The below simplified query worked for me in mysql with cte as (select *, row_number() over (partition by username order by startdate) as rn, count(*) over (partition by username) as ct from useractivity) select * from cte where rn = case when ct = 1 then ct else ct - 1 end ;
@hardas81 Жыл бұрын
I'm getting the same results by running the following: with cte as (select *, row_number() over (partition by username order by startdate) as rn, count(1) over (partition by username) as cnt from useractivity) select * from cte where cnt = 1 or rn = cnt - 1;
@user-qn5yf8bc1c6 ай бұрын
i will go with this one much simpler
@dwaipayansaha4443 Жыл бұрын
My solution with the script:- create table activity1( username varchar(20), acitivity varchar(20), startdate date, enddate date); insert into activity1 (username,acitivity,startdate,enddate) values ('Amy','Travel','2020-02-12','2020-02-20'), ('Amy','Dancing','2020-02-21','2020-02-23'), ('Amy','Travel','2020-02-24','2020-02-28'), ('Joe','Travel','2020-02-11','2020-02-18'), ('Adam','Travel','2020-02-12','2020-02-20'), ('Adam','Dancing','2020-02-21','2020-02-23'), ('Adam','Singing','2020-02-24','2020-02-28'), ('Adam','Travel','2020-03-01','2020-03-28'); with t1 as(select username,activity,startdate,enddate,row_number() over(partition by username order by enddate desc) sorted_date, count(enddate) over(partition by username) count from activity1) select username,activity,startdate,enddate from t1 where (sorted_date=2 and count >1) or (sorted_date=1 and count=1)
@AbhishekSharma-vm7tr4 ай бұрын
why count-1 not because there is one record with count 4 for this second most should be 2 but it is adam 3
@AdinaAdina-mg5bz Жыл бұрын
IT'S ALWAYS THE UNDERRATED VID THAT'S LEGIT! THANK YOU!
@hilarylomotey7051 Жыл бұрын
Genius as always. Thanks for sharing. Am sure u are overwhelmed with so many emails now. I am sure mine is missing in your inbox. Anyway love your videos. Wish you could locate my mail tho 😂😂😂. Keep it up and we love you
@kanwalhemant Жыл бұрын
Same if we mention DESC in order by clause of row number & then replace 'cnt-1' to '2'
@RaviKumar-pn8uw Жыл бұрын
First Like... I'm eagerly waiting for your videos sir... Thanks 👍❤️
@arslansohail6254 Жыл бұрын
Thanks, TFQ and love you boss the way you explain the query and solve it's up to the mark ♥
@Sharmasurajlive Жыл бұрын
My attempt with CTE and Window functions :- sample data: create table useractivity (username text, activity text, startdate date, enddate date); insert into useractivity values ('Amy','Travel','2020-02-12','2020-02-20'); insert into useractivity values ('Amy','Dancing','2020-02-21','2020-02-23'); insert into useractivity values ('Amy','Travel','2020-02-24','2020-02-28'); insert into useractivity values ('Joe','Travel','2020-02-11','2020-02-18'); insert into useractivity values ('Adam','Travel','2020-02-12','2020-02-20'); insert into useractivity values ('Adam','Dancing','2020-02-21','2020-02-23'); insert into useractivity values ('Adam','Singing','2020-02-24','2020-02-28'); insert into useractivity values ('Adam','Travel','2020-03-01','2020-03-28'); solution : with cte as (select *, rank() over (partition by username order by startdate desc) as rnk from useractivity), cte2 as (select username, activity, startdate, enddate, rank() over (partition by username order by startdate) as rnk from cte where rnk
@akiemcameron53038 ай бұрын
I need to work a lot of similar problems so partition and case becomes second nature. This is a great tutorial
@maheshmmr9543 Жыл бұрын
Hii, your video are more useful, so please upload video about the topics like cluster, and indexex, thnq
@SantiagoZuluaga Жыл бұрын
Thank you so much for sharing! In your solution, lines 4 & 5 could be simplified using: , COUNT(*) OVER(partition by username) AS cnt
@imyours777 Жыл бұрын
how would you order them then? by default ordering is asc
@SantiagoZuluaga Жыл бұрын
