December 10, 2014
Distinct, AVG , COUNT how all behaves if null value exist in table
These are all basics but when asked in interview we make mistake.
select * from #temp
select AVG(id) from #temp [5/3 or 5/5]
select COUNT(*) from #temp [5 or 3]
select distinct(id) from #temp [1,2 or 1,2,null]
select * from #temp
select AVG(id) from #temp [5/3 or 5/5]
select COUNT(*) from #temp [5 or 3]
select distinct(id) from #temp [1,2 or 1,2,null]
![HOW TO FIND Nth HIGHEST NUMBER THE COMMON INTERVIEW QUESTION. [OFFSET/FETCH USE TO FIND THE SAME]](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhwH2YIGCemIzMO467RzSpbE8dL2J9lPMFMUrkD7E2rzkDkiIEXPOzF98lJA9pdR3uMydxI7EUgro7xEG5TfLwSO4BbGunLplVsrBrL7pUGOjkaJ2ZUybINsZQK9n3MYFYPIatxAMW8vaQN/s72-c/1.PNG)
Hello, my name is Jack Sparrow. I'm a 50 year old self-employed Pirate from the Caribbean.