7.test

来自「嵌入式Linux应用开发详解的所有相应源码」· TEST 代码 · 共 43 行

# OK : Simple join
select staff.first_name, staff.last_name, users.uname from staff, users 
	where staff.staff_id = users.staff_id \p\g


# OK : Simple join with aliases
select S.first_name, S.last_name, U.uname 
	from staff=S, users=U
	where S.staff_id = U.staff_id \p\g


# OK : 3 way 
select staff.first_name, staff.last_name, users.uname, contact.phone
	from staff, users, contact
	where staff.staff_id = users.staff_id 
		and users.staff_id = contact.staff_id\p\g


# OK : 3 way with table aliases
select S.first_name, S.last_name, U.uname, C.phone
	from staff = S, users = U, contact = C
	where S.staff_id = U.staff_id and U.staff_id = C.staff_id\p\g



# OK : Join with sysvar being carried through the result table
select S._rowid, S.first_name, S.last_name, U.uname, C.phone
	from staff = S, users = U, contact = C
	where S.staff_id = U.staff_id and U.staff_id = C.staff_id\p\g

# OK : do some ordering as well
select S.first_name, S.last_name, U.uname, P.pay_date, P.pay_amount
	from staff = S, users = U, pay = P
	where S.staff_id = U.staff_id and U.staff_id = P.staff_id
	order by S.last_name\p\g


# OK : ordering and distinct
select distinct S.first_name, S.last_name, U.uname, P.pay_amount
	from staff = S, users = U, pay = P
	where S.staff_id = U.staff_id and U.staff_id = P.staff_id
	order by S.last_name\p\g