@@imyours777 You're COUNTING, so you don't need to order them. Of course, you need to order them but in function ROW_NUMBER() so it's gonna stay as it is: --using mysql WITH t AS ( select U.*, ROW_NUMBER() OVER (PARTITION BY username ORDER BY endDate DESC) as order_activity, COUNT(*) OVER (PARTITION BY username) as total_activities from useractivity AS U ) SELECT username, activity, startDate, endDate FROM t WHERE (order_activity = 1 AND total_activities = 1) OR (order_activity = 2 AND total_activities >= 2);
@imyours777 Жыл бұрын
@@SantiagoZuluaga ya your logic doesnt need to order but in his logic we need to order to use cnt-1 great 👍
@harshavardhanachyuta2055 Жыл бұрын
@@imyours777 with cte as ( select *, row_number() over(partition by username order by startdate) as rn, count(*) over(partition by username) as cnt from useractivity ) select username,activity,startdate,enddate from cte where rn = (case when cnt > 1 then cnt-1 else 1 end)
@sayantabarik4252 Жыл бұрын
with cte as ( select *, row_number() over(partition by username order by enddate) as rn, count(*) over(partition by username order by username) as cnt from activity1) select username,acitivity,startdate,enddate from cte where rn= case when cnt=1 then 1 else cnt-1 end
@gnataeee Жыл бұрын
Thanks TFQ. Very helpful I have a basic question Leaving the rows with count =1 aside for now, would it be correct to sort the partition by DESCENDING order and then select the rows with row number rn=2 instead of the cnt-1?
@aagastya7752 Жыл бұрын
It can be..
@achyuthbudi9838 Жыл бұрын
yes
@MJ-gu6uu Жыл бұрын
Yes. And then union those rows with count=1. 👍
@pavanareddy4536 Жыл бұрын
Thanks T! heading to learn the frame clause next. I tried out using your logic just ordered the result by start date in descending order. with cte as (select * ,row_number() over (partition by username order by startdate desc) rn ,count(*) over (partition by username order by startdate desc range between unbounded preceding and unbounded following) cnt from useractivity order by username, startdate desc) select username, activity, startdate, enddate from cte where rn= case when rn=cnt then 1 else 2 end;
@FaisalAli-ps7th9 ай бұрын
I don't understand why do we need to have count at all when the question concerns about activity and time sorting. We could just sort the data by start_date and use nth_value() to get the second most recent activity. My solution is below: with cte as ( select *, coalesce(nth_value(activity,2) over (partition by username order by start_date desc range between unbounded preceding and unbounded following),activity) as second_most_recent_activty from activity1) select username,activity,start_date,end_date from cte where activity=second_most_recent_activty; Coalesce is used because for Joe null will be returned as it has only one row.
@SANDATA764 Жыл бұрын
Thank you thoufiq
@techTFQ Жыл бұрын
Thank you for alll the support Ahmed 🙏🏼 Appreciate it
@SANDATA764 Жыл бұрын
@@techTFQ looking forward your upcoming python playlist InshaAllah
@Krishna48784 Жыл бұрын
@@techTFQ thank you for sharing queries and explaining . Can you do a video on postgresql with json .
@dasoumya Жыл бұрын
Hello! This is my simple solution: With cte as(Select *, row_number()over(partition by username order by startdate desc) as rn,count(*)over(partition by username ) as total_count From User activity) Select username, activity, startdate,enddate From cte Where rn=2 or total_count=1
@Manishkumar-iw1cy Жыл бұрын
I think simplest one select username,activity,startdate,enddate from (select *, row_number() over(partition by username order by startdate desc) as rn from Useractivity) where rn=2 union select username,activity,startdate,enddate from Useractivity group by username having count(username)=1
@adrie9522 Жыл бұрын
Thank you so much this helped a lot!!!! You saved my life
@vikkynarepagul2568 ай бұрын
sir a kind request is that always share a data set so that we can also practice it
@fathimafarahna2633 Жыл бұрын
Awesome mate👌🏻 God bless you
@arunadharmasena9236 Жыл бұрын
Hey man, It works great and without any problems.
