I was eagerly waiting for this video. Thank you sir for providing such good content at zero cost ❤.
@ousmanesidibe951
Ай бұрын
I'm start seeing Chandoo as the better version of me, now how am i gonna achieved this? in the mean time thank you for all knowledge shared.
@godismyway7305
Ай бұрын
How the hell in your 15 yeras on youtube , you don't have least 1M Subscribers is beyond me. I personaly admir how fast , skilled and good teahcer you are but I beieieve to get the most out of your effort , you should jump in to Technical fileds like WEB Technologies from easy ones like HTML CSS JS and framworks REACT & Angular , and some Backend Technolgies like Node.jS Express Phyton with some Database design courses like NoSQL(MongoDB) and relational the one you are teaching in this video. It is time to evalute our progress. This is from your day one subscriber. Keep it up!
@bhaweshjoshi3309
Ай бұрын
First of all a big thank you for providing such kind of content . But i have a humble request please dont stop this series at any cost also don't make it paid or for only members in the end . This playlist is the only one which i am following from my heart so that i can understand and work well with data . Once again thank you for all of this ❤
@namrataghosh111
Ай бұрын
Select count(distinct c.last_name) as Distinct_Last_Name from customer c; (Answer -599) Select date(rental_date), count(*) as rental_count from rental group by date(rental_date) order by rental_count limit 1; ( Answer - 2005-07-31 /Count -679) Thank you Chandoo for this amazing tutorials and so easily explained !
@dailyuploads3959
Ай бұрын
Your teaching method is great. I really appreciate your work and content. Personally i give you 5 stars ⭐⭐⭐⭐⭐ rating. Thanks again continue. Love from pakistan
@HaileysHomes-ix5yu
Ай бұрын
watched this when it dropped yesterday, back here today to practice. wish me luck🍀
@giladyair7007
Ай бұрын
HW2: with Rentals_s1 as (select date(r.rental_date) 'Rental Day' , r.rental_id, r.staff_id from rental r where r.staff_id = 1 group by date(r.rental_date), r.rental_id ) select date(pa.payment_date) 'Rental day', count(*) 'Rental Count' , sum(amount) 'Total Payment' from Rentals_s1 rs1 join payment pa on pa.rental_id = rs1.rental_id group by date(pa.payment_date) order by sum(amount) desc This was my first time experimenting with SQL, and the lesson was really easy to understand and enjoyable. Thank you so much for your contribution to the community.
@aaronpencil6009
29 күн бұрын
😮😮😮
@thePaulCode
29 күн бұрын
Great!
@muhammadsadiq2330
28 күн бұрын
You are so good at writing SQL queries.
@rashidkhan8161
Ай бұрын
In query no.9 you should filter out film_id instead of inventory_id because each film has multiple copies and each copy has its own inventory_id
@chandoo_
Ай бұрын
Good find. I have updated the webpage with correct query. Please use this. with low_rentals as (select i.film_id, count(*) from rental r join inventory i on i.inventory_id = r.inventory_id group by i.film_id having count(*)
@maciejkopczynski55
Ай бұрын
@@chandoo_ This will unfortunately not show us films that were not rented at all. We would have to go with your suggestion from the video and start the groupping from film - inventory tables instead of inventory - rental tables.
@chandoo_
Ай бұрын
Both of the queries won't show the "never" rented films. For that you need a left-anti-join on films to rentals (to see which films have null rental record).
@veeras6412
7 күн бұрын
@@chandoo_ I think each rental operation have unique rental_id so we have to count how many rental_ids generated for each inventory id, then to film_id. I have tried the folowing query, please justify it. SELECT f.title, i.film_id, count(r.rental_id) rental_count FROM sakila.inventory i join rental r on r.inventory_id=i.inventory_id join film f on f.film_id=i.film_id group by i.film_id having rental_count
@veeras6412
6 күн бұрын
@@chandoo_please justify this for 'low rentals query' SELECT f.title, i.film_id, count(r.rental_id) rental_count FROM sakila.inventory i join rental r on r.inventory_id=i.inventory_id join film f on f.film_id=i.film_id group by i.film_id having rental_count
@prasad1686
29 күн бұрын
Thank you for the excellent and straightforward yet professional explanation. I appreciate your valuable efforts and looking forward to more videos from you. 🙏
@remuslupinhp
Ай бұрын
Hey Chandoo, this is super good practice. I have been learning SQL from other sources also because I want to be able to absorb the maximum out of your videos, so if possible, please cover Left, Right, and outer joins also as we have only done inner joins till now in this video... Looking forward to your upcoming videos, I am not able to focus on anything else waiting for your videos and then when they come, I have to watch and practice along instantly. Keep up the great work... I know a little bit of Excel, Power BI and SQL so this is where everything comes together. Thanks a lot! ❤❤
@harishsree2901
Ай бұрын
Thank you Anna... I really loved the way you are explaining.... Waiting for the next video...
