Hand in: the database picture (png or jpeg or ) for question 2, the queries (as SQL statements) for question 4

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Bertina O’Brien’
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1 Using PostgreSQL, USDA database: 1. Create the version of the USDA database that follows on subsequent pages. Note this has been taken from a site that lists this database as a PostgreSQL sample database. 2. Develop a database picture similar to Figure 5.7 that shows tables, primary keys, and foreign keys (and references to PKs). Include at least one non-key attribute from each table. You do not have to include all attributes, but PKs and FKs are a must. Using PostgreSQL, Company database: 3. Create database, insert data 4. Queries see last page Hand in: the database picture (png or jpeg or ) for question 2, the queries (as SQL statements) for question 4 via to 3902@acs.uwinnipeg.ca 1

2 USDA Food Database R18 Name: data_src; Type: TABLE; Schema: public; Owner: -; Tablespace: CREATE TABLE data_src ( datasrc_id character(6) NOT NULL, authors text, title text NOT NULL, "year" integer, journal text, vol_city text, issue_state text, start_page text, end_page text Name: datsrcln; Type: TABLE; Schema: public; Owner: -; Tablespace: CREATE TABLE datsrcln ( ndb_no character(5) NOT NULL, nutr_no character(3) NOT NULL, datasrc_id character(6) NOT NULL Name: deriv_cd; Type: TABLE; Schema: public; Owner: -; Tablespace: CREATE TABLE deriv_cd ( deriv_cd text NOT NULL, derivcd_desc text NOT NULL 2

3 Name: fd_group; Type: TABLE; Schema: public; Owner: -; Tablespace: CREATE TABLE fd_group ( fdgrp_cd character(4) NOT NULL, fddrp_desc text NOT NULL Name: food_des; Type: TABLE; Schema: public; Owner: -; Tablespace: CREATE TABLE food_des ( ndb_no character(5) NOT NULL, fdgrp_cd character(4) NOT NULL, long_desc text NOT NULL, shrt_desc text NOT NULL, comname text, manufacname text, survey character(1), ref_desc text, refuse integer, sciname text, n_factor double precision, pro_factor double precision, fat_factor double precision, cho_factor double precision Name: footnote; Type: TABLE; Schema: public; Owner: -; Tablespace: 3 CREATE TABLE footnote ( ndb_no character(5) NOT NULL, footnt_no character(4) NOT NULL,

4 footnt_typ character(1) NOT NULL, nutr_no character(3), footnt_txt text NOT NULL Name: nut_data; Type: TABLE; Schema: public; Owner: -; Tablespace: CREATE TABLE nut_data ( ndb_no character(5) NOT NULL, nutr_no character(3) NOT NULL, nutr_val double precision NOT NULL, num_data_pts double precision NOT NULL, std_error double precision, src_cd integer NOT NULL, deriv_cd text, ref_ndb_no character(5), add_nutr_mark character(1), num_studies integer, min double precision, max double precision, df integer, low_eb double precision, up_eb double precision, stat_cmt text, cc character(1) Name: nutr_def; Type: TABLE; Schema: public; Owner: -; Tablespace: 4 CREATE TABLE nutr_def ( nutr_no character(3) NOT NULL, units text NOT NULL, tagname text, nutrdesc text,

5 num_dec smallint, sr_order integer Name: src_cd; Type: TABLE; Schema: public; Owner: -; Tablespace: CREATE TABLE src_cd ( src_cd integer NOT NULL, srccd_desc text NOT NULL Name: weight; Type: TABLE; Schema: public; Owner: -; Tablespace: CREATE TABLE weight ( ndb_no character(5) NOT NULL, seq character(2) NOT NULL, amount double precision NOT NULL, msre_desc text NOT NULL, gm_wgt double precision NOT NULL, num_data_pts integer, std_dev double precision Name: data_src_pkey; Type: CONSTRAINT; Schema: public; Owner: -; Tablespace: ALTER TABLE ONLY data_src ADD CONSTRAINT data_src_pkey PRIMARY KEY (datasrc_id 5

6 Name: datsrcln_pkey; Type: CONSTRAINT; Schema: public; Owner: -; Tablespace: ALTER TABLE ONLY datsrcln ADD CONSTRAINT datsrcln_pkey PRIMARY KEY (ndb_no, nutr_no, datasrc_id Name: deriv_cd_pkey; Type: CONSTRAINT; Schema: public; Owner: -; Tablespace: ALTER TABLE ONLY deriv_cd ADD CONSTRAINT deriv_cd_pkey PRIMARY KEY (deriv_cd Name: fd_group_pkey; Type: CONSTRAINT; Schema: public; Owner: -; Tablespace: ALTER TABLE ONLY fd_group ADD CONSTRAINT fd_group_pkey PRIMARY KEY (fdgrp_cd Name: food_des_pkey; Type: CONSTRAINT; Schema: public; Owner: -; Tablespace: ALTER TABLE ONLY food_des ADD CONSTRAINT food_des_pkey PRIMARY KEY (ndb_no Name: nut_data_pkey; Type: CONSTRAINT; Schema: public; Owner: -; Tablespace: ALTER TABLE ONLY nut_data ADD CONSTRAINT nut_data_pkey PRIMARY KEY (ndb_no, nutr_no 6

