Google SQL Interview Problem | Solving SQL Interview Query

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  • Опубліковано 31 лип 2024

КОМЕНТАРІ • 149

  • @faitusjeline
    @faitusjeline Рік тому +19

    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
    ;

  • @AdinaAdina-mg5bz
    @AdinaAdina-mg5bz Рік тому +4

    IT'S ALWAYS THE UNDERRATED VID THAT'S LEGIT! THANK YOU!

  • @fathimafarahna2633
    @fathimafarahna2633 Рік тому

    Awesome mate👌🏻 God bless you

  • @hardas81
    @hardas81 Рік тому +6

    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-qn5yf8bc1c
      @user-qn5yf8bc1c 5 місяців тому

      i will go with this one much simpler

  • @RaviKumar-pn8uw
    @RaviKumar-pn8uw Рік тому +4

    First Like... I'm eagerly waiting for your videos sir... Thanks 👍❤️

  • @arslansohail6254
    @arslansohail6254 Рік тому +1

    Thanks, TFQ
    and love you boss the way you explain the query and solve
    it's up to the mark ♥

  • @dwaipayansaha4443
    @dwaipayansaha4443 Рік тому +9

    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-vm7tr
      @AbhishekSharma-vm7tr 3 місяці тому

      why count-1 not because there is one record with count 4 for this second most should be 2 but it is adam 3

  • @hilarylomotey7051
    @hilarylomotey7051 Рік тому +2

    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

  • @tonysun203
    @tonysun203 Рік тому

    Really appreciate for your sharing. 👍

  • @adrie9522
    @adrie9522 Рік тому

    Thank you so much this helped a lot!!!! You saved my life

  • @reddaiahreddymallu
    @reddaiahreddymallu Рік тому

    Thank you, Thoufiq.

  • @avi8016
    @avi8016 Рік тому

    Thankyou very much sir for bringing this 🙏

  • @arunadharmasena9236
    @arunadharmasena9236 Рік тому

    Hey man, It works great and without any problems.

  • @akiemcameron5303
    @akiemcameron5303 7 місяців тому

    I need to work a lot of similar problems so partition and case becomes second nature. This is a great tutorial

  • @kanwalhemant
    @kanwalhemant 11 місяців тому +1

    Same if we mention DESC in order by clause of row number & then replace 'cnt-1' to '2'

  • @maheshmmr9543
    @maheshmmr9543 Рік тому +1

    Hii, your video are more useful, so please upload video about the topics like cluster, and indexex, thnq

  • @sravankumar1767
    @sravankumar1767 Рік тому

    Superb bro 👌 👏 👍

  • @lankelajagadeesh2155
    @lankelajagadeesh2155 Рік тому

    Very nice explanation

  • @birthdaycake8438
    @birthdaycake8438 Рік тому

    Fast download, thank you brother))

  • @shadabsiddiqui28
    @shadabsiddiqui28 Рік тому

    thanks for sharing.. love from india

  • @abhipoornisnsadvocate5158
    @abhipoornisnsadvocate5158 Рік тому

    Good to know about frame clause. Range one

  • @venkatgandam3573
    @venkatgandam3573 Рік тому

    Thank you sir

  • @girishggirishg1083
    @girishggirishg1083 Рік тому

    VERY INTERESTING

  • @Sharmasurajlive
    @Sharmasurajlive Рік тому

    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

  • @fouzanak9162
    @fouzanak9162 Рік тому

    Awesome program

  • @vp4673
    @vp4673 Рік тому

    @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

  • @jogerilely2828
    @jogerilely2828 Рік тому

    super! cheatwas easily installed

  • @pravaskumar4937
    @pravaskumar4937 Рік тому

    Thanx

  • @aryakurniasandi9671
    @aryakurniasandi9671 Рік тому +1

    what application do you use to create the source code database?

  • @pavanareddy4536
    @pavanareddy4536 Рік тому

    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;

  • @sayantabarik4252
    @sayantabarik4252 Рік тому +1

    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

  • @vishalsonawane.8905
    @vishalsonawane.8905 6 місяців тому

    Continue.....

  • @SantiagoZuluaga
    @SantiagoZuluaga Рік тому +2

    Thank you so much for sharing! In your solution, lines 4 & 5 could be simplified using:
    , COUNT(*) OVER(partition by username) AS cnt

    • @imyours777
      @imyours777 Рік тому +1

      how would you order them then? by default ordering is asc

    • @SantiagoZuluaga
      @SantiagoZuluaga Рік тому +2

      @@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
      @imyours777 Рік тому +1

      @@SantiagoZuluaga ya your logic doesnt need to order but in his logic we need to order to use cnt-1
      great 👍

    • @harshavardhanachyuta2055
      @harshavardhanachyuta2055 Рік тому

      @@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)

  • @muditmishra9908
    @muditmishra9908 Рік тому

    hi , I think for windows function: row_number(), it should be-> order by end_date desc.