@SheikImranMohamedAribu
Ай бұрын
Dear Chandoo, thank you for the knowledge, I am so grateful for this video, I have learnt a lot. Please teach us more of the queries using "WITH" function and more of using the subqueries. Please enlighten us on these more.
@ArifKhan-qo9jw
Ай бұрын
This is really helpful for beginners, we want to see some example advanced sql quires. Thanks Chandoo.!
@SathishKumar-oq7tz
29 күн бұрын
Thank you Chandu anna for making such nice content.
@canirmalchoudhary8173
19 күн бұрын
Good tutorial on PQ. I am still learning.
@sseemm
Ай бұрын
THANK YOU CHANDOO. YOU'RE THE BEST ❤❤
@user-uy7xw4jp3m
Ай бұрын
How many distinct last names do we have in the data? 599 rows
@cafemsoffice9171
Ай бұрын
"Whenever I visit places of worship, I always pray for you, Chandoo. Thank Y🤗U ! For sharing this with us; it means the world to us."
@alexrosen8762
26 күн бұрын
Very very useful, super thanks 🙏
@mustaphamuhammed1054
Ай бұрын
Thank you for what you do Sir 🙏
@wasifsaleem597
Ай бұрын
Hi. It is wonderful video. Please let me know will you further cover SQL or it is completed
@aparnapandravada5073
Ай бұрын
Thank you… it is awesome
@madhulika60
14 күн бұрын
H1 : select count(distinct last_name) as 'Number of distinct last names' from customer;
@ankitmishra2998
Ай бұрын
You're the bestestttttttttttttttttttttttttttttttttttttttttttt
@tifanycantillo3760
29 күн бұрын
Nice video! Thanks! I would for sure purchase a course by you maybe on Udemy platform, totally worth it
@ravi__negi__4333
Ай бұрын
thank u so muchc channdo sir,,,and my humble request that if u add your perosnal experince diffultes u face in sql in these video aslo then veru helpful
@muhammadtasiu5533
25 күн бұрын
Great
@shaiksabdar6
Ай бұрын
Awesome My SQL Queries explanation with examples...💥👌👍
@MirzanFawas
19 күн бұрын
count of active customers = 584. select count(active) from customer where active=1;
@suryasreemanth3387
23 күн бұрын
# H1) How many distinct last names are present in the data ? select count(distinct last_name) as last_names from customer; # H2) What is the busiest day in the rentals date ? select date(rental_date) as Date,count(rental_date) as rentals from rental group by date(rental_date) order by rentals desc limit 1; # H3) How much rentals we make on each day ? select date(r.rental_date) as Date , sum(p.amount) 'Amount Produced' from rental r join payment p on p.rental_id=r.rental_id group by Date order by Date asc; # H4) What are the three top earning days so far? select date(r.rental_date) as Date , sum(p.amount) 'Amount Produced' from rental r join payment p on p.rental_id=r.rental_id group by Date order by SUM(amount) desc LIMIT 3;
@lakshmanvajjakeshavula5380
Ай бұрын
Thank you
@xeeshanahmad8757
Ай бұрын
Thank you Sir.
@mrpeterhacks
10 күн бұрын
Hello Chandoo, thanks for these tutorials, but my com is showing compatibility error, it is saying I try workbench 6.3, but it's not available on mysql website, any help?
@user-ht3mg3jc4n
13 күн бұрын
@chandoo sir how to answer for the homework problems in notebook or in any option to answer those
@AruljeraldJ-b3o
19 күн бұрын
For counting distinct last name Select count(distinct(lastname))) from customer
@stytricks
Ай бұрын
Sir, in your Telugu channel too there is an in-depth video on sql right. Does that covers all concepts what you explained here
@seanetho5133
Ай бұрын
Not sure if I missed something but for Q8) 31:55, shouldn't we be counting rental ID not customer ID? My thoughts would be that one customer could rent more than one movie (one customer ID could have more than one rental ID) and therefore, counting rental ID would give a better understanding of how many movies each customer rented?
