thank you

thank you

I stated my SQL journey pretty recently and of all the people who provide solutions on internet YOU ARE THE BEST!!! Your videos help build the intuition and thinking while writing the queries rather than just cramming up code. Thanks Frederik.

thank you

thank you

thank you

thank you

thank you

Repeatedly was getting errors, figured out I missed the important point that product_key can be repeated for a particular customer as it's a foreign key

thank you

Great content, im trying to get better at sql and this channel is showing me exactly what not to do (just kidding folk, you great!

thank you

Thank you! Learned something new 😃

Great idea! Thanks for the motivation to get this done!

There is an extra parathesis.

month is an alias.

Hi Frederik..very good explanation..

🚀

yo

In case of a tie the rank window function would have the same output for the rows. Since the question asked that the ranks be unique you should use Row_number function instead

Let's go

You're back ! Yay! Happy to support you if u have a Patreon

Happy to see you back after 5 months

thank you

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-- efficient code select c.name as "Customers" from Customers as c left join Orders as o on c.id = o.customerId where o.id is null ---- simple to understand but not efficient code select name as "Customers" from Customers where id not in (select distinct Customerid from Orders)

-- Write your PostgreSQL query statement below select max(salary) as SecondHighestSalary from Employee where salary < (select max(salary) from Employee)

select firstName, lastName, city, state from Person as p left join Address as a on p.personId = a.personId

select event_day as day, emp_id, sum(out_time - in_time) as total_time from Employees group by 1,2

thank you please continue making videos

thank you please continue making videos

thank you

this is very helpful, thanks for sharing

thanks for the video but i solved the problem with if select sum(boxes.apple_count + if(chests.apple_count is null,0,chests.apple_count)) as apple_count, sum(boxes.orange_count + if(chests.orange_count is null, 0,chests.orange_count )) as orange_count from boxes left join chests on boxes.chest_id = chests.chest_id

I’m a programmer (not doing any data stuff) and it was really nice to get an insight to what people choose and why data-analysis wise. Thanks!

why are we selecting event_date in subquery?

What is issue in this Ms SQL server (output is same except order) >> Select v.customer_id,count(v.visit_id) as count_no_trans from Visits as v right join Transactions as t on t.Visit_id=v.Visit_id where t.transaction_id is null group by v.customer_id

you videos are very helpful, kindly continue making videos .

Why sum(b.weight) instead of (a.weight)?

thanks

thank you

why can't i use this solution : gives correct answer but not accepted select product_id, min(year) as first_year, quantity, price from Sales group by product_id what is wrong with it?

ure amazing

Hi! Can anyone help me understand why the following code returns 2 products? SELECT p.product_id, p.product_name FROM product p JOIN sales s ON s.product_id=p.product_id WHERE s.sale_date NOT BETWEEN '2019-04-01' AND '2019-12-31' GROUP BY p.product_id

i didn't hear about this page yet

WOW This is an efficient code😍

This is the first question in SQL 50 that I wasn't able to solve right away. Thanks for your helpful video. Now I know how to approach these kinds of problems

Why is it a sum and not count in your query? Aren’t we counting number of cancelled rides ??

here is my approach in mysql : with cte as(select order_id,customer_id,order_date,a.item_id,item_category,quantity,dayname(order_date) as dayname from amazon_orders a left join amazon_items i on a.item_id=i.item_id) select item_category, sum(case when item_category in ('book','phone','glasses') and dayname='Monday' then quantity else 0 end) as 'Monday', sum(case when item_category in ('book','phone','glasses') and dayname='Tuesday'then quantity else 0 end) as 'Tuesday', sum(case when item_category in ('book','phone','glasses') and dayname='Wednesday' then quantity else 0 end) as 'Wednesday', sum(case when item_category in ('book','phone','glasses') and dayname='Thursday' then quantity else 0 end) as 'Thursday', sum(case when item_category in ('book','phone','glasses') and dayname='Friday' then quantity else 0 end) as 'Friday', sum(case when item_category in ('book','phone','glasses') and dayname='Saturday' then quantity else 0 end) as 'Saturday', sum(case when item_category in ('book','phone','glasses') and dayname='Sunday' then quantity else 0 end) as 'Sunday' from cte group by 1;