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SQL Interview Problem asked by Product Based Company | Solving SQL Interview Query 

techTFQ
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21 авг 2024

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Комментарии : 149   
@angmai3543
@angmai3543 Год назад
Thanks for your video. I try to solve this problem first and here are my code. select id_name + ', ' + name2 as result from (select *, lead(id_name) over (order by id) as name2 from (select *, cast(id as varchar) + ' ' + name as id_name from emp_input) x) y where id % 2 0
@smurfguy5035
@smurfguy5035 Год назад
Table name is input SELECT concat(a.id,' ',a.name,' ',b.id,' ',b.name) from input as a JOIN input as b On a.id < b.id WHERE a.id%2 = 1 and b.id%2 = 0 and b.id- a.id = 1
@LoveIndia3
@LoveIndia3 Год назад
I learn and learned all tricky SQL part from your videos only..request you to make video on "sql Index" with practical examples
@techTFQ
@techTFQ Год назад
thank you and noted. I'll get back to making tutorial videos after couple of months
@LoveIndia3
@LoveIndia3 Год назад
@@techTFQ thanks a lot..will wait..
@shrabanchakmathisanimename8683
​@@techTFQbro sql install error in my pc
@pavanareddy4536
@pavanareddy4536 Год назад
Loved your solution! It is a superior solution for the problem. I tried solving it before watching your solution and came up with this. If record is odd, combine text from this row and leading row. So much to improve! thanks for everything! select * from (select case when id%20 then id||' '||name||', '|| lead(id) over (order by id) || ' '|| lead(name) over (order by id) end as res from input) x where x.res IS NOT null;
@ayazahamed8254
@ayazahamed8254 2 месяца назад
Very Well Explained. MySql Solution: SELECT GROUP_CONCAT( CONCAT_WS(', ', CONCAT(id,' ',name))) as Result FROM ( SELECT id, name, ntile(20) over(order by name) as grp FROM `workers` ) as temp GROUP BY grp;
@mijailmariano7846
@mijailmariano7846 Год назад
Applicable, concise, and practical. Thank you 👏🏽
@SwethaNandyala-sf9lt
@SwethaNandyala-sf9lt Год назад
with odd as ( select * from emp_input where id%20 ), even as ( select * from emp_input where id%2=0 ) select concat(odd.id," ",odd.name,",", even.id," ",even.name)as output from odd,even where even.id = odd.id+1 this is my solution
@sowmi3vinay583
@sowmi3vinay583 Год назад
Logic is this with cte1 as (select * from emp_input where id%2=0), cte2 as (select * from emp_input where id%20) select * from cte1 a inner join cte2 b on a.id=b.id+1
@avi8016
@avi8016 Год назад
Great explanation, got to learn about the ntile function today, thanks a lot sir!
@partymaschine92
@partymaschine92 Год назад
I haven‘t really verified my thoughts but I‘d rather go with the mod function to create buckets. In this case you won’t have to speficy how many buckets you need (you just need to think of the size of each bucket and use the row number as numerator). For this exact problem though, is your way of solving it, just perfect! Thanks for sharing 👍
@rajattalnikar6507
@rajattalnikar6507 Год назад
My Simple Solution with Logic :- Simply assign group number to all the records such that Emp1 and Emp2 go to group 1, Emp3 and Emp4 go to group 2 and so on. This can be simply done by dividing their serial number by 2 and then doing a ceil on it to ensure it is an integer. Now you can apply group concat based on this column. create table data( ID int, Name varchar(20) ); insert into data values (1,'Emp1'), (2,'Emp2'), (3,'Emp3'), (4,'Emp4'), (5,'Emp5'), (6,'Emp6'), (7,'Emp7'), (8,'Emp8'); select grp_no,group_concat(res_col) from( select concat(ID,' ',Name) as res_col,ceil(ID/2) as grp_no from data ) A group by grp_no;
@techTFQ
@techTFQ Год назад
i just executed your solution and the output is incorrect
@rajattalnikar6507
@rajattalnikar6507 Год назад
@@techTFQ How I just ran it again and its coming as expected. You might be seeing one extra column for grp_no which can be eliminated easily by using the query inside a subquery. select output from (select grp_no, group_concat(res_col) as output from (select concat(ID, ' ', Name) as res_col, ceil(ID/2) as grp_no from data) A group by grp_no) B;
@ArthurCittaAguiar1
@ArthurCittaAguiar1 Год назад
I have learned a lot from your videos! A lot of things that I thought were complex and that were actually simple to understand. Thank you so much for everything and congratulations for the channel!!
