| Version 15 (modified by , 4 years ago) ( diff ) |
|---|
Напредни извештаи од базата (SQL и складирани процедури)
Број на извршени прегледи од секој доктор, број на издадени упати од секој доктор и број на закажани термини кај секој доктор, за секое тримесечје, за тековната година
select d2.doktor_id, c.ime, c.prezime, b.naziv as bolnica, o.naziv as oddel, b.grad,
case when pregledi_quarter.trimesecje = 1 then pregledi_quarter.pregledi else 0 end as quarter_1_pregledi,
case when pregledi_quarter.trimesecje = 2 then pregledi_quarter.pregledi else 0 end as quarter_2_pregledi,
case when pregledi_quarter.trimesecje = 3 then pregledi_quarter.pregledi else 0 end as quarter_3_pregledi,
case when pregledi_quarter.trimesecje = 4 then pregledi_quarter.pregledi else 0 end as quarter_4_pregledi,
case when zakazani_quarter.trimesecje = 1 then zakazani_quarter.zakazani_termini else 0 end as quarter_1_zakazani_termini,
case when zakazani_quarter.trimesecje = 2 then zakazani_quarter.zakazani_termini else 0 end as quarter_2_zakazani_termini,
case when zakazani_quarter.trimesecje = 3 then zakazani_quarter.zakazani_termini else 0 end as quarter_3_zakazani_termini,
case when zakazani_quarter.trimesecje = 4 then zakazani_quarter.zakazani_termini else 0 end as quarter_4_zakazani_termini,
case when upati_quarter.trimesecje = 1 then upati_quarter.upati_od_pregled else 0 end as quarter_1_upati_od_pregled,
case when upati_quarter.trimesecje = 2 then upati_quarter.upati_od_pregled else 0 end as quarter_2_upati_od_pregled,
case when upati_quarter.trimesecje = 3 then upati_quarter.upati_od_pregled else 0 end as quarter_3_upati_od_pregled,
case when upati_quarter.trimesecje = 4 then upati_quarter.upati_od_pregled else 0 end as quarter_4_upati_od_pregled
from bolnica b
join oddel o on b.bolnica_id = o.bolnica_id
join doktor d2 on (o.oddel_id, o.bolnica_id) = (d2.oddel_id, d2.bolnica_id)
join covek c on d2.doktor_id = c.covek_id
left join (
select extract (quarter from p.vreme) as trimesecje, count (p.pregled_id) as pregledi,
d.doktor_id , o.naziv as oddel, b.naziv as bolnica
from pregled p
join doktor d on p.doktor_id = d.doktor_id
join oddel o on (d.oddel_id, d.bolnica_id) = (o.oddel_id, o.bolnica_id)
join bolnica b on o.bolnica_id = b.bolnica_id
where extract (year from p.vreme) = extract (year from current_date)
group by 1, 3, 4, 5
) pregledi_quarter on d2.doktor_id = pregledi_quarter.doktor_id
left join (
select r.doktor_id, count(r.rezervacija_id) zakazani_termini,
extract (quarter from t.vreme) as trimesecje
from rezervacija r
join termin t on (r.termin_id, r.doktor_id) = (t.termin_id, t.doktor_id)
where extract (year from t.vreme) = extract (year from current_date)
group by 1, 3
) zakazani_quarter on d2.doktor_id = zakazani_quarter.doktor_id
left join (
select p.doktor_id , count (u.upat_id) as upati_od_pregled,
extract (quarter from p.vreme) as trimesecje
from pregled p
join upat u on p.pregled_id = u.pregled_id
where extract (year from p.vreme) = extract (year from current_date)
group by 1, 3
) upati_quarter on d2.doktor_id = upati_quarter.doktor_id ;
Број на пациенти кои биле на преглед во даден оддел и број на упати кои биле издадени за во оддел после извршен преглед, за секое тримесечје, за тековната година
select distinct o.naziv as oddel, b.naziv as bolnica,
case when pregled_quarter.trimesecje = 1 then pacienti_quarter.pacienti else 0 end as quarter_1_pacienti,
case when pregled_quarter.trimesecje = 2 then pacienti_quarter.pacienti else 0 end as quarter_2_pacienti,
case when pregled_quarter.trimesecje = 3 then pacienti_quarter.pacienti else 0 end as quarter_3_pacienti,
case when pregled_quarter.trimesecje = 4 then pacienti_quarter.pacienti else 0 end as quarter_4_pacienti,
case when pregled_quarter.trimesecje = 1 then upati_quarter.upati else 0 end as quarter_1_upati,
case when pregled_quarter.trimesecje = 2 then upati_quarter.upati else 0 end as quarter_2_upati,
case when pregled_quarter.trimesecje = 3 then upati_quarter.upati else 0 end as quarter_3_upati,
case when pregled_quarter.trimesecje = 4 then upati_quarter.upati else 0 end as quarter_4_upati
from oddel o
join bolnica b on o.bolnica_id = b.bolnica_id
join (
select o.oddel_id, o.bolnica_id, extract (quarter from p.vreme) as trimesecje
from pregled p
join doktor d on p.doktor_id = d.doktor_id
join oddel o on (d.oddel_id, d.bolnica_id) = (o.oddel_id, o.bolnica_id)