@jmhall1962 Жыл бұрын
It's not necessary to complicate the COUNT function with RANGE: partitioning by username is adequate. The following query is simpler: WITH activity_recency AS (SELECT *, ROW_NUMBER() OVER (PARTITION BY username ORDER BY startdate DESC) AS recency, COUNT(*) OVER (PARTITION BY username) AS activity_count FROM user_activity) SELECT username, activity, startdate, enddate FROM activity_recency WHERE recency = 2 OR activity_count = 1; This query can also easily be enhanced to return any nth most recent or the most recent activity if there aren't n activities.
@jcwynn4075 Жыл бұрын
Would this return extra records though? For example where activity_count is 1 and recency is also 1
@jmhall1962 Жыл бұрын
@@jcwynn4075, the WHERE clause filters existing rows, it can never create addition rows.
@jcwynn4075 Жыл бұрын
@@jmhall1962 I mean without the RANGE clause, the most recent activity for each person would have activity_count = 1, so would those get returned even if they have more than one activity?
@jmhall1962 Жыл бұрын
@@jcwynn4075 In my query, COUNT partitions by username but does include an ORDER BY clause. This approach results in each row having activity_count equal to the total number of activities for the associated user. This is simpler than adding a RANGE clause to undo the undesired behavior caused by including ORDER BY in the COUNT function.
@jmhall1962 Жыл бұрын
If the database supports the LEAST function, then changing the WHERE clause to "WHERE recency = LEAST(2, activity_count)" permits querying for nth recency by changing a single number.
@abhipoornisnsadvocate5158 Жыл бұрын
Good to know about frame clause. Range one
@venkatareddykunduru18287 ай бұрын
easiest solution: with cte as (select *, row_number() over(partition by username order by startdate desc) rn, count() over(partition by username) as cnt from activity1) select * from cte where (rn=1 and cnt=1) or (rn=2 and cnt>1)
@tonysun203 Жыл бұрын
Really appreciate for your sharing. 👍
@vp4673 Жыл бұрын
@TechTfq -I could see that you use window function effectively in most of the solutions. Just want to know if window function is good performancewise also... Thanks
@shibinjohn2792 Жыл бұрын
Sir my solution in MySQL - select * from (SELECT uc.*, ROW_NUMBER() OVER(PARTITION BY username ORDER BY username,startDate) as row_count,count(*) OVER(PARTITION BY username) as total_count FROM `user_activity` uc) as x where x.row_count = 2 or (x.row_count = 1 and x.total_count = 1);
@abhinasneupane2392 Жыл бұрын
Thanks for sharing TFQ, The question says the table does not contain Primary Key! Does that mean there could be two person with same name, there could be different records for same name, eg there could be two differrent person named Amy? Do we need to cosider this as an edge case?
@ea_naseer Жыл бұрын
I think that you could sniff out start date collisions.
@avi8016 Жыл бұрын
Thankyou very much sir for bringing this 🙏
@reddaiahreddymallu Жыл бұрын
Thank you, Thoufiq.
@angaddeepsingh7483 Жыл бұрын
with t2 as(select username, activity, startdate, enddate from(select *, rank() over (partition by username order by enddate desc) as recency from useractivity u) t1 where recency=2), t3 as (select * from useractivity u where username not in (select username from t2)) select * from t2,t3
@sjsheikDawood Жыл бұрын
I feel this is straight forward.. Kindly let me know if this approach has any drawbacks.... with intr as (select *,count(*) over (partition by username ) as cnt, rank() over (partition by username order by startDate desc ) as rnk from UserActivity), flag as (select *, case when cnt =1 then 'valid' when rnk =2 then 'valid' else 'invalid' end as flg from intr) select * from flag where flg='valid'
@shadabsiddiqui28 Жыл бұрын
thanks for sharing.. love from india
@cemanivannan272 ай бұрын
I'm getting the same results by running the following: if it's wrong, please let me know with cte as(select *,row_number() over(partition by username order by startdate desc) rn from activity1 qualify rn=2) select * exclude rn from cte union all select * from activity1 where username in (select username from activity1 group by username having count(*)=1);
@CassStevens Жыл бұрын
Using rank instead of row_number select username, activity, startDate, endDate from (select *, case when max(rnk) over (partition by username) = 1 then 'x' when rnk = 2 then 'x' end as slct from (select *, rank() over (partition by username order by endDate desc) as rnk from UserActivity ) x ) y where y.slct is not null
@vishalsonawane.89057 ай бұрын
Please add the Data set for the practice
@muditmishra9908 Жыл бұрын
hi , I think for windows function: row_number(), it should be-> order by end_date desc.
