If question is not readable or incomplete , I'm not gonna answer that question🙄

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Except one😉
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By S S

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Duplicate

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Now correct
That time I just checked one test case😅

Very cool number

🔰 Dunzo Answers

1. select id, name
from student
order by score desc, id asc
limit 3;



2. SELECT IFNULL(customer_name, "N/A") customer_name, IFNULL(product_name, "N/A") product_name, IFNULL(quantity, 0) quantity
FROM ((customer LEFT OUTER JOIN invoice ON customer.id = invoice.customer_id) LEFT OUTER JOIN invoice_item ON invoice_item.invoice_id = invoice.id) LEFT OUTER JOIN product ON invoice_item.product_id = product.id
ORDER BY customer.id, product.id;



3. SELECT
p.product_name,
(COALESCE (SUM(CASE WHEN i.time_paid is null and i.time_canceled is null and i.time_refunded is null IS NOT NULL THEN it.quantity * it.price ELSE 0 END ),0)) as due,
(COALESCE (SUM(CASE WHEN i.time_paid IS NOT NULL THEN it.quantity * it.price ELSE 0 END ),0)) as paid,
(COALESCE (SUM(CASE WHEN i.time_canceled IS NOT NULL THEN it.quantity * it.price ELSE 0 END ),0)) as canceled,
(COALESCE (SUM(CASE WHEN i.time_refunded IS NOT NULL THEN it.quantity * it.price ELSE 0 END ),0)) as refunded
FROM
PRODUCT p
LEFT OUTER JOIN invoice_item it ON p.id = it.product_id
LEFT OUTER JOIN invoice i ON it.invoice_id = i.id
GROUP BY
product_name
ORDER BY
p.id ASC



4. SELECT c.name,
c.phone,
SUM(CASE WHEN h.direction = 'in' THEN h.duration END) as IncomingCost,
SUM(CASE WHEN h.direction = 'out' AND h.duration > 120 THEN 500 + 2*(h.duration-120)
ELSE 2*(h.duration-120)
END) as OutgoingCost,
SUM(CASE WHEN h.direction = 'in' THEN h.duration END +
CASE WHEN h.direction = 'out' AND h.duration > 120 THEN 500 + 2*(h.duration-120)
ELSE 2*(h.duration-120)
END) as TotalCost
FROM customer c
JOIN (SELECT 'out' as directon, duration, dialed_on, outgoing_phone as phone
FROM history
WHERE YEAR(dialed_on) = 1995
AND MONTH(dialed_on) = 1
UNION ALL
SELECT 'in' as direction, duration, dialed_on, incoming_phone as phone
FROM history
WHERE YEAR(dialed_on) = 1995
AND MONTH(dialed_on) = 1
) h ON c.phone = h.phone
GROUP BY c.name,
c.phone



5. select c.company_code, c.founder,
count(distinct l.lead_manager_code), count(distinct s.senior_manager_code),
count(distinct m.manager_code),count(distinct e.employee_code)
from Company c, Lead_Manager l, Senior_Manager s, Manager m, Employee e
where c.company_code = l.company_code
and l.lead_manager_code=s.lead_manager_code
and s.senior_manager_code=m.senior_manager_code
and m.manager_code=e.manager_code
group by c.company_code order by c.company_code;