7 Name: nutr_def_pkey; Type: CONSTRAINT; Schema: public; Owner: -; Tablespace: ALTER TABLE ONLY nutr_def ADD CONSTRAINT nutr_def_pkey PRIMARY KEY (nutr_no Name: src_cd_pkey; Type: CONSTRAINT; Schema: public; Owner: -; Tablespace: ALTER TABLE ONLY src_cd ADD CONSTRAINT src_cd_pkey PRIMARY KEY (src_cd Name: weight_pkey; Type: CONSTRAINT; Schema: public; Owner: -; Tablespace: ALTER TABLE ONLY weight ADD CONSTRAINT weight_pkey PRIMARY KEY (ndb_no, seq Name: datsrcln_datasrc_id_idx; Type: INDEX; Schema: public; Owner: -; Tablespace: CREATE INDEX datsrcln_datasrc_id_idx ON datsrcln USING btree (datasrc_id Name: food_des_fdgrp_cd_idx; Type: INDEX; Schema: public; Owner: -; Tablespace: CREATE INDEX food_des_fdgrp_cd_idx ON food_des USING btree (fdgrp_cd Name: footnote_ndb_no_idx; Type: INDEX; Schema: public; Owner: -; Tablespace: 7

8 CREATE INDEX footnote_ndb_no_idx ON footnote USING btree (ndb_no, nutr_no Name: nut_data_deriv_cd_idx; Type: INDEX; Schema: public; Owner: -; Tablespace: CREATE INDEX nut_data_deriv_cd_idx ON nut_data USING btree (deriv_cd Name: nut_data_nutr_no_idx; Type: INDEX; Schema: public; Owner: -; Tablespace: CREATE INDEX nut_data_nutr_no_idx ON nut_data USING btree (nutr_no Name: nut_data_src_cd_idx; Type: INDEX; Schema: public; Owner: -; Tablespace: CREATE INDEX nut_data_src_cd_idx ON nut_data USING btree (src_cd Name: datsrcln_datasrc_id_fkey; Type: FK CONSTRAINT; Schema: public; Owner: - ALTER TABLE ONLY datsrcln ADD CONSTRAINT datsrcln_datasrc_id_fkey FOREIGN KEY (datasrc_id) REFERENCES data_src(datasrc_id Name: datsrcln_ndb_no_fkey; Type: FK CONSTRAINT; Schema: public; Owner: - ALTER TABLE ONLY datsrcln 8

9 ADD CONSTRAINT datsrcln_ndb_no_fkey FOREIGN KEY (ndb_no, nutr_no) REFERENCES nut_data(ndb_no, nutr_no Name: food_des_fdgrp_cd_fkey; Type: FK CONSTRAINT; Schema: public; Owner: - ALTER TABLE ONLY food_des ADD CONSTRAINT food_des_fdgrp_cd_fkey FOREIGN KEY (fdgrp_cd) REFERENCES fd_group(fdgrp_cd Name: footnote_ndb_no_fkey; Type: FK CONSTRAINT; Schema: public; Owner: - ALTER TABLE ONLY footnote ADD CONSTRAINT footnote_ndb_no_fkey FOREIGN KEY (ndb_no) REFERENCES food_des(ndb_no Name: footnote_nutr_no_fkey; Type: FK CONSTRAINT; Schema: public; Owner: - ALTER TABLE ONLY footnote ADD CONSTRAINT footnote_nutr_no_fkey FOREIGN KEY (nutr_no) REFERENCES nutr_def(nutr_no Name: nut_data_deriv_cd_fkey; Type: FK CONSTRAINT; Schema: public; Owner: - ALTER TABLE ONLY nut_data ADD CONSTRAINT nut_data_deriv_cd_fkey FOREIGN KEY (deriv_cd) REFERENCES deriv_cd(deriv_cd 9