@putulsaini6788
@putulsaini6788 6 месяцев назад
sir, make more videos on solving interview question , best teaching method and the way you explain each and every line is very much understandable. Thank you sir.
@QuestKnow
@QuestKnow Год назад
I have tried this..! Select Convert(varchar,X.ID)+' '+X.name+', '+ Convert(varchar,X.a) +' '+ b From ( Select ID, name, Lead(ID,1) Over(order by ID asc) a, Lead(Name,1) Over(order by Name asc) b From Input )X Where X.ID%20 Order by 1 asc
@vablestory2.0
@vablestory2.0 Год назад
I solved it using self join with cte as( select e1.id id1,e2.id id2 ,e1.name name1 ,e2.name name2 from emp_input e1 left join emp_input e2 on e1.id+1 = e2.id where e2.name is not null and e1.id % 2 0 ) select concat(id1,name1,',',id2,name2) output from cte
@riasingh11
@riasingh11 3 месяца назад
your videos are the most helpful ones
@Alexpudow
@Alexpudow 7 месяцев назад
select concat(a.id,' ',a.name,', ',b.id,' ',b.name) from (select * from emp_input where id%20) a join (select * from emp_input where id%2=0) b on a.id=b.id-1
@sujaa1000
@sujaa1000 6 месяцев назад
Simply one of the best channels!!!
@vikramsolanki1750
@vikramsolanki1750 Год назад
WITH CTE_1 (col_id,Name, LeadID, LeadName) as ( select col_id, Name, lead(col_id) over (order by col_id) LeadID , lead(Name) over (order by col_id) LeadName from demo2 order by col_ID ), CTE_2 (CIS) as -- since CASE CIS can not be used in where clause ( select (case when MOD(col_id,2) =1 then col_id || Name || ' , ' || LeadID || LeadName Else NULL END) as CIS from CTE_1 ) -- WHERE is processed before SELECT. It doesn't know what DerviedRegion is at that point. select * from CTE_2 where CIS IS NOT NULL
@pramodkumar-kw7rj
@pramodkumar-kw7rj Год назад
I saw one video of yours and I completely understood Sir I want to learn SQL in a proper manner can please suggest me your playlist which one I need to refer . Hope you will reply 😊
@udhaybhaskarbellamkonda1678
@udhaybhaskarbellamkonda1678 9 месяцев назад
Your Videos are very helpful, gained more knowledge through your videos and I like the way you are explaining step by step and providing every dataset for practise ,Thanks a lot
@techTFQ
@techTFQ 8 месяцев назад
You're welcome
@rayees_thurkki
@rayees_thurkki Год назад
Please upload every week solve SQL projects, also could you please start a tutorial series for SQL 0 to 360 Degree❤
@Sharmasurajlive
@Sharmasurajlive Год назад
My approach using CTE, concat, MOD and join :- create table data (id int, name text); insert into data values (1,'Emp1'); insert into data values (2,'Emp2'); insert into data values (3,'Emp3'); insert into data values (4,'Emp4'); insert into data values (5,'Emp5'); insert into data values (6,'Emp6'); insert into data values (7,'Emp7'); insert into data values (8,'Emp8'); solution :- with set1 as (select *,row_number() over() as rec, concat(id,' ',name) as id_name from data where id%2 0), set2 as (select *,row_number() over() as rec, concat(id,' ',name) as id_name from data where id%2 = 0) select concat(s1.id_name,', ',s2.id_name) as Result from set1 s1 join set2 s2 on s1.rec=s2.rec; Thanks @techTFQ for such efforts 👍👍
@aquibaslam5897
@aquibaslam5897 Год назад
