SQL PLUS QUESTIONS

Question

According to the tables and attributes below I need to solve 25 questions. I answered some of them and some of them I couldn't solve. I need your help please.
school (sch_code, sch_name, sch_phone, sch_dean)
advisor (adv_code, adv_fname, adv_lname, adv_phone, sch_code)
major (maj_code, maj_desc, sch_code)
maj_adv (maj_code, adv_code)
student (std_code     , std_lname, std_fname, std_gend, maj_code, std_dob)
grade (std_code,     gr_lname, gr_fname, gr_t1, gr_t2, gr_hw, gr_pr)
create table school
(sch_code varchar2(8) constraint school_sch_code_pk primary key,
sch_name varchar2(50),
sch_phone varchar2(12),
sch_dean varchar2(20));
insert into school values (‘BUS’, ‘School of Business’, ‘281-283-3100’, ‘Ted Cummings’);
insert into school values (‘EDU’, ‘School of Education’, ‘281-283-3600’,’Dennis Spuck’);
insert into school values (‘HSH’, ‘School of Humanities and Human Sciences’, ‘281-283-3333’, ‘Bruce Palmer’);
insert into school values (‘SCE’, ‘School of Science and Computer Engineering’, ‘281-283-3700’,’Sadegh Davari’);
create table advisor
(adv_code     varchar2(08)     constraint advisor_adv_code_pk primary key,
adv_lname     varchar2(15),
adv_fname     varchar2(15),
adv_phone     varchar2(12),
sch_code     varchar2(8) constraint advisor_sch_code_fk references school (sch_code));
insert into advisor values (‘A1’, ‘Porter’, ‘Mattie’, ‘281-283-3163’, ‘BUS’);
insert into advisor values ( ‘A2’, ‘Grady’, ‘Perdue’ ,‘281-283-3400’, ‘BUS’);
insert into advisor values (‘A3’, ’Tran’, ’Van’, ‘281-283-3203’, ‘BUS’);
insert into advisor values (‘A4’, ‘Saleem’, ‘Naveed’, ‘281-283-3202’, ‘BUS’);
insert into advisor values (‘A5’, ‘Kwok-Bon’, ‘Yue’, ‘281-283-3864’, ‘SCE’);
insert into advisor values (‘A6’, ‘Jones’, ‘Lisa’, ‘281-283-3551’, ‘EDU’);
insert into advisor values (‘A7’, ‘Palmer’, ‘Bruce’, ‘281-283-3445’, ‘HSH’);
create table major
(maj_code     varchar2(10)     constraint major_maj_code_pk primary key,
maj_desc      varchar2(30),
sch_code     varchar2(8)      constraint major_sch_code_fk references school(sch_code));
insert into major values (‘ACCT’, ‘Accounting’, ‘BUS’);
insert into major values (‘FINC’, ‘Finance’, ‘BUS’);
insert into major values (‘ISAM’, ‘Management Information Systems’, ‘BUS’);
insert into major values (‘CSCI’, ‘Computer Science’, ‘SCE’);
insert into major values (‘HIST’, ‘History’, ‘HSH’);
insert into major values (‘INST’, ‘Instructional Technology’, ‘EDU’);
create table maj_adv
(maj_code      varchar2(10)     constraint maj_adv_maj_code_fk references major (maj_code),
adv_code     varchar2(08)     constraint maj_adv_adv_code_fk references advisor (adv_code),
constraint maj_adv_maj_code_adv_code_cpk primary key (maj_code, adv_code));
insert into maj_adv values (‘ACCT’, ’A1’);     
insert into maj_adv values (‘ACCT’, ‘A2’);
insert into maj_adv values (‘FINC’, ’A2’);
insert into maj_adv values (‘ISAM’, ’A3’);
insert into maj_adv values (‘ISAM’, ’A4’);
insert into maj_adv values (‘CSCI’, ‘A5’);
insert into maj_adv values (‘INST’, ‘A6’);
insert into maj_adv values (‘HIST’, ‘A7’);
create table student
(std_code     varchar2(9),
std_lname     varchar2(15)     constraint student_std_lname_nn not null,
std_fname     varchar2(15)     constraint student_std_fname_nn not null,
std_gend     varchar2(8),     
maj_code     varchar2(10)     constraint student_maj_code1_fk references major (maj_code),
std_dob     date,
constraint student_std_code_pk primary key (std_code));
insert into student values (‘S1’, ‘Jordan’, ‘Michael’, ‘F’, ‘FINC’, to_date(‘10-Mar-1962’, 'DD-Mon-YYYY'));
insert into student values (‘S2’, ‘Barkley’, ‘Charles’, ‘M’, null, to_date(‘12-Sep-1964’, 'DD-Mon-YYYY'));
insert into student values (‘S3’, ‘Johnson’, ’Magic’, ‘M’, ‘ACCT’, to_date(‘13-Sep-1960’, 'DD-Mon-YYYY'));
insert into student values (‘S4’, ‘Williams’, ‘Serena’, ‘F’,‘ISAM’, to_date(‘23-Oct-1980’, 'DD-Mon-YYYY'));
insert into student values (‘S5’, ‘Duncan’, ‘Tim’, ‘M’, ‘ISAM’, to_date(‘07-Aug-1972’, 'DD-Mon-YYYY'));
insert into student values (‘S6’, ‘Graff’, ’Steffi’, ‘F’, ‘CSCI’, to_date(‘30-Apr-1962’, 'DD-Mon-YYYY'));
insert into student values (‘S7’, ‘Navratilova’, ’Martina’, ‘F’, ’ACCT’, to_date(‘18-Dec-1972’, 'DD-Mon-YYYY'));
REMARK insert into student (std_code, std_lname, std_fname, std_dob)
REMARK values (‘S2’, ‘Barkley’, ‘Charles’, to_date(‘12-Sep-1964’,'DD-Mon-YYYY'));
create table grade
(std_code     varchar2(9)     constraint grade_std_code_pk primary key
                    constraint grade_std_code_fk references student (std_code),
gr_lname     varchar2(15)     constraint grade_gr_lname_nn not null,
gr_fname     varchar2(15)     constraint grade_gr_fname_nn not null,
gr_t1          number(5),     constraint grade_gr_t1_cc check (gr_t1 between 0 and 100),
gr_t2          number(5),     
gr_hw          number(5),     
gr_pr          number(5));
insert into grade values (‘S1’, ‘Jordan’, ‘Michael’, 90, 80, 98, 90);
insert into grade values (‘S2’, ‘Barkley’, ‘Charles’, 60, 100, 100, 60);
insert into grade values (‘S3’, ‘Johnson’, ’Magic’, 88, 98, 96, 98);
insert into grade values (‘S4’, ‘Williams’, ‘Serena’, 92, 92, 92, 92);
insert into grade values (‘S5’, ‘Duncan’, ‘Tim’, 94, 90, 94, 96);
insert into grade values (‘S6’, ‘Graff’, ’Steffi’, 80, 84, 83, 72);
insert into grade values (‘S7’, ‘Navratilova’, ’Martina’, 91, 88, 94, 95);
the questions are:
Give the commands for the following:
1.     Display the names of students who don't have a major.
2.     Display the names of students whose last name has two "o" in it.
3.     Display the names of students whose first name includes an "e" in it.
4.     Display the names of students whose first name has exactly five characters in it.
5.     Display the names of students whose test 1 grade is higher than test 2 grade.
6.     Display the names of students whose project grade is between tests 1 and test 2 grades.
7.     Give the name of student whose last name starts with the character "B" or who never turned in homework.
8.     Give the command that will list two columns and one record. The first column will have the heading "Major" and "ACCT" as the value. The second column will have the heading “MIS_Count” and will display the number of accounting advisors.
9.     Display the name of the student whose test 1 score is higher than the average on test 1.
10.     Give the first and last names of students who scored above average on test 1 as well as above average on test 2.
11.     Display the system date variable using the dual table.
12.     Display the system date variable using the grade table making sure that the date is displayed only once.
13.     Give the name of the student who is the oldest student and scored the highest on the first test.
14.     List different majors and number of advisors for these majors.
15.     Display the names of students who don't have grade for any item.
16.     Display the names of the students who don't have grade for at least one item.
17.     Display names of students who have the same major as “Michael Jordan.”
18.     Give the names of students whose average on test 1 and test 2 is greater than the average on homework and project.
19.     Display the name and average on test 1 and test 2 for each student.
20.     Display the number of students in student table using three different methods.
21.     Display the number accounting students in the student table.
22.     Display the number of accounting and finance students in the student table.
23.     Display the average on test 1 and test 2 for female students.
24.     Display the number of accounting and finance students whose test 1 grades are higher than the average on test 1.
25.     Display the name of the major that has more than two students.
I am stuck with the following questions(1,12,13, 15,16,17,24) and the rest I finished them but I am not quite sure.
Thank you and looking forward to hear your suggestions.

