Will give best answer points and thumbs up !

Must be able to use on Oracle*

Script:

drop table workon; drop table employee; drop table project; drop table division; create table division (did integer, dname varchar (25), managerID integer, constraint division_did_pk primary key (did) ); create table employee (empID integer, name varchar(30), salary float, did integer, constraint employee_empid_pk primary key (empid), constraint employee_did_fk foreign key (did) references division(did) ); create table project (pid integer, pname varchar(25), budget float, did integer, constraint project_pid_pk primary key (pid), constraint project_did_fk foreign key (did) references division(did) ); create table workon (pid integer references project(pid), empID integer references employee(empID), hours integer, constraint workon_pk primary key (pid, empID) ); /* loading the data into the database */ insert into division values (1,'engineering', 2); insert into division values (2,'marketing', 1); insert into division values (3,'human resource', 3); insert into division values (4,'Research and development', 5); insert into division values (5,'accounting', 4); insert into project values (1, 'DB development', 8000, 2); insert into project values (2, 'network development', 6000, 2); insert into project values (3, 'Web development', 5000, 3); insert into project values (4, 'Wireless development', 5000, 1); insert into project values (5, 'security system', 6000, 4); insert into project values (6, 'system development', 7000, 1); insert into employee values (1,'kevin', 32000,2); insert into employee values (2,'joan', 42000,1); insert into employee values (3,'brian', 37000,3); insert into employee values (4,'larry', 82000,5); insert into employee values (5,'harry', 92000,4); insert into employee values (6,'peter', 45000,2); insert into employee values (7,'peter', 68000,3); insert into employee values (8,'smith', 39000,4); insert into employee values (9,'chen', 71000,1); insert into employee values (10,'kim', 46000,5); insert into employee values (11,'smith', 46000,1); insert into workon values (3,1,30); insert into workon values (2,3,40); insert into workon values (5,4,30); insert into workon values (6,6,60); insert into workon values (4,3,70); insert into workon values (2,4,45); insert into workon values (5,3,90); insert into workon values (3,3,100); insert into workon values (6,8,30); insert into workon values (4,4,30); insert into workon values (5,8,30); insert into workon values (6,7,30); insert into workon values (6,9,40); insert into workon values (5,9,50); insert into workon values (4,6,45); insert into workon values (2,7,30); insert into workon values (2,8,30); insert into workon values (2,9,30); insert into workon values (1,9,30); insert into workon values (1,8,30); insert into workon values (1,7,30); insert into workon values (1,5,30); insert into workon values (1,6,30); insert into workon values (2,6,30); 

1). Among all employees who work on DB development, list the name of employee who makes least salary.

2) List the name of the division that has more projects than division of marketing

3)List the name of employees who do not work in the same division with chen but has more salary than all employees of chens division

4)Using update statement to increase the budget of a project if there are more than one employee working on it (hint: you may use structure ... where 1 > (select count().)) (show data before and after update)