we can try like this also with cte as ( select *, concat(id,+' '+ name ) as nm from emp_input) ,final as ( select *, lead (nm,1)over (order by id) as new from cte ) select concat(nm,+','+new), id from final WHERE id%2 0;
@jaycorner3996
@jaycorner3996 Год назад
easy method use case statement for oracle select*from(select case when id=1 then '1 emp1,2 emp2' when id=2 then '3 emp3,4 emp4' when id=3 then '5 emp5,6 emp6' when id=4 then '7 emp7,8 emp8' end finalre from emp_input) where finalre is not null;
@tenysebastian8925
@tenysebastian8925 7 месяцев назад
Hi Thoufiq, I came up with this solution without using any window functions or cte SELECT x.empo||', '||y.empe AS RESULT FROM (SELECT id, id||' '||name empo FROM input i1 WHERE MOD(id, 2) != 0) x , (SELECT id, id||' '||name empe FROM input i2 WHERE MOD(id, 2) = 0) y WHERE y.id - x.id = 1
@arturoramirez712
@arturoramirez712 Год назад
You didn't use the within group order by clause in the string aggregate function. It's possible to get 2 Emp2, 1 Emp1, etc. I know because that is what I got when I left out the order clause select string_agg (concat (ID,' ', Name) , ', ' ) within group (order by ID) as output from emp_input group by round(ID/2.0, 0) /* create buckets */ What about creating one long string and then cutting it into distinct pieces...is that possible with no grouping or window clause?
@kandavel0588
@kandavel0588 Год назад
Extremely good instruction, bro
@techTFQ
@techTFQ Год назад
Glad you think so!
@kaushalmzp
@kaushalmzp Год назад
Liked your solution only ;-)
@techTFQ
@techTFQ Год назад
Thank you brother 🙏🏼
@thesunfamily7762
@thesunfamily7762 Год назад
The query below works in MySQL. with a1 as ( select CONCAT(ID,'',NAME,', ') as nlist, row_number() over (order by ID) as RN from emp_input where id%2=1 ), a2 as ( select CONCAT(ID,'',NAME) as nlist, row_number() over (order by ID) as RN from emp_input where id%2=0 ) select concat(a1.nlist,a2.nlist) from a1 join a2 on a1.RN=a2.RN order by a1.RN; This also works in MySQL based on one comment under this video: SELECT CONCAT(A.id, ' ', A.name, ', ',B.id, ' ', B.name) FROM emp_input AS A INNER JOIN emp_input AS B ON A.id + 1 = B.id and A.id % 2 0;
@manis2831
@manis2831 Год назад
with cte as (select *, ntile(4) over() result1 from input) select group_concat(id," ",name) Result from cte group by result1
@arunny9430
@arunny9430 Год назад
As an absolute beginner who started learning simple select statements, my solution was: select concat(id,name, ',' , id+1,' Emp',id+1) as Output from tbl where mod(id,2) 0; of course, this fails miserably if all names are not simply Emp and number. So, if we want to stick to very basic syntax, is it possible to concat outputs from two select statements (one with mod(id,2) 0 and the other mod(id,2) =0 in anyway? I love this site and I am sure I will be spending a lot of time here.
@rushikeshfargade6984
@rushikeshfargade6984 Год назад
I started watching your videos , very excellent explanation . will wait for more such videos.
@Vishal_19
@Vishal_19 Год назад
with cte as (select concat(id,' ',name) as result from input) , cte2 as (select result, lead(result,1) over(order by result) end from cte) select concat(result,', ',end) as result from cte2 where result%2 != 0
@shaikusman536
@shaikusman536 Год назад
Awsome that was easyily done...thanks brother...