) pregled_quarter on (o.oddel_id, o.bolnica_id) = (pregled_quarter.oddel_id, pregled_quarter.bolnica_id)
left join (
select o.oddel_id, o.bolnica_id, count (p.covek_pacient_id) as pacienti
from pregled p
join doktor d on p.doktor_id = d.doktor_id
join oddel o on (d.oddel_id, d.bolnica_id) = (o.oddel_id, o.bolnica_id)
join bolnica b on o.bolnica_id = b.bolnica_id
where extract (year from p.vreme) = extract (year from current_date)
group by 1, 2
) pacienti_quarter on (o.oddel_id, o.bolnica_id) = (pacienti_quarter.oddel_id, pacienti_quarter.bolnica_id)
left join (
select u.oddel_id, u.bolnica_id, count(u.upat_id) as upati
from upat u
join pregled p on u.pregled_id = p.pregled_id
where extract (year from p.vreme) = extract (year from current_date)
group by 1, 2
) upati_quarter on (o.oddel_id, o.bolnica_id) = (upati_quarter.oddel_id, upati_quarter.bolnica_id)
Пациент на кој му се извршени највеќе прегледи
create view pacient_najmnogu_pregledi as select pacient_id ,covek.ime,covek.prezime, count (pregled.pregled_id) pacientot_bil_na_pregled from pacient as p join pregled as pregled on pregled.covek_pacient_id=p.covek_pacient_id join covek as covek on covek.covek_id=p.covek_pacient_id group by p.pacient_id,covek.covek_id; select pnp.ime, pnp.prezime ,pnp.pacient_id , pnp.pacientot_bil_na_pregled as broj_pregledi from pacient_najmnogu_pregledi as pnp where pnp.pacientot_bil_na_pregled= (select max(pacientot_bil_na_pregled) from pacient_najmnogu_pregledi);
Број на издадени упати за оддел
create view broj_na_upati as select o.oddel_id ,o.naziv , count(u.oddel_id) as broj_upati from oddel as o join upat as u on u.oddel_id=o.oddel_id group by o.naziv ,o.oddel_id select * from broj_na_upati - --пребарување за сите оддели. select * from broj_na_upati where naziv='Kardiologija' - -- пребарување според назив на одделот.
Број на издадени рецепти за пациентите
create view broj_recepti as select covek_pacient_id ,c.ime ,c.prezime ,count(recepta_id) as broj_recepti from recepta as r join covek as c on c.covek_id=r.covek_pacient_id group by recepta_id ,c.covek_id select covek_pacient_id ,ime ,prezime ,broj_recepti from broj_recepti as br -- за сите пациенти. select covek_pacient_id ,ime ,prezime ,broj_recepti from broj_recepti as br where broj_recepti=(select max(broj_recepti) from broj_recepti) --пациент со најголем број на издадени рецепти. select covek_pacient_id ,ime ,prezime ,broj_recepti from broj_recepti as br where covek_pacient_id=9 --може да се пребарува според пациент.
Доктор кој извршил највеќе прегледи
create view br_pregledi_za_sekoj_doktor as
select p.doktor_id, d.br_licenca, count(p.pregled_id) br_pregledi_od_doktor
from pregled p
join doktor d on p.doktor_id = d.doktor_id
group by p.doktor_id, d.doktor_id ;
select br_p.doktor_id, br_p.br_licenca, c.ime, c.prezime, br_p.br_pregledi_od_doktor as najvekje_pregledi
from br_pregledi_za_sekoj_doktor br_p
join covek c on br_p.doktor_id = c.covek_id
where br_p.br_pregledi_od_doktor = (
select max(br_pregledi_od_doktor) from br_pregledi_za_sekoj_doktor
) ;
Просечна возраст на пациенти за во даден оддел во болница за претходната година
select oo.bolnica_id, bb.naziv as bolnica, oo.oddel_id, oo.naziv as oddel,
(
select avg(distinct extract (year from age(((case when substring(c.embg, 5, 1) in ('9')
then '1'|| substring(c.embg, 5, 3) else '2'|| substring(c.embg, 5, 3) end)
|| '-' || substring(c.embg, 3, 2) || '-' || substring(c.embg, 1, 2))::date))) as godini
from oddel o
join bolnica b on o.bolnica_id = b.bolnica_id
join upat u on (o.bolnica_id, o.oddel_id) = (u.bolnica_id, u.oddel_id)
join rezervacija r on u.upat_id = r.upat_id
join termin t on (r.doktor_id, r.termin_id) = (t.doktor_id, t.termin_id)
join pacient p on u.covek_pacient_id = p.covek_pacient_id
join covek c on p.covek_pacient_id = c.covek_id
where extract (year from t.vreme) = extract (year from now() - interval '1 year')
group by o.bolnica_id, o.oddel_id
having (o.bolnica_id, o.oddel_id) = (oo.bolnica_id, oo.oddel_id)
) prosek_godini_pacienti
from oddel oo
join bolnica bb on oo.bolnica_id = bb.bolnica_id ;
Број на доктори во секој оддел
select o.bolnica_id, b.naziv as bolnica, b.grad as bolnica_grad, o.naziv as oddel, count (d.doktor_id) broj_doktori_vo_oddel from oddel o join doktor d on (o.bolnica_id, o.oddel_id) = (d.bolnica_id, d.oddel_id) join bolnica b on o.bolnica_id = b.bolnica_id group by (o.bolnica_id, o.oddel_id), b.naziv, b.grad ;
Note:
See TracWiki
for help on using the wiki.