  • @KISHOREKUMAR-mu3xo
    @KISHOREKUMAR-mu3xo Рік тому

    @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;

  • @dasoumya
    @dasoumya Рік тому +1

    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

  • @gnataeee
    @gnataeee Рік тому +5

    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?

  • @mohammadarslanjaved
    @mohammadarslanjaved Рік тому

    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

  • @abhilasharya9435
    @abhilasharya9435 Рік тому

    Hi, i have shared one query over mail .it is complex one can you please make a video on that one??

  • @SANDATA764
    @SANDATA764 Рік тому +2

    Thank you thoufiq

    • @techTFQ
      @techTFQ  Рік тому +1

      Thank you for alll the support Ahmed 🙏🏼
      Appreciate it

    • @SANDATA764
      @SANDATA764 Рік тому

      @@techTFQ looking forward your upcoming python playlist InshaAllah

    • @Krishna48784
      @Krishna48784 Рік тому

      @@techTFQ thank you for sharing queries and explaining . Can you do a video on postgresql with json .

  • @Manishkumar-iw1cy
    @Manishkumar-iw1cy Рік тому

    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

  • @FaisalAli-ps7th
    @FaisalAli-ps7th 8 місяців тому +1

    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.

  • @Mihiret468
    @Mihiret468 Рік тому

    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

  • @asavlogs8446
    @asavlogs8446 Рік тому

    Sir.
    Trigger, cursor aur user defined function ki video banayiye

  • @venkatareddykunduru1828
    @venkatareddykunduru1828 6 місяців тому

    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)

  • @theresepraisethelord201
    @theresepraisethelord201 Рік тому

    Hi.when it's 4 it has returned the third row but we still need only the second?

  • @vikkynarepagul256
    @vikkynarepagul256 7 місяців тому

    sir a kind request is that always share a data set so that we can also practice it

  • @srishtijain642
    @srishtijain642 Рік тому

    please make video on USER DEFINED FUNCTIONS in sql

  • @abhinasneupane2392
    @abhinasneupane2392 Рік тому +2

    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
      @ea_naseer Рік тому +1

      I think that you could sniff out start date collisions.

  • @obaiahlakkineni6065
    @obaiahlakkineni6065 Рік тому

    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?

  • @ammar0466
    @ammar0466 Рік тому

    Can you make video about writing sql using github copilot

  • @vishalsonawane.8905
    @vishalsonawane.8905 6 місяців тому

    Please add the Data set for the practice

  • @CassStevens
    @CassStevens Рік тому

    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

  • @netpolun-ltd.7267
    @netpolun-ltd.7267 Рік тому

    group by/having/count/limit/offset - probably 3 times shorter :D

  • @shibinjohn2792
    @shibinjohn2792 Рік тому +1

    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);

  • @cemanivannan27
    @cemanivannan27 Місяць тому

    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);

  • @narasimhamurty859
    @narasimhamurty859 Рік тому

    Brother , sorting family members problem could u slove in MS SQL server.

  • @rishabhgupta1029
    @rishabhgupta1029 11 місяців тому

    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)
    )

  • @nehalahmad5678
    @nehalahmad5678 Рік тому

    Hi Taufeeq,
    I think if we go directly with original question then we can achieve result by using only row_number () , Right?

  • @narasimhayadavmadduluri7129

    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);

  • @shilashm5691
    @shilashm5691 Рік тому

    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(*)

  • @Mihiret468
    @Mihiret468 Рік тому

    One more thing I added one row for Joe having recent activity than the given one. To make the given activity second recent.

  • @007SAMRATROY
    @007SAMRATROY Рік тому

    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);

  • @karthikeyasoft
    @karthikeyasoft Рік тому

    I think if one user has only 2 records. we need to consider another case also. cnt =2

  • @angaddeepsingh7483
    @angaddeepsingh7483 Рік тому

    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

  • @nawalambavkar7543
    @nawalambavkar7543 Рік тому

    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

  • @yashsaxena7754
    @yashsaxena7754 Рік тому +3

    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)

  • @shubhampokar2532
    @shubhampokar2532 Рік тому

    I think simple or clause would have done the job, like 'where rn=2 OR cnt=1'

  • @Naru_NNN
    @Naru_NNN Рік тому

    Hi TFQ
    May i know
    What is the difference b/w count (*) and count(1) and count (column)?