@Mihiret468 Жыл бұрын
Thank you Tofiq, based on your explanation I can solve the problem as below, what do you think? Is it possible or? With cte_secondActivity As (Select *, row_number() over(partion by userName order by start date desc) as row_nr From user Activity) Select username, activity,startdate,start date, From cte_secondActivity Where row_nr=2
@lankelajagadeesh2155 Жыл бұрын
Very nice explanation
@mohammadarslanjaved Жыл бұрын
Respected sir I hope you are well .kinldy make a video for beginner to expert which course start for DB and which DBMS use. Which DBMS have a scope in future .plsease shear your experience in video .and tell step by step which course should first then second then third etc. And how we apply for job and which compney should apply through linked-in because each company required 2-3 year experiences.But we have don't experience. Kindly tell us a plate form or you tube channel link. Thanks a lot sir
@rishabhgupta1029 Жыл бұрын
have solved by : select * from ( select *, row_number()over(PARTITION by username order by endDate desc) as rn, count(*) over (PARTITION by username) as total_records from Table_1 ) where ( (rn = 1 and total_records =1) or (rn = 2 and total_records 1) )
@yashsaxena7754 Жыл бұрын
Sharing an alternative! with cte as (select username,activity,startDate,endDate,rank() over(partition by username order by startDate desc) as rnk from user_activity), table1 as (select username,activity,startDate,endDate,rnk,max(rnk) over(partition by username) as max_rnk from cte) select username,activity,startDate,endDate from table1 where rnk = 2 or (max_rnk = 1)
@vishalsonawane.89057 ай бұрын
Continue.....
@kamalmukhi3993 Жыл бұрын
Hello Thoufiq, Thanks for the video!! I have one question. In the problem statement it is written that a user cannot perform two activities at the same time. Suppose if table contains records with overlapping time periods, shouldn't that condition be checked as well and those records be discarded? Thanks Kamal
@girishggirishg1083 Жыл бұрын
VERY INTERESTING
@birthdaycake8438 Жыл бұрын
Fast download, thank you brother))
@KISHOREKUMAR-mu3xo Жыл бұрын
@techtfq,, why dont we use u this way based on rowid,,, select * from (select username,activity,startsate,count(*) over (partition by username) as cnt,row_number() over (partition by username order by rowid asc) rn from useractivity) where rn=2 or cnt=1;
@shilashm5691 Жыл бұрын
Easier Solution: with cte as (select *, row_number() over (partition by username order by CURDATE() - endDate) as recent_day_number from activity_table) select * from cte where recent_day_number in (2) or username in (select username from activity_table group by username having count(*)
@narasimhayadavmadduluri7129 Жыл бұрын
with narsi as (select a.*,row_number()over(partition by username order by startdate desc) rn from activity1 a) select username,acitivity,startdate,enddate from narsi where rn=2 or username in(select username from narsi group by username having max(rn)=1);
@asavlogs8446 Жыл бұрын
Sir. Trigger, cursor aur user defined function ki video banayiye
@Mihiret468 Жыл бұрын
One more thing I added one row for Joe having recent activity than the given one. To make the given activity second recent.
@abhilasharya9435 Жыл бұрын
Hi, i have shared one query over mail .it is complex one can you please make a video on that one??
