| Version 40 (modified by , 15 months ago) ( diff ) |
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Напредни извештаи од базата (SQL и складирани процедури)
Прикажување на совет чија што дата на побарување не е постара од 6 месеци гледано од денешната дата, прикажување на корисник кој го побарал советот и прикажување на број на одговори на прашањето.
SELECT
bpc.id AS consultation_id,
bpc.date_askes AS claim_date,
bpc.title AS consultation_title,
bpc.description AS consultation_description,
u.id AS user_id,
u.name AS user_name,
COUNT(bpa.id) AS answers_count,
ARRAY_AGG(bpa.reply) AS all_answers
FROM
project.blog_post_for_consultations bpc
LEFT JOIN
project.users u ON bpc.users_id = u.id
LEFT JOIN
project.blog_post_answers bpa ON bpc.id = bpa.blogpostconsid
WHERE
bpc.date_askes >= current_date - interval '6 months'
GROUP BY
bpc.id, bpc.date_askes, bpc.title, bpc.description, u.id, u.name
ORDER BY
answers_count DESC;
Прикажување на вработен (доктор) кој што има напишено најмногу извештаи за едно исто милениче.
За да го тестираме овој прашалник ги додадовме следниве податоци:
INSERT INTO project.reports
(description, usersid, petsid)
VALUES('Za isto zivotno', 8, 1);
WITH DoctorPetReportsCount AS (
SELECT
u.id AS doctor_id,
u.name AS doctor_name,
r.petsID AS pet_id,
COUNT(r.id) AS reports_count
FROM
project.reports r
LEFT JOIN
project.users u ON r.usersID = u.id
GROUP BY
u.id, u.name, r.petsID
),
MaxReportsPerPet AS (
SELECT
pet_id,
MAX(reports_count) AS max_reports_count
FROM
DoctorPetReportsCount
GROUP BY
pet_id
),
PetsWithMaxReports AS (
SELECT
pet_id,
max_reports_count
FROM
MaxReportsPerPet
WHERE
max_reports_count = (SELECT MAX(max_reports_count) FROM MaxReportsPerPet)
)
SELECT
dprc.doctor_id,
dprc.doctor_name,
dprc.pet_id,
dprc.reports_count
FROM
DoctorPetReportsCount dprc
JOIN
PetsWithMaxReports pwmr ON dprc.pet_id = pwmr.pet_id AND dprc.reports_count = pwmr.max_reports_count
JOIN
project.users u ON dprc.doctor_id = u.id;
За секое прашање кое е поставено во Blog Post For Consultations колку одговори има, од кои корисници се дадени одговорите и каква е улогата на корисниците кои одговориле(admin, manager, employee, standard).
SELECT
bpc.ID AS question_id,
bpc.title AS question_title,
COUNT(bpa.ID) AS number_of_answers,
u."name" AS user_name,
r.type AS user_role
FROM
project.blog_post_for_consultations bpc
JOIN
project.blog_post_answers bpa ON bpc.ID = bpa.blogpostconsid
JOIN
project.users u ON bpa.usersID = u.ID
JOIN
project.roles r ON u.role_id = r.ID
GROUP BY
bpc.ID, u.ID, r.ID -- Include the role ID in the GROUP BY clause
ORDER BY
bpc.ID;
Прикажување на продукт кој е прв додаден во системот за продажба и за него постојат најмалку нарачки кои биле направени од корисниците.
WITH ProductSales AS (
SELECT
p.id AS product_id,
p.name AS product_name,
p.description AS product_description,
p.price AS product_price,
MIN(o.quantity) AS sold_quantity,
p.dateadded AS date_added
FROM
project.products p
LEFT JOIN
project.product_are_made_orders pamo ON p.id = pamo.id_products
LEFT JOIN
project.orders o ON pamo.id_orders = o.id
GROUP BY
p.id, p.name, p.description, p.price, p.dateadded
),
RankedProducts AS (
SELECT
*,
ROW_NUMBER() OVER (ORDER BY sold_quantity, date_added) AS rn
FROM
ProductSales
)
SELECT
product_id,
product_name,
product_description,
product_price,
sold_quantity,
date_added
FROM
RankedProducts
WHERE
rn = 1;
Секој ветеринарен центар колку вкупно миленичиња примил и колку е просечната возраст во секој ветеринарен центар.
SELECT
vc.name AS vet_center_name,
-- vc.description AS vet_center_description,
COUNT(DISTINCT pvvc.id_pets) AS number_of_pets_treated,
AVG(EXTRACT(YEAR FROM age(current_date, p.dateofbirthday))) AS average_pet_age
FROM
project.vet_centers vc
JOIN
project.pets_visit_vet_centers pvvc ON vc.id = pvvc.id_vet_centers
JOIN
project.pets p ON pvvc.id_pets = p.id
GROUP BY
vc.id, vc.name, vc.description
ORDER BY
number_of_pets_treated DESC;
Koja e најновата терапија според датум која е препишена, за која дијагноза е препишена, и во кој ветеринарен центар е препишена.