@mankaadda1061
Ай бұрын
SIR JI PLEASE COMPLETE DATA ANALYIST COURSE
@syedtahirbukhari8377
Ай бұрын
Good was waiting
@abdulbasital-sufyani6828
Ай бұрын
I will press LIKE first Then watch the content
@Durgada_l
22 күн бұрын
Select active, count(*) from customers group by active;
@aaronpencil6009
Ай бұрын
Halo Chandoo, please interrogate my opinion on query #7. this is how I did it, following your thought process but going my own way: # 7) All Sci-fi films in our catalogue select c.category_id, c.name, fc.film_id, f.title from category c join film_category fc on fc.category_id=c.category_id join film f on f.film_id=fc.film_id having category_id='14';
@X4Xubi
Ай бұрын
Query#9 Sir its a good thing U used CTE method, I will need to understand the concept of it bcox its still not clear to me so I did it in a simplest way. Here's the query; SELECT f.title, COUNT(r.rental_id) AS rental_count FROM film f LEFT JOIN inventory i ON f.film_id = i.film_id LEFT JOIN rental r ON i.inventory_id = r.inventory_id GROUP BY f.title HAVING COUNT(r.rental_id) < 1;
@mahnoorzahid9967
6 күн бұрын
homework 2 answer select date(rental_date),count(*) from rental group by date(rental_date) order by count(*)desc;
@sindhus5585
Ай бұрын
I have tried to do so many attempts but I couldn't how to do please upload the vedio of how to add the resources to view the Page
@Professor-G
25 күн бұрын
SELECT count(distinct c.last_name) FROM sakila.customer c;
@dhanavenkatachennakeswarar944
Ай бұрын
table can also be called as Entity, So that we can understand it as Table Relation Diagram.
@pratikdoshi8177
Ай бұрын
Hi sir I am a finance background student. can I learn this data analysis without code and the benefits of learning,please tell
@mkgeidam
Ай бұрын
Thanks for the video @Chandoo, i clicked the Sample data and completed queries files, but couldn't download it, pls need your help.
@prithivi_n
Ай бұрын
Please recheck the 9th query. A single film has multiple inventory_id’s.
@chandoo_
Ай бұрын
Great catch. I will update the webpage with correct query soon. But this one should work. The condition reverts to less than 5 life time rentals. with low_rentals as (select i.film_id, count(*) from rental r join inventory i on i.inventory_id = r.inventory_id group by i.film_id having count(*)
@ankurkeshri7062
Ай бұрын
Thanks for all you are doing🎉
@comedygang9792
24 күн бұрын
select date(rental_date),count(*) as counto from rental group by date(rental_date) order by counto desc;
@Durgada_l
22 күн бұрын
Select count(last_name) from customers.
@kingjamesjc2007
29 күн бұрын
What is the difference between data visualization and Data analysis
@Durgada_l
21 күн бұрын
Select * from customers Order by customers Limit 3;
@zaa4260
Ай бұрын
Sir for me i meet errors in some of these quires. and i copy as yours but I dont know whats wrong
@arshdeepnival7974
Ай бұрын
Hi Chandoo, can you share these concepts using postgresql please because i have learnt postgresql but not mysql, there are queries which are similar in both but such things aren't there in postgresql
@AmanRaut-u3b
Ай бұрын
please start power bi also sir
@Musiclove-hd3tk
27 күн бұрын
#H1) How many distinct last names we have in the data? select distinct count(c.last_name) from customer c;
@sakshiawadhiya7267
Ай бұрын
I didn't got this databass during installation.. would you provide this database
@Professor-G
24 күн бұрын
Query no. 6: What is the busiest day? A simple solution is to add an ORDER BY as follows: SELECT date(r.rental_date), count(*) rentals FROM sakila.rental r group by date(r.rental_date) order by rentals desc ; The answer is: 2005-07-31 was the busiest day with 679 rentals.
@khalidansari3201
Ай бұрын
Power bi start scretch
@emmanueloladipojohnson890
27 күн бұрын
Query for the HW 2 select date(rental_date) d, count(*) as rental_count from sakila.rental r group by date(rental_date) order by rental_count desc limit 1;
@valasalasivakumar7815
Ай бұрын
Please make more Telugu videos
@richaagrawal376
Ай бұрын
Please suggest good books for learning sql
@Name-pj3dv
Ай бұрын
Question busiest day ans: select max(rental_date), count(*) from rental order by rental_date desc;
@GIRISHKUMAR-w2m
Ай бұрын
Unable to solve the h2 problem, can you please help me out
@bhargavchowdary9833
Ай бұрын
Hi Chandu bro ❤ Sub queries and stored procedure gurinchi detailed Video chesthaaraa
@chandoo_
Ай бұрын
In 90% of data analyst work, you don't need either of those. I cover CTE & Joins in the video. Subquery is a similar concept and can be easily learnt. I've never written a sproc in my life. So as of now no plans to expand the SQL section. If in future, I make a course, I will add these topics to it.
