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) ;

Тренд на препишани лекови по тримесечја за тековната година

select distinct l.ime_lek,
        case when recepta_quarter.trimesecje = 1 then recepti_za_lekovi.recepti_za_lek else 0 end as quarter_1_recepti_za_lek,
        case when recepta_quarter.trimesecje = 2 then recepti_za_lekovi.recepti_za_lek else 0 end as quarter_2_recepti_za_lek,
        case when recepta_quarter.trimesecje = 3 then recepti_za_lekovi.recepti_za_lek else 0 end as quarter_3_recepti_za_lek,
        case when recepta_quarter.trimesecje = 4 then recepti_za_lekovi.recepti_za_lek else 0 end as quarter_4_recepti_za_lek
from recepta r
join lekovi l on r.lek_id = l.lek_id
join (
        select r.recepta_id, extract (quarter from r.datum_izdavanje) as trimesecje
        from recepta r 
) recepta_quarter on r.recepta_id = recepta_quarter.recepta_id
join (
        select r.lek_id, count(r.recepta_id) as recepti_za_lek
        from recepta r
        where extract (year from r.datum_izdavanje) = extract (year from current_date)
        group by 1
) recepti_za_lekovi on r.lek_id = recepti_za_lekovi.lek_id
order by 2 desc, 3 desc, 4 desc, 5 desc ;

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 ;