didi iska notes v share kr do

-- Assignment No.2 -- 1. Fetch data From Customer Table where either the customer pincode is among (1245,54321,98765)or the country's name usa. select distinct(pin_code) from customer where country = "USA"; -- 2.fetch Customer_id , customer_name and their emails for customers who live in either the USA or Canada. select customer_id , customer_name , email_id from customer where country in ('USA' , 'Canada'); -- 3. Replace Country Name from UK to United Kingdom in Customer table. select * from customer; update customer set country = 'United Kingdom' where country = 'uk'; -- 4. Remove all Customer Who Lives in Either the Spain or Germany. select * from Customer; delete from customer where country in (spain , Germany); -- 5. Get All Customer who lives at Pincode 12345 in India. select * from customer where pin_Code = '12345' and country = 'India'; -- 6. Get all the Customer whose last is Adams. select * from customer where customer_name like '% Adams'; -- 7. delete all the Customer whose last is Adams. delete from customer where customer_name like '% Adams'; -- 8. delete all the Customer whose last is smith. delete from customer where customer_name like '% smith'; -- 9. fetch all the distnict countries from customer table select distinct(country) from customer; -- 10. Get all customer who lives at pincode 23456. select * from customer where pin_code = 23456;

Question No.7 and 13 Both Same Questions.

didi , one request from my end that can u take window function question more as much as possible. because is more imp and often asked in interviews. so ,if possible 👃.(ranking related ques.)

when the part 2 will come?

Mam please make playlist of TREE DATA STRUCTURE in java Implementation of tree , binary tree, binary search tree, N-ary tree their questions

thank you ma'am, ekdm mast clear ho ja rha ek ek topic, thanks a lot

Hello ma'am Can you please provide That Notes PPT ?

I watched 2,3 video's of windows functions from different channels, But here I get simplest explaination. Now I understood the concept Thank uu didi 😊 After you cover all this topics take pratice questions asked in interviews and "project" Also so as to get the better understanding.

Ma'am foreign key vs join. Qki foreing key b toh two tables ko join krte hai

Ma'am great way of explaining ❤

Thanks for the questions 🙏I am a new subscriber. I have done all question in one time. i am a beginner.

Mam such a good teaching

Diii subquery, cte, windows functions ye sab ni karenge kya😪😪😪

Can u take window functions first?

Hello viewers , Please Correct SELF JOIN syntax . It should be 'ON' in place of 'WHERE' . Both Commands will result in same output but different in process. When we use 'ON' with joins, it checks the condition and only connects rows from both tables that meet this condition. But HOW exactly is 'WHERE' result in same output as 'ON'? Initially, the JOIN is performed between the Employee and Manager tables, resulting in a cross join. This means that every row of the Employee table is connected with every row of the Manager table. For example, if the Employee table has 10 rows and the Manager table also has 10 rows, this will result in 10 * 10 = 100 rows. Then, we filter these rows to keep only those where the Employee ID matches the Manager ID, which will reduce the result to 10 rows.

why we cant 'on' in place of "where"

very informative didi. very briefly explained each and everything...

Luved the explanation ..pls continue the playlist & upload regular basics ... I am gonna share it in my network