@Sttuey
@Sttuey Год назад
select string_agg(Concat(id, ' ', name), ', ') from emp_input group by id - case when id % 2 = 0 then 1 else 0 end;
@venkatareddykunduru1828
@venkatareddykunduru1828 7 месяцев назад
This is easiest solution you will ever find for this question: select concat(row1,' ',name1,',',row2,' ',name2) as result from (select e1.id as row1,e1.name as name1,e2.id as row2,e2.name as name2 from emp_input e1 join emp_input e2 on e1.id+1=e2.id where row2%2==0)
@subodhthore6454
@subodhthore6454 Год назад
My Approach before watching the solution: with cte as (select *, concat(id ,name) as hello from productBased), cte1 as (select id,name, lag(hello) over(order by name) as hello2 ,hello from cte) select concat(hello2,",",hello) from cte1 where id%2=0;
@TheVaibhavdang
@TheVaibhavdang Год назад
Nice Solution Brother Can't we do it using case statements and then using concat and max ? Please Suggest
@techTFQ
@techTFQ Год назад
there can be multiple diff soln for this. may be you try and share solution
@p10rambo
@p10rambo Год назад
@@techTFQ how about this? select case id % 2 != 0: concat( concat(id, name), concat(lead(id), lead(name) ) from INPUT
@zeeshanahmed2594
@zeeshanahmed2594 Год назад
My solution (applicable only when total number of records in input table is even) with cte as (select id, concat(id, name, ', ', lead(concat(id, name)) over (order by id)) as output from input) select output from cte where cte.id%2 = 1;
@muthukamalan.m6316
@muthukamalan.m6316 Год назад
thanks a lot sir. it would be great if you post list of interview based sal questions and solutions as you did before. it really helped me
@sreenucnu4641
@sreenucnu4641 Год назад
Excellent explanation bro.... 👍👍👍 Loved it
@rishikatakam7034
@rishikatakam7034 Год назад
Thanks for these kind of videos
@SANDATA764
@SANDATA764 Год назад
You are amazing bhai, thank you
@techTFQ
@techTFQ Год назад
Thank you so much Ahmed 😀
@swapnilsolanki8595
@swapnilsolanki8595 Год назад
My solution SELECT id || ' ' || name || ',' || id_lead || ' ' || name_lead from ( SELECT id, lead (id,1,id) over (order by id) as id_lead, name, lead (name,1,name) over (order by id) as name_lead, mod(id,2) as flag from tablename ) as X where X.flag 0 ;
@Sandeep_ThesatisfiedHuman
@Sandeep_ThesatisfiedHuman Год назад
That’s grt bro , 🎉
@sravankumar1767
@sravankumar1767 Год назад
Nice explanation bro 👍 👌 👏
@techTFQ
@techTFQ Год назад
Thank you so much 🙂
@arijitnayak1593
@arijitnayak1593 Год назад
with a as (select x.id as id1,x.name as name1,y.id as id2,y.name as name2, ROW_NUMBER() over (partition by x.id order by x.id) as rn from p3_tab as x inner join p3_tab as y on (x.id%2)!=0 and x.id
@Rizwan_Siddique
@Rizwan_Siddique Год назад
Thanks a lot, Great detailed video. Please make video detail video on Match_recognize() function.
@kishoriredekar8436
@kishoriredekar8436 Год назад
You are awesome!
@user-pm2sz7lc6q
@user-pm2sz7lc6q 3 месяца назад
It will work with any dataset as long as the total count is in even. .. .. WITH cte1 AS ( SELECT (COUNT(*)/2)::INT AS total FROM input), cte2 AS ( SELECT (id || ' ' || name) AS result, NTILE(total) OVER(ORDER BY id ASC) AS pairs FROM input, cte1) SELECT STRING_AGG(result, ', ') AS RESULT FROM cte2 GROUP BY pairs ORDER BY 1 ASC;
@anumehaayushi7998
@anumehaayushi7998 Год назад
Awesome explanation ❤️
@amandeepmahlawat5785
@amandeepmahlawat5785 Год назад
for oracle sql select listagg(full_name,',') from (select employee_id||first_name as full_name, ntile(4) over(order by employee_id) as bucket from emps) group by bucket;
@dwarakanath2756
@dwarakanath2756 Год назад
Thank you so much Bro!!
@techTFQ
@techTFQ Год назад
You're welcome!
@echodelta7680
@echodelta7680 Год назад
Good evening. I am teaching myself MySQL on Workbench to apply for Data Analyst profile later on. Do you have any compilation of relevant SQL problems that I can use to practice along? Thanks.