@jogerilely2828 Жыл бұрын
super! cheatwas easily installed
@fouzanak9162 Жыл бұрын
Awesome program
@aryakurniasandi9671 Жыл бұрын
what application do you use to create the source code database?
@sravankumar1767 Жыл бұрын
Superb bro 👌 👏 👍
@srishtijain642 Жыл бұрын
please make video on USER DEFINED FUNCTIONS in sql
@netpolun-ltd.7267 Жыл бұрын
group by/having/count/limit/offset - probably 3 times shorter :D
@007SAMRATROY Жыл бұрын
with cte as ( Select a.*, rank() over (partition by username order by startdate, enddate desc) as a_rank from activity1 a ) Select username,acitivity,startdate,enddate from cte where a_rank = 2 union all select username,acitivity,startdate,enddate from cte where username not in (Select username from cte where a_rank > 1);
@pravaskumar4937 Жыл бұрын
Thanx
@karthikeyasoft Жыл бұрын
I think if one user has only 2 records. we need to consider another case also. cnt =2
@KrishNa-vp8un Жыл бұрын
select * from (select *,row_number()over(partition by username order by (select 0) ) as rownum ,count (*) over (partition by username order by (select 0) ) as count from useractivity)useractivity where rownum= case when count =1 then 1 else count-1 end
@nawalambavkar7543 Жыл бұрын
I use count and row_number logic : WITH t1 AS (SELECT *, Row_number() OVER( partition BY username ORDER BY enddate ) rn, Count(username) OVER( partition BY username) cc FROM activity1 ORDER BY username, enddate) SELECT * FROM t1 WHERE cc < 2 OR rn = 2
@srinivasn415 Жыл бұрын
My solution to the problem -> 1. Get ranks based on start date partitioned by user name ordered by username and rank - store it as x, 2. select all rows from x that have have ranks < 3 and create a case when using lead() to see if the next row has the same username as current_row -> 1 if Yes, 2 if No - store it as X, 3. Select all rows from X with lead = 0
@venkatgandam3573 Жыл бұрын
Thank you sir
@nehalahmad5678 Жыл бұрын
Hi Taufeeq, I think if we go directly with original question then we can achieve result by using only row_number () , Right?
@sidharthadaggubati438 Жыл бұрын
No. You still need count of records per username.
@obaiahlakkineni6065 Жыл бұрын
Sir, Will you be able to provide solution for below question? We have two records and five columns. Actually there is Duplicate record in that two records but one column is having different name for two records. So, my question is how to write a SQL query to remove that duplicate record in the Dashboard?
@shubhampokar2532 Жыл бұрын
I think simple or clause would have done the job, like 'where rn=2 OR cnt=1'
@AbhishekSharma-vm7tr4 ай бұрын
i am not getting why count - 1 because if i want second most value it should be (adam-Dancing Not Adam-singing) cnt - 1 is correct for amy but not for adam
@yashkanojiya1146 Жыл бұрын
for sql workbench users with cte as (SELECT concat(id,' ', name) as concat,ntile(4) over(order by id) as 'ntiles' FROM interviewquestion.emp) select group_concat(concat) as result from cte group by ntiles order by 1
@theresepraisethelord201 Жыл бұрын
Hi.when it's 4 it has returned the third row but we still need only the second?
@gamingwithhemend9890 Жыл бұрын
If count = 2, then cnt-1 would return the wrong value !!!, so you need to write a case for that one too. Pls reply
@basavapn6487 Жыл бұрын
That is correct , question is mostly second recent that means from last 2nd record needs to be fetched , if we have count=2, second recent record is first record
@jcwynn4075 Жыл бұрын
If there are only 2 records, then the first record is the second most recent. How's it wrong?
@gamingwithhemend9890 Жыл бұрын
Ohh okay!!!... I understand, thanks❤
@rameshthanikonda7027 Жыл бұрын
Hi Toufiq, How to connect with you over telegram. Thanks
@pavipatil007 Жыл бұрын
Hi TFQ, i do have 6 years of experience in SQL ,which role is good ? DATA science or Analyst ,kindly make video on how to find remote jobs for SQL
@ukkashs444 Жыл бұрын
Refer me any SQL related job. I am from Chennai
@akshayvn14 Жыл бұрын
Data Engineer!