1) select e.employee_id , e.first_name || ' ' || e.last_name as full_name ,e.email ,d.department_name from employees as e left join departments as d on e.department_id = d.department_id ----------------------------------------------------------------------------------------------------- 2) select e.employee_id , e.first_name || ' ' || e.last_name as full_name ,e.email ,j.job_title from employees as e left join jobs as j on e.job_id = j.job_id ----------------------------------------------------------------------------------------------------- 3) select e.first_name, e.employee_id , o.order_name, count(o.order_id) as "total orders" from employees as e left join orders as o ON e.employee_id = o.employee_id group by e.employee_id,e.first_name,o.order_name order by e.first_name ----------------------------------------------------------------------------------------------------- 4) select e.*,count(o.order_id) from employees as e left join orders as o ON e.employee_id = o.employee_id group by e.employee_id HAVING COUNT(o.order_id)>0; select e.*,count(o.order_id) from employees as e left join orders as o on e.employee_id = o.employee_id group by e.employee_id,o.order_id having count(o.order_id)>1 select e.first_name, e.employee_id , o.order_name, count(o.order_id) as "total orders" from employees as e left join orders as o ON e.employee_id = o.employee_id group by e.employee_id,e.first_name,o.order_name order by e.first_name ------------------------------------------------------------------------------------------------------- 5) select d.department_name,count(e.employee_id) as "total employees" from departments as d left join employees as e ON d.department_id = e.department_id group by d.department_name -------------------------------------------------------------------------------------------------------- 6) SELECT e.employee_id, e.first_name, COUNT(o.order_id) AS order_count FROM employees AS e LEFT JOIN orders AS o ON e.employee_id = o.employee_id GROUP BY e.employee_id, e.first_name HAVING COUNT(o.order_id) > 1; ------------------------------------------------------------------------------------------------ 7) select e.first_name,d.department_name,j.job_title from employees as e left join departments as d on e.department_id = d.department_id left join jobs as j on e.job_id = j.job_id where (d.department_name = 'IT' and j.job_title = 'Software Developer') or (d.department_name = 'Sales' and j.job_title = 'Sales Executive') ------------------------------------------------------------------------------------------------- 8) SELECT e.employee_id, e.first_name, COUNT(o.order_id) AS order_count FROM employees AS e LEFT JOIN orders AS o ON e.employee_id = o.employee_id GROUP BY e.employee_id, e.first_name HAVING COUNT(o.order_id) > 1; --------------------------------------------------------------------------------------------------------- 9) SELECT e.first_name, SUM(o.order_price * o.quantity) AS "total price" FROM employees AS e LEFT JOIN orders AS o ON e.employee_id = o.employee_id GROUP BY e.first_name; select e.first_name,sum(order_price * quantity) as "total price" from employees as e left join orders as o ON e.employee_id = o.employee_id group by e.first_name having e.first_name ilike 'linda' select e.*,o.* from employees as e left join orders as o ON e.employee_id = o.employee_id where e.first_name ilike 'linda'

Very good and informative explanation by you all the topics but explain other topic like - procedure, views, window function for complete sql.

Hello Ma'am. This is Harshal . Hope you are doing great. We as students Love your videos it's a great platform to learn mysql it's a great asset to us as a fresher. We would appreciate if you can share us more assignments question so that it can help us to practice more and more.. once again thanks a lot Ma'am

luved ur explanation sister , pls continue

Thanks for such an informative and efforts you do to help us ☺️

In your Insert data a error is showing (cannot add or update child row)

Appear Two unavailable videos

Behtreen hai bus all topics cover kariye ,sabhi ko define kariye

I got 7 out 10 in 1st and last ques o got confused i did not use distinct function

I'm learning SQL for data analyst. I request u please take daily class

Hello mam

Very informative

mam, please provides pdf nots..if possible

subscribed cuz i want more questions from sql 👏👏👏

👏👏👏👏👏 nice video

Thank you. Pls keep going.

last question ki query me aap ne sirf count nikala, but actual me distinct categories id nikalni this naaa?

Thanks ma'am, your video or assignment is very helpful for me, I have watched many videos on KZbin but your videos are different from others, keep the video going on ma'am and making more videos and assignments.

❤❤

Hamne ans vector ko locally kyu lia? Why did we not pass it by reference or keep it global?

You are good mentor mam

Dear Mam, Maine SQL k bahut se Lectures attend kiye but Is platform pe kafi acchha explnation mila with examples and practice session k sath. Ab ise hi continue dekh raha hu. Please keep on going. I personally find it very helpful for me and hope sbko acchhi lage apki lectures. Thanks a lot making video on SQL.

Alright, guyz, time to flex those brain muscles on the next assignment! Link is here - kzbin.info/www/bejne/d5TFmZSLm9yCm7c

How did you think of this? Was it by drawing a recursive tree or you learnt and then implemented or you could come up with this approach?

I envy your handwriting haha! Good explanation.

nice

Great

Excellent job you have done sister may Allah bless you and grant you health wealth and success ameen regards.

great teaching style...easy to understand..we want more technology as JS,Html etc..i hope u understand maam..thanks for making leacture

Will you cover advance SQL like cte functions and subquery and joins and widows function?

Hi Mam ...Good afternoon Hope you are doing well