@maximilianodelatorre2513
@maximilianodelatorre2513 Год назад
AWSOMEEEEEEEEE
@akash4517
@akash4517 Год назад
HI Taufiq , Sharing my solution %sql WITH CTE AS (select *,concat(id,' ',name) as new_val from emp_input) ,CTE1 AS (select *,concat(new_val,',',lead(new_val,1,'NothingtoADD') over(order by id)) as new_val2 from CTE) select * from CTE1 where id%2!=0
@bharatarora2006
@bharatarora2006 Год назад
select concat(B.Id -1,B.Name,",",A.Id, A.Name) from (select * FROM Employee where Mod(Id,2)=0 ) A inner join (select Id+1 as Id,Name FROM Employee where Mod(Id,2)!=0 ) B on A.Id = B.Id
@user-lm1ft4yd3t
@user-lm1ft4yd3t Год назад
select case when id%2 =1 then out else null end as result from ( select *,CONCAT(id,' ',name) +','+ lead (CONCAT(id,' ',name),1) over (order by id) as out from emp_input) a where case when id%2 =1 then out else null end is not null;
@nikhilavemuri955
@nikhilavemuri955 Год назад
Awesome
@AbhinavKumar-mm1ys
@AbhinavKumar-mm1ys Год назад
All I did differently was use ceil(id/2) as buckets
@techTFQ
@techTFQ Год назад
nice 👍
@haleynguyen5721
@haleynguyen5721 Год назад
Here my solution :)) select concat(x.id, x.name, ',', x.lead_id, x.lead_name) as result from ( select * , lead(id,1) over() as lead_id , lead(name,1) over() as lead_name from emp_input e1 ) x where mod(x.id, 2) 0;
@PRIYANKASharma-uv8lx
@PRIYANKASharma-uv8lx Год назад
Thanks
@techTFQ
@techTFQ Год назад
your welcome
@rameshthanikonda5089
@rameshthanikonda5089 Год назад
Hi Toufiq, what is the difference between order by(column_name) and order by 1. along with the difference between count(column_name), count(*) and count 1. kindly do one separate video. Thanks
@techTFQ
@techTFQ Год назад
i have explained this a few times in different videos but may be will consider your suggestion for a future video
@rameshthanikonda5089
@rameshthanikonda5089 Год назад
@@techTFQ thank you..!! Toufiq. for your quick response
@rameshthanikonda5089
@rameshthanikonda5089 Год назад
also please consider the below as well. Hi Toufiq, I attended sql interview and the interviewer asked me below sql query. Kindly help me with a short video so that i'll build some confidence and will crack the next interviews. Thanks SQL: Table A with column C1 has 7 records I. E 1,1,1,1,1,Null, Null and Table B with column C2 has 5 records I. E 1,1,1,2,Null. How records we will fetch by using inner join, left join, right join and full join.
@ankushsharma3965
@ankushsharma3965 Год назад
@@rameshthanikonda5089 is it pwc question?
@rameshthanikonda7027
@rameshthanikonda7027 Год назад
@@ankushsharma3965 I don't remember at the moment. If you know the answer please comment here. Thanks
@mrtharunprao
@mrtharunprao Год назад
Sir I have learned Oracle SQL, so my question is that learning PL/SQL will give me a added advantage or it's not that necessary
@techTFQ
@techTFQ Год назад
depends on your job. some jobs like sql developers will need PL/SQL but a data analyst role will not need it
@mrtharunprao
@mrtharunprao Год назад
@@techTFQ Thank you sir...
@mrtharunprao
@mrtharunprao Год назад
@@techTFQ Sir can you please tell me how to practice SQL question for Analyst role...
@techTFQ
@techTFQ Год назад
Check out StrataScratch and LeetCode
@mrtharunprao
@mrtharunprao Год назад
@@techTFQ Sure sir Sir is that free or paid platform.. Just had last query.