WITH LatestTherapies AS (
SELECT
ptvc.id_pets,
MAX(t.appoitmentDate) AS latest_therapy_date
FROM
project.pets_visit_vet_centers ptvc
JOIN
project.therapy t ON ptvc.id_pets = t.diagnosticsID
GROUP BY
ptvc.id_pets
)
SELECT
vc.name AS vet_center_name,
p.id AS pet_id,
--p.description AS pet_description,
d.description AS diagnosis_description,
t.description AS latest_therapy_description,
lt.latest_therapy_date
FROM
project.vet_centers vc
JOIN
project.pets_visit_vet_centers ptvc ON vc.id = ptvc.id_vet_centers
JOIN
project.pets p ON ptvc.id_pets = p.id
JOIN
project.diagnostics d ON p.id = d.usersID
LEFT JOIN
LatestTherapies lt ON p.id = lt.id_pets
LEFT JOIN
project.therapy t ON lt.latest_therapy_date = t.appoitmentDate AND p.id = t.diagnosticsID
ORDER BY
latest_therapy_date DESC;
Прикажување на мобилен оператор на кој што се најчесто корисниците претплатници. Доколку мобилниот оператор Telecom започнува на 072 или 071, мобилниот оператор VIP започнува на 078 или 075.
SELECT
CASE
WHEN SUBSTRING(u.number, 1, 3) IN ('072', '071') THEN 'Telecom'
WHEN SUBSTRING(u.number, 1, 3) IN ('078', '075') THEN 'VIP'
ELSE 'Other Operator'
END AS mobile_operator,
COUNT(u.id) AS number_of_users
FROM
project.users u
where
u.number LIKE '072%' OR
u.number LIKE '071%' OR
u.number LIKE '078%' OR
u.number LIKE '075%'
GROUP BY
mobile_operator
ORDER BY
number_of_users DESC;
За секој ветеринарен центар, колку пари одвојува за нови вработени и која е просечната плата по вработен.
WITH SalaryData AS (
SELECT
v.id AS vet_center_id,
v.name AS vet_center_name,
j.predictedsalery AS salary
FROM
vet_centers v
JOIN
jobs j ON v.id = j.vetCentersID
),
SalaryAggregates AS (
SELECT
vet_center_id,
vet_center_name,
COUNT(salary) AS number_of_employees,
SUM(CAST(REPLACE(salary, ' EUR', '') AS DECIMAL)) AS total_salary_expenses,
AVG(CAST(REPLACE(salary, ' EUR', '') AS DECIMAL)) AS average_salary
FROM
SalaryData
GROUP BY
vet_center_id, vet_center_name
)
SELECT
sa.vet_center_name,
sa.number_of_employees,
sa.total_salary_expenses,
sa.average_salary
FROM
SalaryAggregates sa
ORDER BY
sa.total_salary_expenses DESC;
Прикажување на ветеринарни центри кои што корисникот може да ги посети доколку живее во Охрид.
SELECT vc.name AS vet_center_name, vc.description AS vet_center_description, vc.adress AS vet_center_address, vc.phoneNumber AS vet_center_phone_number FROM project.vet_centers vc JOIN project.cities c ON vc.citiesID = c.ID WHERE c.name = 'Ohrid';
Прикажување која терапија за која дијагноза е препишена и од кој доктор е препишена.
SELECT
t.description AS therapy_description,
d.description AS diagnosis_description,
u.name AS doctor_name,
u.lastname AS doctor_lastname
FROM
project.therapy t
JOIN
project.diagnostics d ON d.id = t.diagnosticsID
JOIN
project.users u ON u.id = d.usersID
JOIN
project.pets p ON p.id = d.usersID
ORDER BY
t.id;
Прикажување на продукти кои што би можело да се потрошат најбрзо бидејќи има побарано најголема количина за нив.
SELECT p.name AS product_name, p.description AS product_description, AVG(o.quantity) AS avg_sales_quantity FROM project.products p JOIN project.product_are_made_orders po ON p.ID = po.id_products JOIN project.orders o ON po.id_orders = o.ID GROUP BY p.ID ORDER BY avg_sales_quantity DESC;
За секој миленик кои болести се дијагностицирани, какви лекови и терапија е препишена.
WITH PetDiagnosis AS (
SELECT
p.id AS pet_id,
p.name AS pet_name,
d.description AS diagnosis_description
FROM
pets p
JOIN
diagnostics_established_pets dep ON p.id = dep.id_pets
JOIN
diagnostics d ON dep.id_diagnostics = d.id
),
PetMedications AS (
SELECT
p.id AS pet_id,
m.name AS medicine_name,
m.description AS medicine_description
FROM
pets p
JOIN
therapy_takes_pets ttp ON p.id = ttp.id_pets
JOIN
therapy t ON ttp.id_therapy = t.id
JOIN
medecines m ON t.diagnosticsID = m.diagnosticsID
),
PetTherapy AS (
SELECT
p.id AS pet_id,
t.description AS therapy_description,
t.appoitmentDate AS therapy_date
FROM
pets p
JOIN
therapy_takes_pets ttp ON p.id = ttp.id_pets
JOIN
therapy t ON ttp.id_therapy = t.id
)
SELECT
pd.pet_id,
pd.pet_name,
pd.diagnosis_description,
pm.medicine_name,
pm.medicine_description,
pt.therapy_description,
pt.therapy_date
FROM
PetDiagnosis pd
LEFT JOIN
PetMedications pm ON pd.pet_id = pm.pet_id
LEFT JOIN
PetTherapy pt ON pd.pet_id = pt.pet_id
ORDER BY
pd.pet_id, pt.therapy_date;
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