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
@LoveIndia3 Жыл бұрын
I learn and learned all tricky SQL part from your videos only..request you to make video on "sql Index" with practical examples
@techTFQ Жыл бұрын
thank you and noted. I'll get back to making tutorial videos after couple of months
@LoveIndia3 Жыл бұрын
@@techTFQ thanks a lot..will wait..
@shrabanchakmathisanimename8683 Жыл бұрын
@@techTFQbro sql install error in my pc
@pavanareddy4536 Жыл бұрын
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;
@mijailmariano7846 Жыл бұрын
Applicable, concise, and practical. Thank you 👏🏽
@smurfguy5035 Жыл бұрын
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
@avi8016 Жыл бұрын
Great explanation, got to learn about the ntile function today, thanks a lot sir!
@ArthurCittaAguiar1 Жыл бұрын
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!!
@sowmi3vinay583 Жыл бұрын
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
@ayazahamed82542 ай бұрын
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;
@partymaschine92 Жыл бұрын
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 👍
@putulsaini67886 ай бұрын
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.
@riasingh113 ай бұрын
your videos are the most helpful ones
@pramodkumar-kw7rj Жыл бұрын
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 😊
@rayees_thurkki Жыл бұрын
Please upload every week solve SQL projects, also could you please start a tutorial series for SQL 0 to 360 Degree❤
@udhaybhaskarbellamkonda167810 ай бұрын
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
@techTFQ8 ай бұрын
You're welcome
@sujaa10007 ай бұрын
Simply one of the best channels!!!
@rushikeshfargade6984 Жыл бұрын
I started watching your videos , very excellent explanation . will wait for more such videos.
@muthukamalan.m6316 Жыл бұрын
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
@arturoramirez712 Жыл бұрын
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?
@kaushalmzp Жыл бұрын
Liked your solution only ;-)
@techTFQ Жыл бұрын
Thank you brother 🙏🏼
@shaikusman536 Жыл бұрын
Awsome that was easyily done...thanks brother...
@TheVaibhavdang Жыл бұрын
Nice Solution Brother Can't we do it using case statements and then using concat and max ? Please Suggest
@techTFQ Жыл бұрын
there can be multiple diff soln for this. may be you try and share solution
@p10rambo Жыл бұрын
@@techTFQ how about this? select case id % 2 != 0: concat( concat(id, name), concat(lead(id), lead(name) ) from INPUT
@kandavel0588 Жыл бұрын
Extremely good instruction, bro
@techTFQ Жыл бұрын
Glad you think so!
@rajattalnikar6507 Жыл бұрын
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 Жыл бұрын
i just executed your solution and the output is incorrect
@rajattalnikar6507 Жыл бұрын
@@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;
@arunny9430 Жыл бұрын
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.
@Rizwan_Siddique Жыл бұрын
Thanks a lot, Great detailed video. Please make video detail video on Match_recognize() function.
@SwethaNandyala-sf9lt Жыл бұрын
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
@vablestory2.0 Жыл бұрын
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
@sreenucnu4641 Жыл бұрын
Excellent explanation bro.... 👍👍👍 Loved it
@rishikatakam7034 Жыл бұрын
Thanks for these kind of videos
@SANDATA764 Жыл бұрын
You are amazing bhai, thank you
@techTFQ Жыл бұрын
Thank you so much Ahmed 😀
@sravankumar1767 Жыл бұрын
Nice explanation bro 👍 👌 👏
@techTFQ Жыл бұрын
Thank you so much 🙂
@kishoriredekar8436 Жыл бұрын
You are awesome!
@rameshthanikonda5089 Жыл бұрын
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 Жыл бұрын
i have explained this a few times in different videos but may be will consider your suggestion for a future video
@rameshthanikonda5089 Жыл бұрын
@@techTFQ thank you..!! Toufiq. for your quick response
@rameshthanikonda5089 Жыл бұрын
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 Жыл бұрын
@@rameshthanikonda5089 is it pwc question?
@rameshthanikonda7027 Жыл бұрын
@@ankushsharma3965 I don't remember at the moment. If you know the answer please comment here. Thanks
@echodelta7680 Жыл бұрын
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.
@Sandeep_ThesatisfiedHuman Жыл бұрын
That’s grt bro , 🎉
@anumehaayushi7998 Жыл бұрын
Awesome explanation ❤️
@mrtharunprao Жыл бұрын
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 Жыл бұрын
depends on your job. some jobs like sql developers will need PL/SQL but a data analyst role will not need it
@mrtharunprao Жыл бұрын
@@techTFQ Thank you sir...
@mrtharunprao Жыл бұрын
@@techTFQ Sir can you please tell me how to practice SQL question for Analyst role...
@techTFQ Жыл бұрын
Check out StrataScratch and LeetCode
@mrtharunprao Жыл бұрын
@@techTFQ Sure sir Sir is that free or paid platform.. Just had last query.