Answer 1

Hi,
user11956564 wrote:
13.     Give the name of the student who is the oldest student and scored the highest on the first test.
I can get the oldest student and the highest score but in two different queries. My question is how can I use them together in one query
List oldest student in class
select std_fname, std_lname from student
where std_dob = (select min(std_dob) from student);
List highest grade in class
select gr_fname, gr_lname from grade
where gr_t1 = (select max(gr_t1) from grade);
because they are two tables. I think this is a confusing question: I'm not sure what the expected output is, or what the lesson you're supposed to learn is.
One way of combining results from different tables is UNION. As long as two queries produce the same number of columns, and the same datatypes, they can be combined with UNION.
For example: the following query displays the name of the employee whose job is 'PRESIDENT' (from the scott.emp table) and the name of department 10 (from the scott.dept table) in one query:
SELECT  ename
,       job
FROM  scott.emp
WHERE   job = 'PRESIDENT'
    UNION
SELECT  dname
,       TO_CHAR (deptno)
FROM    scott.dept
WHERE   deptno = 10
;Both SELECT clauses have 2 columns.
This 1st column in the 1st part is a VARCHAR2, so the 1st column of the 2nd part must also be a VARCHAR2. It is.
The 2nd column in the 1st part is a VARCHAR2, so the 2nd column of the 2nd part must also be a VARCHAR2. However, deptno (which we want to display here) is not a VARCHAR2, but a NUMBER; that's why I used TO_CHAR to put a VARCHAR2 value in that place.
17.     Display names of students who have the same major as “Michael Jordan.”
here is what I did
select std_fname, std_lname, maj_code
from student
where maj_code=(select maj_code
from studentJordan');
I am not sure if my answer is correct or not because in the output I got Michael JordanThis question is unclear, too.
Assuming you do not want Michael Jordan in the results, add a 2nd condition to the WHERE clause, so that you select
people who have the right major (which you say you are doing correctly, but the query seems to have lost something when you posted it)
AND
people who are not named Micheal Jordan
24.     Display the number of accounting and finance students whose test 1 grades are higher than the average on test 1.
It was too confusing for me.Are you sure?
Look at what you tried for 13.
select  std_fname, std_lname from student
where   std_dob = (select min(std_dob) from student);
select  gr_fname, gr_lname from grade
where   gr_t1 = (select max(gr_t1) from grade);The function for average (AVG) is used very much like the other aggregate function, like MIN and MAX.