큰수,내림)
    LIMIT 0,100 (첫번째 부터~100번째 까지)

* INSERT INTO 테이블 VALUES ( 값들,...)
  INSERT INTO 테이블(칼럼들,...) VALUES ( 값들,...)

* UPDATE 테이블 SET 칼럼=값,...
    WHERE 칼럼=값,...

* DELETE FROM 테이블
    WHERE 칼럼=값,...

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http://www.databasedesign.co.uk/sqlselectshortsummary.htm

 

 

SQL SELECT statement - short summary

 

Function	Example
select from	select * from customer select c_no, sname from customer
distinct	select distinct c_no from invoice
order by	select * from customer order by sname select * from customer order by city, balance desc
where	select * from customer where city = ‘London’ and balance <= cred_lim
between	select * from invoice where inv_date between #10-dec-99# and #14-1-00#
like	select * from customer where sname like ‘Dz*’
in	select * from customer where city in (‘London’, ‘Leeds’)
avg,count,max, min,sum,var, stddev	select sum(balance) from customer select count(*) from customer select sum(balance) as TotalBalance from customer select sum(balance), max(cred_lim) from customer
group by	select city, sum(balance) from customer group by city select city, max(balance) as [Highest balance for this city] from customer group by city
having	select city, sum(balance) group by city having sum(balance) > 500
top	select top 2 * from customer order by balance desc select top 1 city, sum(balance) from customer group by city order by sum(balance) desc select top 20 percent * from customer order by balance desc
inner join	select a.c_no, sname, inv_no, amount from customer as a inner join invoice as b on a.c_no = b.c_no where city = ‘London’ and balance > 100
left join	select a.c_no, sname, inv_no, amount from customer as a left join invoice as b on a.c_no = b.c_no where city = ‘London’ and balance > 100
subquery	select * from customer where city = (select city from customer where sname = ‘Sallaway’) select * from customer where c_no not in (select c_no from invoice)
any, all	select * from employee where salary < any (select salary from employee) select * from employee where salary >= all (select salary from employee)
exists, not exists	select * from customer where not exists (select * from invoice where customer.c_no = invoice.c_no)
union	select * from violinplayers union select * from pianoplayers
from a query	select * from query1 where city = ‘London’
select into	select * into temp1 from customer where city = ‘London’ select * into customer in ‘accts1.mdb’ from customer
crosstab query	transform sum(weeklysales.s_value) as sumofs_value select employee.e_name from employee inner join (category inner join weeklysales on category.c_no = weeklysales.c_no) on employee.e_no = weeklysales.e_no group by employee.e_name pivot category.c_name