Version 4 (modified by 3 years ago) ( diff ) | ,
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Напредни извештаи од базата (SQL и складирани процедури)
Извештаи за клиент според тромесечие, купени производи и потрошена сума
create view podatoci_klienti as SELECT DISTINCT u.user_name, u.surname, CASE WHEN tabela.quarter=1 THEN tabela.kupeni_proizvodi ELSE 0 END AS prv_kvartal, CASE WHEN tabela.quarter=2 THEN tabela.kupeni_proizvodi ELSE 0 END AS vtor_kvartal, CASE WHEN tabela.quarter=3 THEN tabela.kupeni_proizvodi ELSE 0 END AS tret_kvartal, CASE WHEN tabela.quarter=4 THEN tabela.kupeni_proizvodi ELSE 0 END AS cetvrt_kvartal, CASE WHEN tabela.quarter=1 THEN tabela.suma ELSE 0 END AS prv_kvartal_suma, CASE WHEN tabela.quarter=1 THEN tabela.suma ELSE 0 END AS vtor_kvartal_suma, CASE WHEN tabela.quarter=1 THEN tabela.suma ELSE 0 END AS tret_kvartal_suma, CASE WHEN tabela.quarter=1 THEN tabela.suma ELSE 0 END AS cetvrt_kvartal_suma FROM shopping_bag AS sb JOIN client AS c ON sb.user_id = c.user_id JOIN users AS u ON c.user_id = u.user_id JOIN orders AS o ON sb.shopping_id = o.shopping_id JOIN product_in_store AS pis ON o.product_in_store_id = pis.product_in_store_id JOIN price AS pr ON pis.product_in_store_id = pr.product_in_store_id JOIN ( SELECT DISTINCT c2.user_id, extract(quarter FROM sb2.order_date) AS quarter, sum(pr.price) AS suma, count(sb2.shopping_id) AS kupeni_proizvodi FROM shopping_bag AS sb2 JOIN orders AS o ON sb2.shopping_id = o.shopping_id JOIN product_in_store AS pis ON o.product_in_store_id = pis.product_in_store_id join price as pr on pis.product_in_store_id=pr.product_in_store_id JOIN client AS c2 ON c2.user_id = sb2.user_id GROUP BY 1,2 ) AS tabela ON c.user_id = tabela.user_id
Извештаи за секоја продавница, бројот на производи во неа, вкупната вредност на производите, просечна цена по производ и вкупно направени нарачки во последните 3 месеци
create view podatoci_prodavnici as select os.names, os.store_id, coalesce(( select count(pis.product_in_store_id) from online_stores as os join product_in_store as pis on os.store_id=pis.store_id ),0) as vkupno_proizvodi, ( select sum(p.price) from online_stores as os join product_in_store as pis on os.store_id=pis.store_id join price as p on pis.product_in_store_id=p.product_in_store_id ) as vkupna_vrednost, ( select avg(p.price) from online_stores as os join product_in_store as pis on os.store_id=pis.store_id join price as p on pis.product_in_store_id=p.product_in_store_id ) as prosek_vrednost, ( select count(sb.shopping_id) from online_stores as os join product_in_store as pis on os.store_id=pis.product_id join orders as ord on pis.product_in_store_id=ord.product_in_store_id join shopping_bag as sb on ord.shopping_id=sb.shopping_id where sb.order_date between now() - interval '3 months' and now() ) as vkupno_naracki from online_stores as os group by 1, 2
Вработен кој извршил достава на најмногу нарачки
create view vkupen_broj_naracki as select se.shipping_employee_id, u.user_name, u.surname, count(s.shipping_id) as broj_naracki from shipping_employee as se join users as u on se.user_id=u.user_id join shipping as s on se.user_id=s.user_id join shopping_bag as sb on s.shopping_id=sb.shopping_id group by 1, 2, 3; select vbn.shipping_employee_id, vbn.user_name, vbn.surname, vbn.broj_naracki as najmnogu_naracki from vkupen_broj_naracki as vbn where vbn.broj_naracki = (select max(broj_naracki) from vkupen_broj_naracki);
Вработените од сите продавници со искуство од 6 или повеќе години и нивната улога во продавницата
create view broj_vraboteni as select os.store_id, os.names, u.user_name, u.surname, wa.works_from, r.role_name, r.role_desc from online_stores as os join store_employee as se on os.store_id=se.store_id join users as u on se.user_id=u.user_id join works_as as wa on se.user_id=wa.user_id join roles as r on wa.id_role=r.id_role where extract(year from now()) - extract(year from wa.works_from) >= 6
Листање на сите продукти според достапност, каталог и продавница, и прикажување на продукт со најниска цена
create view produkti_cena as select p.names, p.in_store, pis.product_in_store_id, c.catalogue_id, c.type_, os.web_address, pr.price from product as p join product_in_store as pis on p.product_id=pis.product_id join existss as e on pis.product_in_store_id=e.product_in_store_id join catalogue as c on e.catalogue_id=c.catalogue_id join online_stores as os on e.store_id=os.store_id join price as pr on pis.product_in_store_id=pr.product_in_store_id group by 1, 2, 3, 4, 5, 6, 7; select pc.names, pc.in_store, pc.product_in_store_id, pc.catalogue_id, pc.type_, pc.web_address, pc.price as najniska_cena from produkti_cena as pc where pc.price = (select min(price) from produkti_cena)create view produkti_cena as
Извештај за вкупно продадени продукти по категорија
create view kategorija_produkti as select distinct c.category_id, c.category_name, case when tabela.naracan_proizvod >= 1 then tabela.naracan_proizvod else 0 end as naracani_proizvodi from category as c join product as p on p.category_id=c2.category_id join product_in_store as pis on p.product_id=pis.product_id join orders as o on o.product_in_store_id=pis.product_in_store_id join( select distinct c2.category_id, c2.category_name, count(o.product_in_store_id) as naracan_proizvod from category as c2 join product as p on p.category_id=c2.category_id join product_in_store as pis on p.product_id=pis.product_id join orders as o on o.product_in_store_id=pis.product_in_store_id group by 1, 2 ) as tabela on c.category_id=tabela.category_id
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