== 1.Извештај за првите 3 валидни прашања кои имаат барем два одговори од професор, од тие три се бира прашањето кое има најмногу одговори во последните 4 месеци \\ {{{ select q3.maximum, p3.* from ( select max(q2.count) as maximum, q2.pra_id from( select distinct q.count, q.pra_id as pra_id from( select count(distinct o.o_id), p.pra_id as pra_id from prasanja p join odgovori o on p.pra_id = o.pra_id where o.p_id is not null and p.pra_datum between now()-interval '4 months' and now() group by p.pra_id ) as q join prasanja p on q.pra_id = p.pra_id join odgovori o on o.pra_id = p.pra_id where q.count >=2 and o.o_validen is true ) as q2 join prasanja p2 on p2.pra_id = q2.pra_id group by q2.pra_id order by maximum desc limit 1 ) as q3 join prasanja p3 on q3.pra_id = p3.pra_id }}} == 2. Извештај за првите три прашања со најмногу одговори од студенти и професори посебно, потоа од тие прашања ја пресметуваме вкупната сума на реакции на одговорите(заедно од професори и студенти) и ги филтритаме само оние кои имаат просечен или надпросечен број на реакции {{{ create view broj_reakcii_na_izbrani_odgovori as ( select sum(q4.br_reakcii) as br_reakcii_total, q4.o_id, q4.pra_id from ( select count(dr.reakcija) as br_reakcii, o.o_id , o.pra_id from ( select * from ( select p.pra_id as prasanje from prasanja p join odgovori o on o.pra_id =p .pra_id where o.s_id is not null group by p.pra_id order by count(o.o_id) desc limit 3 ) as q1 union select q2.prasanje from ( select p.pra_id as prasanje from prasanja p join odgovori o on o.pra_id =p .pra_id where o.p_id is not null group by p.pra_id order by count(o.o_id) desc limit 3 ) as q2 ) as q3 join prasanja p2 on q3.prasanje = p2.pra_id join odgovori o on o.pra_id =p2.pra_id join dava_reakcija dr on o.o_id = dr.o_id group by o.o_id, o.pra_id union select count(rn.reakcija) as br_reakcii, o.o_id , o.pra_id from ( select * from ( select p.pra_id as prasanje from prasanja p join odgovori o on o.pra_id =p .pra_id where o.s_id is not null group by p.pra_id order by count(o.o_id) desc limit 3 ) as q1 union select q2.prasanje from ( select p.pra_id as prasanje from prasanja p join odgovori o on o.pra_id =p .pra_id where o.p_id is not null group by p.pra_id order by count(o.o_id) desc limit 3 ) as q2 ) as q3 join prasanja p2 on q3.prasanje = p2.pra_id join odgovori o on o.pra_id =p2.pra_id join reagira_na rn on o.o_id = rn.o_id group by o.o_id, o.pra_id ) as q4 group by q4.o_id , q4.pra_id ); create view broj_reakcii_na_odgovor as ( select sum(q4.br_reakcii) as br_reakcii_total, q4.o_id, q4.pra_id from ( select count(dr.reakcija) as br_reakcii, o.o_id , o.pra_id from ( select q1.prasanje from ( select p.pra_id as prasanje from prasanja p join odgovori o on o.pra_id =p .pra_id group by p.pra_id ) as q1 union select q2.prasanje from ( select p.pra_id as prasanje from prasanja p join odgovori o on o.pra_id =p .pra_id group by p.pra_id ) as q2 ) as q3 join prasanja p2 on q3.prasanje = p2.pra_id join odgovori o on o.pra_id =p2.pra_id join dava_reakcija dr on o.o_id = dr.o_id group by o.o_id, o.pra_id union select count(rn.reakcija) as br_reakcii, o.o_id , o.pra_id from ( select q1.prasanje from ( select p.pra_id as prasanje from prasanja p join odgovori o on o.pra_id =p .pra_id group by p.pra_id ) as q1 union select q2.prasanje from ( select p.pra_id as prasanje from prasanja p join odgovori o on o.pra_id =p .pra_id group by p.pra_id ) as q2 ) as q3 join prasanja p2 on q3.prasanje = p2.pra_id join odgovori o on o.pra_id =p2.pra_id join reagira_na rn on o.o_id = rn.o_id group by o.o_id, o.pra_id ) as q4 group by q4.o_id , q4.pra_id ); select brnio.br_reakcii_total, p.* from broj_reakcii_na_izbrani_odgovori brnio join prasanja p on brnio.pra_id = p.pra_id where brnio.br_reakcii_total >= (select avg(brno.br_reakcii_total) from broj_reakcii_na_odgovor brno) }}} == 3.Извештај за пронаоѓање на најдобрите 5 професори т.е. професорите кои имаат одговори со најдобри рејтингзи, имаат поставено најмногу материјали и предаваат најмногу курсеви.\\ {{{ select parts.p_korisnicko_ime korisnicko_ime, parts.rejt+parts.mat+parts.kurs as vk from ( select prof.p_korisnicko_ime, case when top_odg.num_rejting is null then 0 else top_odg.num_rejting end as rejt, case when najm_mat.num_mat is null then 0 else najm_mat.num_mat end as mat, case when najm_kurs.num_kurs is null then 0 else najm_kurs.num_kurs end as kurs from profesori prof left join ( select o.p_id, q1.c1+q2.c2 as num_rejting from odgovori o join ( select o.o_id, count(o.o_id) c1 from odgovori o left join dava_reakcija dr on o.o_id = dr.o_id where o.p_id is not null and dr.reakcija is true group by o.o_id ) q1 on o.o_id = q1.o_id join ( select o.o_id, count(o.o_id) c2 from odgovori o left join reagira_na rn on o.o_id = rn.o_id where o.p_id is not null and rn.reakcija is true group by o.o_id ) q2 on o.o_id = q2.o_id order by num_rejting desc ) top_odg on prof.p_id = top_odg.p_id left join ( select p.p_id, count(m.m_id) num_mat from materijali m join profesori p on m.p_id = p.p_id group by p.p_id ) najm_mat on prof.p_id =najm_mat.p_id left join ( select p.p_id, count(epn.p_id) num_kurs from kursevi ku join e_prof_na epn on ku.ku_id=epn.ku_id join profesori p on epn.p_id =p.p_id group by p.p_id ) najm_kurs on prof.p_id = najm_kurs.p_id ) as parts order by vk desc limit 5 }}} == 4. Извештај за секој предмет најпопуларните 3 курсеви во последните 3 месеци кои имаат барем по 2 исти категории за материјали и прашања\\ {{{ select pr.pr_id, pr.pr_ime, q1.ku_ime, q1.br_isti_kategorii, count(distinct q2.pra_id) as br_prasanja from predmeti pr join kursevi ku on pr.pr_id = ku.pr_id join ( select ku.ku_id, ku.ku_ime, count(*) br_isti_kategorii from kursevi ku left join prasanja pra on ku.ku_id = pra.ku_id left join e_oznaceno_so eos on pra.pra_id = eos.pra_id left join materijali m on ku.ku_id = m.ku_id where eos.ka_id = m.ka_id group by ku.ku_id, ku.ku_ime having count(*)>=2 order by br_isti_kategorii desc limit 3 ) q1 on ku.ku_id = q1.ku_id join ( select pras.ku_id as ku_id, pras.pra_id from prasanja pras where now()>=pras.pra_datum and age(now(),pras.pra_datum)<= interval'3 months' ) q2 on ku.ku_id=q2.ku_id group by pr.pr_id, pr.pr_ime,q1.ku_ime, q1.br_isti_kategorii order by br_prasanja desc }}}