@comedygang9792
24 күн бұрын
31st july 2005 was the busiest day
@susanshifali9726
25 күн бұрын
H2 solution: select s.store_id, date(r.rental_date) 'rental date', count(r.rental_id) 'total rentals', sum(p.amount) 'total revenue' from rental r join payment p on p.rental_id = r.rental_id join staff st on st.staff_id = p.staff_id join store s on s.store_id = st.store_id where s.store_id = 1 group by date(r.rental_date) order by date(r.rental_date); I hope this is one of the appropriate methods to answer this query.
@rawol9647
19 күн бұрын
I did the same thing!!
@konthamparimala2406
15 күн бұрын
hi everyone to practice these queries we need a sakila data set right where it is provided . please anyone can answer my question if you already know because it may take time for chandu sir reply. documentation related to sakila is there in description but i am not geting where is the actual data set .please anyone can help me ?
@chandoo_
14 күн бұрын
Refer to video #1 of this (here - kzitem.info/news/bejne/qpmbx6WasmmeeKg ) for instructions on how to install MySQL with Sakila. If you already have MySQL but not Sakila, download the installation scripts from official MySQL page here (in the Example databases section) - dev.mysql.com/doc/index-other.html
@refulancer6759
15 күн бұрын
-- all sci-fi films in our catalgoe SELECT title FROM FILM WHERE film_id in ( SELECT film_id FROM film_category WHERE category_id IN (SELECT category_id FROM category WHERE name = 'Sci-Fi') );
@thePaulCode
29 күн бұрын
# What are the three top earning days so far? SELECT date(p.payment_date) AS 'Payment Date', COUNT(*) AS 'Total Rental', SUM(p.amount) AS Amount FROM payment p GROUP BY date(p.payment_date) ORDER BY Amount DESC LIMIT 3; Can you suggest improvements for this query?
@saianishperuboyina1168
Ай бұрын
Hello Chandoo, while working on join function it showed a Error Code:1046, saying No database selected and to select the DB by try double clicking on the schemas. I tried for a long but still was not working....Until i googled it and used the "USE" function. Could you help me understand where did I go wrong and when do we have to apply this "USE" function.
@chandoo_
Ай бұрын
You need to double click on the word "sakila" in the schemas to use it. Alternatively, you can also write "Use sakila;" as the first line of your SQL file so all the subsequent queries run against sakila.
@saianishperuboyina1168
Ай бұрын
@@chandoo_ I opened the workbench again today to work on homework questions....same error is coming up, i am unable to use Mysql despite double clicking & using USE function.
@chandoo_
Ай бұрын
Please search up online with the error code. There could be other reasons.
@thePaulCode
Ай бұрын
Dear Professor @chandoo # H1) How many distinct last names we have in the data? SELECT COUNT(DISTINCT c.last_name) AS 'Distinct last name' FROM rental r JOIN customer c ON r.customer_id = c.customer_id; Output: 599
@sindhus5585
Ай бұрын
Hii sir how to add the users to view the Pages in Power Bi
@AjitKumar-fs9lc
Ай бұрын
Select f.film,f.title,f.release_year,c.name,c.category_id from film f join film_category fc using(film_id) join category c using(category_id) where c.category_id=14
@yashsharma5042
23 күн бұрын
Hello Chandoo Sir, where can I get the CSV file of Sakila ?
@davui_21
22 күн бұрын
He demonstrated it in the first video of this series. Down below.. kzitem.info/news/bejne/qpmbx6WasmmeeKg
How to connect with Shakila database please guide me
@rudraaa-k3m
Ай бұрын
can you share the data for practice
@sipperkhan9784
Ай бұрын
How many distinct names gave: 599 rows.
