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Explain why count is use here

you used nested query because "it is advanced problem and is very good problem" fr.....you just saw the solution and came her to explain

Actual query requires lil bit of more explanation

Wehere use name = *

os ot possible i n mysql

this is unique code i think i wrote: select distict city from station where city like '[a,e,i,o,u]%' and '%[a,i,e,o,u]';

SELECT DISTINCT (city) FROM station WHERE ID%2=0;

Thanks a lot for uploading all solution

Super super super You teach us very understandable way Thank you so much 👍👍

SELECT CASE WHEN (A+B>C) AND (B+C>A) AND (C+A>B) THEN CASE WHEN (A=B) OR (B=C) OR (C=A) THEN 'Equilateral' WHEN (A=B) OR (B=C) OR (A=C) THEN 'Isosceles' ELSE 'Scalene' END ELSE 'Not A Triangle' END FROM TRIANGLES; Please let me know what's wrong here.

SELECT CONCAT(NAME,"(",SUBSTR(OCCUPATION,1,1),")") FROM OCCUPATIONS ORDER BY NAME ; select concat('There are a total of ', count(occupation)," ", occupation,"s.") from occupations group by occupation order by count(occupation); this query also same n , then why he not working

Why not longitude the condition that given below station table, also you are not ise and statement between city and station

I'm getting runtime Error select[Doctor],[Professor],[Singer],[Actor] from select(ROW_Number() Over (partition by Occupation ORDER BY Name)rn, Name, occupation FROM OCCUPATIONS group by occupation, name) AS tab2 pivot (max(Name) for occupation in ([Doctor],[Professor],[Singer],[Actor])AS tab1 order by rn) can you tell me why..?

bahut mast hai aapko=a channel

Is there any good training material ?

Geez, I was making that a lot more difficult than it needed to be!!!!

SELECT CONCAT(NAME, '(',LEFT(OCCUPATION,1),')') FROM OCCUPATIONS ORDER BY NAME SELECT CONCAT('There are a total of ', COUNT(OCCUPATION), ' ',LOWER(OCCUPATION), 's.') FROM OCCUPATIONS GROUP BY OCCUPATION ORDER BY COUNT(OCCUPATION), OCCUPATION this code you shown is incorrect

why iit is not runing

Hi mam i m writing my code below but there is an error coming error code : 1064 Check the manual that correspondant. Pls help select case when (a+b>c) and (a+C>b) and (b+c>a) then case when (a=b) and(b=c) and (c=a) then 'equilateral' when (a=b) or (b=c) or (c=a) or 'isosceles' else 'scelene' end end else 'not a triangle' end from Triangles

Why put occupation in last?

select distinct city from station where city not regexp '^[aeiou]|[aeiou]$' this works too

select distinct(city) from station where city NOT REGEXP '^[aeiou]' This works too

Very good explanation thank you mam

Try this solution also: select name from students where marks >75 order by substr(name, -3), id;

plz suggest from where i can learn sql advance queries

ur voice is very low and give an clear explanation by graphically

Well done!

super

select distinct CITY from STATION where id % 2=0; why showing error

Thanks for making this video about hackerrank certification.. how can we connect with your linkedin?

Why we use group by command?

nice way to divide drive in system

can we use REGEXP somehow?

I have a query could u please answer me

This is more easy ----The sum of the length of any two sides of a triangle is greater than the length of the third side. SELECT CASE WHEN A >= (B + C) OR B >= (A + C) OR C >= (A + B) THEN 'Not A Triangle' WHEN A = B AND A = C THEN 'Equilateral' WHEN A = B OR B = C OR A = C THEN 'Isosceles' ELSE 'Scalene' END FROM TRIANGLES;

Thanks so much , honestly this is the only query that made sense to me , I have a question why this only works with MS SQL and not Oracle , I'm still learning SQL and learned through some online videos that uses your logic in solving it , and is it essential to learn all syntax possible and how ?

Why we are using 's.' That is giving error

I love your videos but an explanation for this solution is highly required please provide.

right answer is : select city, length(city ) from station order by length(city), city limit 1; select city, length(city ) from station order by length(city) desc, city limit 1

ERROR 1064 (42000) at line 1: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '1 city, len(city) as length from station order by length asc, city asc select t' at line 1

solution for ms sql server

Please tell me what I do... For improving my skill..

Continue I am waiting for next video.

My all doubt clear..

Good teaching experience and i am satisfy with your topic and enjoy it.

Hey! Can you explain this? sachindewangan2000@gmail.com Here's my mail ID, would really appreciate if you could help me out.

nice

rn is for what?

Can the explanation be more elaborate? instead of just typing out the solution

nice