@shahrukhsultan8727
@shahrukhsultan8727 Год назад
This can be done without using CTE: select string_agg(name, ',') as Result from (select concat(id, ' ' , name) as name, ntile(4) over() as bucket from emp_input) as innerTable group by bucket order by bucket
@Sttuey
@Sttuey Год назад
A CTE and a derived table are essentially the same here and will produce an identical execution plan!
@fathimafarahna2633
@fathimafarahna2633 Год назад
👌🏻👌🏻👌🏻👌🏻👌🏻👌🏻
@techTFQ
@techTFQ Год назад
Thank you :)
@asavlogs8446
@asavlogs8446 Год назад
Sir, Trigger, cursor aur user defined function par video banayiye
@srishtijain642
@srishtijain642 Год назад
please make video on USER DEFINED FUNCTIONS in sql
@gowtham4383
@gowtham4383 Год назад
Thanks bro
@chintudg5367
@chintudg5367 Год назад
Please make videos on hive queries bigdata
@techTFQ
@techTFQ Год назад
noted bro
@cvakumart
@cvakumart Год назад
Explaing phase is very fast, some may not catch up your speed, otherwise your explanation is very good.
@rishikatakam7034
@rishikatakam7034 Год назад
string_ agg it can be used in oracle
@rose9466
@rose9466 Год назад
Can you make videos on product metrics
@dedeegal
@dedeegal Год назад
Mine is shorter and more readable: --select t1.id, t1.name, t2.id, t2.name select t1.id || " " || t1.name || ", " || t2.id || " " || t2.name -- sqlite3-syntax from emp_input t1 join emp_input t2 on t1.id+1=t2.id where (t1.id-1)%2=0 But both are rubbish for the general case.
@vishalsonawane.8905
@vishalsonawane.8905 7 месяцев назад
Done
@monasanthosh9208
@monasanthosh9208 3 месяца назад
MySQL Solution (For Freshers) With CTE as (Select *,concat(id," ",Name) as N1 from emp_input), CTE2 as (Select *,Lead(N1,1) Over (Order by id) as N2,row_number() over (Order By id) as RN from CTE) Select concat(N1,",",N2) as Result from CTE2 Where N2 is not null and Rn%2!=0;
@harish7733
@harish7733 Год назад
with temp as ( select id,case when mod(id,2)=1 then id+1 else id end new_id, id||' '||name as new_name from emp) select listagg(new_name,',') from temp group by new_id
@asavlogs8446
@asavlogs8446 Год назад
Make a video on views
@LoveIndia3
@LoveIndia3 Год назад
It is already there ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-cLSxasHg9WY.html
@techTFQ
@techTFQ Год назад
already did. check in my channel
@hasanmougharbel8030
@hasanmougharbel8030 Год назад
Hey there, God bless your efforts. I am still new to sql with a general enquiry. How non-clustered index differs from clustered index? It has anything to do with grouping of data while indexing? Thanks a lot.
@apachekafka773
@apachekafka773 Год назад
WITH t1 as(select *, id||' '||employee as joined FROM sample), t2 as (SELECT id, joined, LEAD(joined,1) over () as result from t1) select joined||', '||result from t2 WHERE (id*1.0)%2 != 0
@capper3360
@capper3360 Год назад
what if you do not know the number of rows are 8, how would you generalize for any number of rows?
@arvindhh3720
@arvindhh3720 Год назад
select full_data from (select concat_ws(" ",id,name, lead(id,1,'empty') over(), lead(name,1,'no_name') over()) as full_data,lead(id,1,'empty') over() as num from emp_input) new_table where mod(num,2) = 0 and num not like 'empty';
@tahakhalid324
@tahakhalid324 14 дней назад
My Code for that problem with combines as ( select id, name, concat(id,' ',name) as concats from emp_input ) , rnking as ( SELECT id, name, concats, row_number()over(order by concats) as rnks from combines ), new_rnking as ( SELECT id, name, concats, rnks, case when rnks%2=0 then rnks - 1 else rnks end as new_rnks from rnking ) SELECT GROUP_concat(concats separator ',') as Result from new_rnking GROUP by new_rnks
@mahendranaik7895
@mahendranaik7895 Год назад
Input id product 1 Choc 2 cadb 3 dairy 4 Choc 5 ecli 6 Choc 7 milky Output id product Sta 1 Choc 1 2 cadb 2 3 dairy 3 4 Choc 1 5 ecli 2 6 Choc 1 7 milky 2 when product choc encountered irteration should reset , can you help to provide a saolution
@life_style819
@life_style819 Год назад
String _agg function is not working in Oracle, it is showing invalid identifier. Why???
