01 SQL Query To Find Nth Highest Salary | SQL Interview Question and answers

In this video I will show you how to write :
SQL Query To Find Nth Highest Salary
Top 3 highest salary in SQL
SQL Interview Question and answers
Download the SQL scripts: github.com/rohanbs1/sql
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[@snowguo1786]

This explain very well on the question using 4 approaches: groupby, row_number, rank, and CTE. Will watch again to review. Thank you!

[@pankajbhagwatkar6720]

Thank you sir for keenly elaborating all those functions and synopsis of queries.... 😇

[@LearningwithRohan]

You’re welcome.. keep learning and growing

[@durgaprasadt7162]

Nice video sir, thank you 💐🙏

[@LearningwithRohan]

You are welcome

[@vru5696]

Can you help in the below questions. 1. Sales(Table) Sale(Column) 10 20 30 -5 -25 Output Profit Loss 60 30 2.Input A 456 467 B 478 478 Output A 456 B 467 b 478 3. Product(Table) id, Name, Product 10,abc,Shirt 20,bcd,Jeans 30,Jkl,Cap 40,Lmn,Shirt 10,abc,Jeans 20,bcd,Shirt Write a query to find out customers who bought both jeans and Shirt?

[@LearningwithRohan]

Sure will do shortly

[@LearningwithRohan]

I mean tomorrow

[@LearningwithRohan]

Here is the SQL for first question with sale as ( select 10 sales union select 20 sales union select 30 sales union select -5 sales union select -25 sales ) select sum(case when sales > 0 then sales else 0 end) profit, sum(case when sales < 0 then sales else 0 end) loss from sale

[@LearningwithRohan]

with cte as ( select 10 id,'abc' as cust_name,'Shirt' as product union select 20 id,'bcd' as cust_name,'Jeans' as product union select 30 id,'Jkl' as cust_name,'Cap' as product union select 40 id,'Lmn' as cust_name,'Shirt' as product union select 10 id,'abc' as cust_name,'Jeans' as product union select 20 id,'bcd' as cust_name,'Shirt' as product ) select id, cust_name , 'Purchased both' sales--, count(product) from cte where product in ('Shirt','Jeans') group by id, cust_name having count(product)>1

[@LearningwithRohan]

Here is the SQL for 2nd one with cte as ( select 'A' names, 456 as value1, 467 as value2 union select 'B' names, 478 as value1, 478 as value2 ) select names, value1 from cte union select names, value2 from cte