@vikramsolanki1750 Жыл бұрын
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
@Sttuey Жыл бұрын
select string_agg(Concat(id, ' ', name), ', ') from emp_input group by id - case when id % 2 = 0 then 1 else 0 end;
@Sharmasurajlive Жыл бұрын
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 👍👍
@tenysebastian89258 ай бұрын
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
@dwarakanath2756 Жыл бұрын
Thank you so much Bro!!
@techTFQ Жыл бұрын
You're welcome!
@jaycorner3996 Жыл бұрын
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;
@aquibaslam5897 Жыл бұрын
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;
@zeeshanahmed2594 Жыл бұрын
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;
@cvakumart Жыл бұрын
Explaing phase is very fast, some may not catch up your speed, otherwise your explanation is very good.
@asavlogs8446 Жыл бұрын
Sir, Trigger, cursor aur user defined function par video banayiye
@srishtijain642 Жыл бұрын
please make video on USER DEFINED FUNCTIONS in sql
@AbhinavKumar-mm1ys Жыл бұрын
All I did differently was use ceil(id/2) as buckets
@techTFQ Жыл бұрын
nice 👍
@chintudg5367 Жыл бұрын
Please make videos on hive queries bigdata
@techTFQ Жыл бұрын
noted bro
@PRIYANKASharma-uv8lx Жыл бұрын
Thanks
@techTFQ Жыл бұрын
your welcome
@gowtham4383 Жыл бұрын
Thanks bro
@maximilianodelatorre2513 Жыл бұрын
AWSOMEEEEEEEEE
@akash4517 Жыл бұрын
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
@QuestKnow Жыл бұрын
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
@rose9466 Жыл бұрын
Can you make videos on product metrics
@amandeepmahlawat5785 Жыл бұрын
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;
@capper3360 Жыл бұрын
what if you do not know the number of rows are 8, how would you generalize for any number of rows?
@rishikatakam7034 Жыл бұрын
string_ agg it can be used in oracle
@nikhilavemuri955 Жыл бұрын
Awesome
@swapnilsolanki8595 Жыл бұрын
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 ;
@arijitnayak1593 Жыл бұрын
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
@thesunfamily7762 Жыл бұрын
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;
@user-lm1ft4yd3t Жыл бұрын
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;
@mahendranaik7895 Жыл бұрын
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
@hasanmougharbel8030 Жыл бұрын
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.
@life_style819 Жыл бұрын
String _agg function is not working in Oracle, it is showing invalid identifier. Why???
@muralikrishnachowdarypolin5601 Жыл бұрын
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 Жыл бұрын
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?
@vishalsonawane.89057 ай бұрын
Done
@Alexpudow7 ай бұрын
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
@tamyers28 Жыл бұрын
Hi, when is your next class?
@fathimafarahna2633 Жыл бұрын
👌🏻👌🏻👌🏻👌🏻👌🏻👌🏻
@techTFQ Жыл бұрын
Thank you :)
@arvindhh3720 Жыл бұрын
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';
@manis2831 Жыл бұрын
with cte as (select *, ntile(4) over() result1 from input) select group_concat(id," ",name) Result from cte group by result1
@asavlogs8446 Жыл бұрын
Make a video on views
@LoveIndia3 Жыл бұрын
It is already there kzfaq.info/get/bejne/mbKDq8Skrc2dh40.html
@techTFQ Жыл бұрын
already did. check in my channel
@abhishekva449 Жыл бұрын
Pls show answers in my sql too
@Vishal_19 Жыл бұрын
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
@user-pm2sz7lc6q4 ай бұрын
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;
@shahrukhsultan8727 Жыл бұрын
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 Жыл бұрын
A CTE and a derived table are essentially the same here and will produce an identical execution plan!
@harish7733 Жыл бұрын
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
@vikashagarwal4305 Жыл бұрын
I think we can do this by concat and self join also
@techTFQ Жыл бұрын
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
@sandhyagajaraj4826 Жыл бұрын
Can you please share notes for SSRS and SSIS
@venkatareddykunduru18287 ай бұрын
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 Жыл бұрын
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;
@saurabhkhare3315 Жыл бұрын
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 Жыл бұрын
The solution is pretty good, One possible optimization is to put that condition in the JOIN clause only instead of in the WHERE clause
@tahakhalid32420 күн бұрын
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
@bharatarora2006 Жыл бұрын
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
@ddvantandar-kw7kl Жыл бұрын
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.
@sayantabarik4252 Жыл бұрын
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;
@haleynguyen5721 Жыл бұрын
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;
@monasanthosh92084 ай бұрын
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;
@avinashgupta8627 Жыл бұрын
sir u are indian i think
@aakashmohan9827 Жыл бұрын
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
@apachekafka773 Жыл бұрын
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
@harshitsalecha221Ай бұрын
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;
@deepakmudigonda4155 ай бұрын
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
@user-hk9rv1qh5m Жыл бұрын
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 Жыл бұрын
SELECT id || name || ‘,’ , LEAD(id) OVER(), LEAD(name) OVER() FROM table_name WHERE MOD(id,2) 1