@ammar0466 Жыл бұрын
Can you make video about writing sql using github copilot
@mindlessscroll Жыл бұрын
Solution - select * from (select username, activity , row_number() over(partition by username order by startdate desc) rn, count(*) over(partition by username order by startdate range between unbounded preceding and current row ) as c from user_activity ) X where rn=2 or c=rn
@monasanthosh92083 ай бұрын
MYSQL Solutions for Freshers With CTE as (Select *,dense_rank() over (Partition by Username order by StartDate) as Rn from Activity1), CTE1 as (Select *, Case When RN=2 then 1 Else 0 end as TRn from CTE), CTE2 as (Select *,Sum(TRN) over (Partition by Username) as TRN_Sum from CTE1) Select Username,Acitivity,Startdate,enddate from CTE2 Where (Case When TRN_SUM=1 then RN=2 Else RN=1 end);
@chaithanyag93 Жыл бұрын
I enrolled for the classes.. I haven’t got back anything yet
@Naru_NNN Жыл бұрын
Hi TFQ May i know What is the difference b/w count (*) and count(1) and count (column)?
@pavanareddy4536 Жыл бұрын
count(*) returns how many records are in the table, same for count(1), count(50), count(-1) doesn't matter the number, for count(number), it just counts total number of records in the table and assigns each record a number given in the count function. Count(column) returns total number of values in the column which are not null. Null is not included here.
@risasa1237 Жыл бұрын
This seems like way too much work for a solution thah can be much simpler than this. Overcomplicating it into
@narasimhamurty859 Жыл бұрын
Brother , sorting family members problem could u slove in MS SQL server.
@AshutoshSingh-vn1kn2 ай бұрын
create table useractivity (username varchar(50), activity varchar(50), startdate date, enddate date); insert into useractivity values ('Amy','Travel','2020-02-12','2020-02-20'); insert into useractivity values ('Amy','Dancing','2020-02-21','2020-02-23'); insert into useractivity values ('Amy','Travel','2020-02-24','2020-02-28'); insert into useractivity values ('Joe','Travel','2020-02-11','2020-02-18'); insert into useractivity values ('Adam','Travel','2020-02-12','2020-02-20'); insert into useractivity values ('Adam','Dancing','2020-02-21','2020-02-23'); insert into useractivity values ('Adam','Singing','2020-02-24','2020-02-28'); insert into useractivity values ('Adam','Travel','2020-03-01','2020-03-28'); select Row_number() over(partition by username order by startdate asc) as row,* into #temp_table from useractivity select username,count(1) as count into #final from #temp_table group by username having count(1)>1 insert into #final select username,count(1) as count from #temp_table group by username having count(1)=1 select ff.username, ff.activity, ff.startdate, ff.enddate from #final f join #temp_table ff on f.username=ff.username and ff.row=1 and f.count=1 union all select ff.username, ff.activity, ff.startdate, ff.enddate from #final f join #temp_table ff on f.username=ff.username and ff.row=2 and f.count1
@srushtiOm2 ай бұрын
Thanks much!
@saisowhitpb24843 ай бұрын
Your output has not matched. Please check it once properly.
@faizan4712Ай бұрын
more simple solution : WITH ranked_activities AS ( SELECT *, ROW_NUMBER() OVER(PARTITION BY username ORDER BY startdate desc) AS rn, COUNT(*) OVER(PARTITION BY username) AS count FROM UserActivity ) SELECT username, activity, startdate, enddate FROM ranked_activities WHERE rn=2 OR count=1
@aaravkumarsingh4018 Жыл бұрын
please provide ddl scripts in video so that we can practise from ourself before watching your solution
@mithunkt1648 Жыл бұрын
Hi, Kindly review my code, improvement suggestions are highly solicited. Thank you. With master as ( Select *, Rank()over(partition by username order by endDate desc) as rank, Lead(endDate,1,0)over(partition by username order by endDate desc) as flag From UserActivity) Select Username, Activity, Startdate, Enddate From master Where rank =2 or flag = 0
@jmhall1962 Жыл бұрын
There are at least two issues with this query. First is that 0 is not a date and the database (e.g., PostgreSQL) might not support casting it to a date. This could be addressed by allowing LEAD to default to NULL and checking for "flag IS NULL" instead of "flag = 0" in the WHERE clause. Second, even with this change, the query returns the wrong results. Both the first and second most recent activities satisfy the criteria for users with more than one activity.