@pankuraju1363
8 күн бұрын
I am using SQL server , how to get the sakila dataset
@chandoo_
8 күн бұрын
There is a github page with Sakila ports for various db servers. See here - github.com/ozzymcduff/sakila-sample-database-ports/tree/master/sql-server-sakila-db
@sidra077
Ай бұрын
Unable to get dataset
@NikkiSingh-s5r
Ай бұрын
How to download/import sakila in my workbench? am not finding any link to download
@aryandubey3141
Ай бұрын
same
@pandubuntty8680
Ай бұрын
H1) How many distinct last names we have in the data? select distinct last_name,c.customer_id,count(last_name) from rental r join customer c on c.customer_id = r.customer_id group by c.customer_id; (or) SELECT COUNT(DISTINCT last_name) AS last_names FROM customer;
@MxStory-ge4fk
27 күн бұрын
-- H1) How many distinct last names we have in the data? select count(distinct last_name) from customer; -- H2) How much money and rentals we make for Store 1 by day? with cte as (select date(r.rental_date) , r.rental_id, r.staff_id from rental r where r.staff_id = 1 group by 1,2 ) select date(p.payment_date) 'rental day', count(*) rental , sum(amount) from cte c join payment p on p.rental_id = c.rental_id group by 1 order by 3 desc; -- H-2 What are the three top earning days so far? select date(payment_date) 'rental day', count(*) rental , sum(amount) earnings, row_number() over(order by sum(amount) desc) rankings from payment group by 1 order by 3 desc limit 3;
@sainathbuthagaddala5327
Ай бұрын
🥰
@HimanshuJagtap
29 күн бұрын
QUESTION #9 9) Which movies should we discontinue from our catalogue (less than 2 lifetime rentals) -- less than or equal to 1 as shown in the video. I wanted to get the film title as well ... The below query returns 0 rows because the minimum number of rentals for the result set is 2 SELECT inventory_id, COUNT(inventory_id) FROM rental r JOIN film f ON r.inventory_id=f.film_id GROUP BY inventory_id HAVING COUNT(*)
@HimanshuJagtap
29 күн бұрын
Note: COUNT(inventory_id) or COUNT(*) gives the same result. Also, film_id and inventory_id are same if you compare the outputted columns.
@chandoo_
29 күн бұрын
You are joining wrong columns. Even if the values are same, you should join inventory_ids on both sides.
@HimanshuJagtap
28 күн бұрын
@@chandoo_ Thanks for the reply. The following query returns the correct/expected output. SELECT f.film_id, i.inventory_id, title, release_year, COUNT(rental_id) AS rentals FROM inventory i JOIN film f ON i.film_id = f.film_id JOIN rental r USING (inventory_id) GROUP BY r.inventory_id HAVING COUNT(rental_id)
@HimanshuJagtap
28 күн бұрын
Solution using subquery: SELECT inventory_id, title FROM inventory i JOIN film f ON i.film_id = f.film_id JOIN rental r USING (inventory_id) WHERE inventory_id IN (SELECT inventory_id FROM rental GROUP BY inventory_id HAVING COUNT(*)
@talhashahzad2314
12 күн бұрын
can anyone please share dataset
@mohdilyas4688
Ай бұрын
show this in sql server bro
@moyemoye-f7o
Ай бұрын
same database same query but different results for query #5, mine does not return distinct values..moye moye
@X4Xubi
Ай бұрын
same case here, mine returns 33 rows as well
@mrbartuss1
Ай бұрын
You must have done something wrong. I got 31: SELECT DISTINCT CONCAT(first_name, last_name) AS full_name FROM customer JOIN rental USING (customer_id) WHERE DATE(rental_date) = '2005-07-26';
@h.narayan
11 күн бұрын
/* How many distinct last names we have in the data? */ SELECT COUNT( distinct last_name) 'Total Number of Last Name ' FROM customer r ;
@ravi__negi__4333
Ай бұрын
views truggers r they use in real life?
@Professor-G
11 күн бұрын
My solution: Homework Query 10 select s.store_id, date(r.rental_date) 'dates', count(r.rental_id) 'rental count', sum(p.amount) 'total payments' from store s join inventory i on i.store_id = s.store_id join rental r on r.inventory_id = i.inventory_id join payment p on p.rental_id = r.rental_id join film f on f.film_id = i.film_id where s.store_id = '1' group by date(r.rental_date) order by date(r.rental_date) limit 3; Looking forward to feedback
@Name-pj3dv
Ай бұрын
Question HOW MANY ACTIVE CUSTOMERS Ans: select count(*) from customer where active = 1;
@mohammadrahil7619
3 күн бұрын
link for the dataset
@chandoo_
2 күн бұрын
It is in the video description or here - chandoo.org/dac
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