@harshitsalecha221
@harshitsalecha221 Месяц назад
WITH cte1 AS (SELECT *, CONCAT(id,name) as mer_name, CASE WHEN id%2=0 THEN id ELSE LEAD(id) OVER(ORDER BY id) END ids FROM emp_input) SELECT GROUP_CONCAT(mer_name) as results FROM cte1 GROUP BY ids;
@ddvantandar-kw7kl
@ddvantandar-kw7kl Год назад
We Hire dedicated teacher. Here is a jsp page and on the other hand asp page now here is a store procedure that need to be invoked. What purpose it will solve this is none of your business. Time will explain you.
@tamyers28
@tamyers28 Год назад
Hi, when is your next class?
@abhishekva449
@abhishekva449 Год назад
Pls show answers in my sql too
@saurabhkhare3315
@saurabhkhare3315 Год назад
Hi Toufiq, Please check this solution : SELECT CONCAT(A.id, ' ', A.name, ', ', B.id, ' ', B.name) AS Req_Result FROM emp_input AS A INNER JOIN emp_input AS B ON A.id + 1 = B.id WHERE (A.id % 2 0) ;
@chintanmistri7747
@chintanmistri7747 Год назад
The solution is pretty good, One possible optimization is to put that condition in the JOIN clause only instead of in the WHERE clause
@muralikrishnachowdarypolin5601
Hi help me on this how to create unique id but I'd contains date-number, numbers is reset when the date is changed. Number start from 1
@sabarinathan6220
@sabarinathan6220 Год назад
Bro from last two days I am trying to make payment for the course... tried with four different cards ...lighthall site simply says card is declined...Is there any other way to make payment?
@vikashagarwal4305
@vikashagarwal4305 Год назад
I think we can do this by concat and self join also
@techTFQ
@techTFQ Год назад
i think we can do this in 10-15 different ways. not only this problem most of sql problems can be solved in multiple different ways
@aakashmohan9827
@aakashmohan9827 Год назад
my approach select (emp_in||','||leadd )as Result from ( select emp_in,lead(emp_in) over()as leadd,row_number() over() as row_no from (select (id||' '||name)as emp_in from emp_input)as a ) as aa where mod(row_no,2)!=0
@sandhyagajaraj4826
@sandhyagajaraj4826 Год назад
Can you please share notes for SSRS and SSIS
@user-hk9rv1qh5m
@user-hk9rv1qh5m Год назад
HI, here is my version of solution for MYSQL with cte1 as (SELECT concat(id," ",name) as result from emp_input), CTE2 as (SELECT result, LEAD (result) over() as result2 FROM cte1) SELECT concat(result, "," " ",result2) as final_result FROM cte2 where mod(concat(result, "," " ",result2),2)=1
@theshanerap
@theshanerap Год назад
SELECT id || name || ‘,’ , LEAD(id) OVER(), LEAD(name) OVER() FROM table_name WHERE MOD(id,2) 1
@sayantabarik4252
@sayantabarik4252 Год назад
Here is my Solution - select * from ( select case when id%2!=0 then concat(id,' ',name,', ',lead(id) over(order by id),' ',lead(name) over(order by id)) end as RESULT from emp_input ) a where a.result is not null;
@deepakmudigonda415
@deepakmudigonda415 5 месяцев назад
The solution you've provided is not viable when the number of rows change. Here is my solution which is not dependent on number of rows. with cte as (select *,case when id % 2 = 0 then 0 else 1 end as num from emp_input), t as( select *, rank() over(partition by num order by id) as rk from cte) select concat(t2.id,' ',t2.name,',',t1.id,' ',t1.name) as result from t t1 join t t2 on t1.rk =t2.rk and t1.id t2.id group by t1.rk
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