@mithunkt1648 Жыл бұрын
@@jmhall1962 Thank you, I reckon i overlooked the conflict between 2nd most recent and 1st activity by user whilst using lead function. Herewith attached new iteration for review. I am finding my paces in the industry and improvement feedback is solicited for undertanding edge cases. With master as ( Select *, Rank()over(partition by username order by endDate desc) as rank, Count(*)over(partition by username) as flag From UserActivity) Select Username, Activity, Startdate, Enddate From master Where rank =2 or flag = 1
@jmhall1962 Жыл бұрын
@@mithunkt1648 That query produces the correct results. Consider using more descriptive names. "Rank" and "flag" don't convey sufficient meaning. In another comment I posted a similar version of the query using "recency" and "activity_count" so that the meaning of the derived values can be easily understood by others.
@mithunkt1648 Жыл бұрын
@@jmhall1962 Thank you for the valuable inputs, I will bear in mind while writing queries now onwards. I am working on my data skills as I am working towards a pivot into the tech. Shout out to @techtfq team and taufeeq for this wonderful platform.
@Leo-qo5hk5 ай бұрын
select * from(select *, dense_rank() over(partition by username order by startDate desc) as rank, count(*) over(partition by username) as count from Google_UserActivity)x where rank=2 or count=1
@indergaming305311 ай бұрын
with cte as ( select *,LEAD(rn,1,0) over (partition by username order by (select null) ) as ld from ( select * from ( select *, row_number() over (partition by username order by (select null) ) as rn from useractivity) A)B where rn in (1,2)) select username,activity,startdate,enddate from cte where ld=0 /*please rate this query out of 10*/
@raviteja8593 Жыл бұрын
i didnt understand wat is this cte
@jordanbentarghi-jeffers Жыл бұрын
google sql query docs public
@user-pm2sz7lc6q3 ай бұрын
WITH cte1 AS ( SELECT *, RANK() OVER(PARTITION BY username ORDER BY startdate DESC) AS rnk, COUNT(*) OVER(PARTITION BY username ORDER BY startdate DESC RANGE BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING) AS total FROM activity) SELECT username, acitivity, startdate, enddate FROM cte1 WHERE rnk = 2 OR total = 1;
@ecoprint3540 Жыл бұрын
For once, the software is actually really useful
@user-lm1ft4yd3t Жыл бұрын
select username,ld as activity from( select *,ROW_NUMBER() over (partition by username order by startdate desc ) rnk FROM ( select *,lead(activity,1,activity) over (partition by username order by startdate desc) ld from [dbo].[UserActivity] )a) b where rnk =1
@memesmacha61 Жыл бұрын
I think for adam 's second most activity is dancing right??
@jacksparrow3595 Жыл бұрын
I have the same doubt,
@jcwynn4075 Жыл бұрын
Second most RECENT is singing. Same as saying second to last
@shreymehta4795 Жыл бұрын
WITH cte AS(SELECT *, dense_rank()over (partition by username ORDER by enddate )rn, count(*) over (partition by username)cnt from useractivity) SELECT username,activity,startdate,enddate FROM cte WHERE rn = 2 or cnt =1
@shivamsingla7596 Жыл бұрын
select username,activity,startdate,enddate from (select *,row_number() over(partition by username order by startdate desc) as rnk from useractivity) as z where rnk=2 union select * from useractivity where username in(select username from useractivity group by username having count(1)=1) solution given by SQL KING nikhil anand