    • @pavanareddy4536
      @pavanareddy4536 Рік тому +3

      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.

  • @yashkanojiya1146
    @yashkanojiya1146 Рік тому

    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

  • @chaithanyag93
    @chaithanyag93 Рік тому

    I enrolled for the classes.. I haven’t got back anything yet

  • @kamalmukhi3993
    @kamalmukhi3993 Рік тому +1

    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

  • @KrishNa-vp8un
    @KrishNa-vp8un Рік тому

    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

  • @pavipatil007
    @pavipatil007 Рік тому

    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
      @ukkashs444 Рік тому

      Refer me any SQL related job. I am from Chennai

    • @akshayvn14
      @akshayvn14 Рік тому

      Data Engineer!

  • @raviteja8593
    @raviteja8593 Рік тому

    i didnt understand wat is this cte

  • @ecoprint3540
    @ecoprint3540 Рік тому

    For once, the software is actually really useful

  • @shabanashabana4603
    @shabanashabana4603 Рік тому

    👍🏼

  • @monasanthosh9208
    @monasanthosh9208 3 місяці тому

    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);

  • @mindlessscroll
    @mindlessscroll Рік тому

    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

  • @jmhall1962
    @jmhall1962 Рік тому +1

    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
      @jcwynn4075 Рік тому

      Would this return extra records though? For example where activity_count is 1 and recency is also 1

    • @jmhall1962
      @jmhall1962 Рік тому

      @@jcwynn4075, the WHERE clause filters existing rows, it can never create addition rows.

    • @jcwynn4075
      @jcwynn4075 Рік тому

      @@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
      @jmhall1962 Рік тому

      @@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
      @jmhall1962 Рік тому

      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.

  • @AbhishekSharma-vm7tr
    @AbhishekSharma-vm7tr 3 місяці тому

    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

  • @sjsheikDawood
    @sjsheikDawood Рік тому

    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'

  • @aaravkumarsingh4018
    @aaravkumarsingh4018 Рік тому

    please provide ddl scripts in video so that we can practise from ourself before watching your solution

  • @sridevim4302
    @sridevim4302 Рік тому

    liked

  • @srinivasn415
    @srinivasn415 Рік тому +1

    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

  • @AshutoshSingh-vn1kn
    @AshutoshSingh-vn1kn Місяць тому

    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

  • @gamingwithhemend9890
    @gamingwithhemend9890 Рік тому +2

    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
      @basavapn6487 Рік тому +1

      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
      @jcwynn4075 Рік тому +1

      If there are only 2 records, then the first record is the second most recent. How's it wrong?

    • @gamingwithhemend9890
      @gamingwithhemend9890 Рік тому +1

      Ohh okay!!!... I understand, thanks❤

  • @Leo-qo5hk
    @Leo-qo5hk 4 місяці тому

    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

  • @memesmacha61
    @memesmacha61 Рік тому

    I think for adam 's second most activity is dancing right??

    • @jacksparrow3595
      @jacksparrow3595 Рік тому

      I have the same doubt,

    • @jcwynn4075
      @jcwynn4075 Рік тому +1

      Second most RECENT is singing. Same as saying second to last

  • @faizan4712
    @faizan4712 28 днів тому +1

    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

  • @risasa1237
    @risasa1237 Рік тому +2

    This seems like way too much work for a solution thah can be much simpler than this. Overcomplicating it into

  • @indergaming3053
    @indergaming3053 10 місяців тому

    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*/

  • @saisowhitpb2484
    @saisowhitpb2484 2 місяці тому

    Your output has not matched. Please check it once properly.

  • @mithunkt1648
    @mithunkt1648 Рік тому +5

    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
      @jmhall1962 Рік тому

      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
      @mithunkt1648 Рік тому

      @@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
      @jmhall1962 Рік тому

      @@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
      @mithunkt1648 Рік тому

      @@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.

  • @user-pm2sz7lc6q
    @user-pm2sz7lc6q 3 місяці тому

    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;

  • @user-lm1ft4yd3t
    @user-lm1ft4yd3t Рік тому

    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

  • @shreymehta4795
    @shreymehta4795 Рік тому

    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

  • @jordanbentarghi-jeffers
    @jordanbentarghi-jeffers Рік тому

    google sql query docs public

  • @shivamsingla7596
    @shivamsingla7596 Рік тому

    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

  • @owaowanothing1233
    @owaowanothing1233 Рік тому

    man I missed this kind of tutorials lol. Great work here, thanks!!!