10 Name: nut_data_ndb_no_fkey; Type: FK CONSTRAINT; Schema: public; Owner: - ALTER TABLE ONLY nut_data ADD CONSTRAINT nut_data_ndb_no_fkey FOREIGN KEY (ndb_no) REFERENCES food_des(ndb_no Name: nut_data_nutr_no_fkey; Type: FK CONSTRAINT; Schema: public; Owner: - ALTER TABLE ONLY nut_data ADD CONSTRAINT nut_data_nutr_no_fkey FOREIGN KEY (nutr_no) REFERENCES nutr_def(nutr_no Name: nut_data_src_cd_fkey; Type: FK CONSTRAINT; Schema: public; Owner: - ALTER TABLE ONLY nut_data ADD CONSTRAINT nut_data_src_cd_fkey FOREIGN KEY (src_cd) REFERENCES src_cd(src_cd Name: weight_ndb_no_fkey; Type: FK CONSTRAINT; Schema: public; Owner: - ALTER TABLE ONLY weight ADD CONSTRAINT weight_ndb_no_fkey FOREIGN KEY (ndb_no) REFERENCES food_des(ndb_no 10

11 11

For each employee list their last name and the number of people they supervise. All employees must be listed (even those who do not supervise anyone).

12 Queries to develop for the Company database 1. Retrieve the last name and address of all employees who earn more than the average salary. Present the results in sequence by last name. SELECT lname, address FROM public.employee where salary > (select avg(salary) from employee) order by lname; 2. For each employee list their last name and the number of people they supervise. All employees must be listed (even those who do not supervise anyone). List the employees in sequence by the number of employees supervised. SELECT s.lname, count(e.ssn) FROM public.employee s left outer join employee e on (s.ssn = e.super_ssn) group by s.ssn, s.lname order by count(e.ssn 12

3. List all first names (fname in employee, dependent_name in dependent) with no duplicates. There is one column in the result list. SELECT fname FROM public.

13 3. List all first names (fname in employee, dependent_name in dependent) with no duplicates. There is one column in the result list. SELECT fname FROM public.employee union select dependent_name from dependent; 4. List the last name (no duplicates) of employees who work on both project 1 and project 2. SELECT distinct lname from employee where ssn in (select essn FROM works_on where pno = 1) and ssn in (select essn from works_on where pno = 2 13

5. Retrieve the last name of each employee who has a spouse. List the employees in sequence by last name. Do this in 3 different ways: a. Use the IN operator.

14 5. Retrieve the last name of each employee who has a spouse. List the employees in sequence by last name. Do this in 3 different ways: a. Use the IN operator. SELECT lname from employee where ssn in (select essn from dependent where relationship = 'Spouse') order by lname; b. Use a correlated subquery. SELECT lname from employee e where exists (select * from dependent d where e.ssn = d.essn and relationship = 'Spouse') order by lname; c. Use the EXISTS operator. SELECT lname from employee e where exists (select * from dependent d where e.ssn = d.essn and relationship = 'Spouse') order by lname; 6. List the names of projects that have 3 or more employees working on them. List the projects alphabetically. SELECT pname FROM project p inner join works_on w on (pnumber=pno) group by pnumber, pname having count(*)>=3 order by pname; 14

7. For each employee, who has a dependent, list their last name

Display the results in sequence by dependent s name.

15 7. For each employee, who has a dependent, list their last name and the name of their oldest dependent. Display the results in sequence by dependent s name. SELECT lname, dependent_name FROM employee e inner join dependent d on (ssn = essn) where d.bdate = (select min(bdate) from dependent y where e.ssn=y.essn Results: Employee last names and children in sequence oldest to youngest 15

bdate) group by e.ssn, e.lname, e.bdate order by e.bdate; 9.

16 8. For each employee list their last name and the number of other employees who are older. All employees must be listed, even the oldest. Display the results in sequence by employee s birth date. SELECT e.lname, count(older.ssn) FROM employee e left join employee older on (e.bdate>older.bdate) group by e.ssn, e.lname, e.bdate order by e.bdate; 9. For each project list the project name and the total number of hours employees have contributed. The list must be sequenced by the total hours. SELECT pname, sum(hours) FROM project left join works_on on (pnumber = pno) group by pnumber, pname order by sum(hours) 16

10. For each employee list their last name, the number of dependents they have, the name of the department they work in, where the department is located, and the last name of their supervisor.

17 10. For each employee list their last name, the number of dependents they have, the name of the department they work in, where the department is located, and the last name of their supervisor. All employees must be listed. The results must be in sequence by department name and then by employee last name. select e.lname, count(d.essn), dname, s.lname from (employee e left join dependent d on (e.ssn = d.essn)) inner join department dp on (dno=dp.dnumber) left join employee s on (s.ssn = e.super_ssn) group by e.ssn, e.lname, dname, s.lname order by dname, e.lname 17

Database design process

Database technology Lecture 2: Relational databases and SQL Jose M. Peña jose.m.pena@liu.se Database design process 1 Relational model concepts... Attributes... EMPLOYEE FNAME M LNAME SSN BDATE ADDRESS