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Wow

Thanks for the video

with cte1 as( select letter_word as single_letter from Alphabetwords where LEN(letter_word)=1), cte2 as( select letter_word as multiple_letter from Alphabetwords where LEN(letter_word)>1) select distinct single_letter,multiple_letter from cte1,cte2 where SUBSTRING(single_letter,1,1) = SUBSTRING(multiple_letter,1,1);

this concept was new to me, thanks man

Nice explanation🙏

Excellent Explanation. Thanks a lot! 🙏

Thank u sir

Thanks for making this series🙏

SELECT DISTINCT CASE WHEN r1.source<r1.destination THEN r1.source ELSE r1.destination END as source, CASE WHEN r1.destination>r1.source THEN r1.destination ELSE r1.source END destination, r1.distance FROM route_distance as r1 INNER JOIN route_distance as r2 ON r1.source=r2.destination AND r1.destination=r2.source AND r1.distance=r2.distance UNION SELECT r1.* FROM route_distance as r1 INNER JOIN route_distance as r2 ON r1.source=r2.destination AND r1.destination=r2.source AND r1.distance<>r2.distance;

Very nice

with cte as( select * , case when runs_scored >= 50 then 1 else 0 end as flag from matches where player_id not in (select player_id from matches where runs_scored = 0) ) select p.name, sum(c.runs_scored) as Runs from cte c join players p on c.player_id = p.id group by p.name having sum(flag) >= 2

Excellent explained

Thnxxx 💯

Thnxx 💯

Select Player Name, sum( Runs_scored) as runs from matches m join players p on m.player_id=p.id where not exists ( select 1 from matches where runs=0 and m.player_id=player_id) group by player_id, Player_name having count( case when runs_scored>=50 then 1 end) >=2

when we write a.source<a.destination on what basis it is filtering on ascii value?

Very nice experience.. excellent

with cte as(select player_id,substr(name,1,instr(name,' ')-1) as first_name,substr(name,instr(name,' ')+1) as last_name from player) select case when mod(cte.player_id,2)=0 then first_name else last_name end as name from cte order by (case when mod(cte.player_id,2)=0 then first_name else last_name end);

SELECT player_name FROM (SELECT CASE WHEN player_id%2<>0 THEN substring_index(name," ",-1) ELSE substring_index(name," ",1) END player_name FROM player_information) as a ORDER BY player_name;

sir i have a question what type of sql questions they ask for fresher role in data analytics interview ? please guide me

with cte as( select player_id, SUBSTRING(name, 1, CHARINDEX(' ', name)-1) first_name, SUBSTRING(name, CHARINDEX(' ', name)+1, LEN(name) - CHARINDEX(' ', name)) last_name from player_information ) select case when player_id%2 = 0 then first_name else last_name end Name from cte order by (case when player_id%2 = 0 then first_name else last_name end)

Good explanation

with cte as(SELECT *, case when length(substring(letter_word,1))> 1 then substring(letter_word,1,1) else '' end as stat FROM alphabetwords) select c.letter_word as Alphabet, c1.letter_word as Words from cte as c join cte as c1 on c.letter_word = c1.stat order by 1;

select a1.letter_word as l1, a2.letter_word l2 from Alphabetwords a1 join Alphabetwords a2 on a1.letter_word != a2.letter_word where l1 = substring(l2, 1,1)

My approach select * from words w1 join words w2 on w1.letter_word=left(w2.letter_word,1) and length(w1.letter_word)!=length(w2.letter_word) order by w1.letter_word;

Thanks sir

with t1 as ( select *,dense_rank()over(order by match_id) as rn from wc_matches) select * from wc_players as a join (select player_id, count(distinct match_id) from t1 where rn<=3 group by player_id having count(distinct match_id)=3)b on a.id=b.player_id

here is my approch using self join :- select a.letter_word as Alphabet,b.letter_word as Words from Alphabetwords as a inner join Alphabetwords as b on a.letter_word = left(b.letter_word,1) and len(b.letter_word) <> 1 order by Alphabet;

with cte as ( select *, case when length(letter_word) = 1 then letter_word end as letter, case when length(letter_word)>1 then letter_word end as word from Alphabetwords order by case when length(letter_word) = 1 then letter_word end ) select c1.letter, c2.word from cte c1 inner join cte c2 on c1.letter = substring(c2.word,1,1) OR SIMPLE SOLUTION: select * from Alphabetwords A join Alphabetwords B on A.letter_word = substring(B.letter_word,1,1) and length(A.letter_word)<length(B.letter_word)

Nice sir

nice explanation sir...

Its an excellent db open source tool....i used this for 3 years...

nice

SELECT c1.letter_word, c2.letter_word FROM alphabetwords as c1 INNER JOIN alphabetwords as c2 ON c1.letter_word=LEFT(c2.letter_word,1) AND length(c2.letter_word)>1 ORDER BY c1.letter_word;

bro your voice is very low

My solution: SELECT a.letter_word AS Alphabet, b.letter_word AS Word FROM Alphabetwords a JOIN Alphabetwords b ON a.letter_word = SUBSTRING(b.letter_word,1,1) AND LEN(a.letter_word) < LEN(b.letter_word) ORDER BY a.letter_word

I think u have used joined tables from 3 corrsponding tables to use over here

can we do it buy partition of productid instead of customerid . I was trying found different results as against customerid. I suppose table is not in 2nf form thats why confusing

Bro why your video has hell of 1000 adds in every video

using subquery ✅- select user_id , (100.0 * followers/(select count(*) as total_users from ( select distinct user_id as users from famous union select distinct follower_id as users from famous)) ) as followers_perc from ( select user_id , count(*) as followers from famous group by user_id order by user_id) x

this was tough