Complex SQL 2 | find new and repeat customers | SQL Interview Questions
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- Опубликовано: 14 янв 2022
- This video is about finding new and repeat customers .using SQL. In this video we will learn following concepts:
how to approach complex query step by step
how to use CASE WHEN with SUM
how to use common table expression (CTE)
Here is the script :
create table customer_orders (
order_id integer,
customer_id integer,
order_date date,
order_amount integer
);
select * from customer_orders
insert into customer_orders values(1,100,cast('2022-01-01' as date),2000),(2,200,cast('2022-01-01' as date),2500),(3,300,cast('2022-01-01' as date),2100)
,(4,100,cast('2022-01-02' as date),2000),(5,400,cast('2022-01-02' as date),2200),(6,500,cast('2022-01-02' as date),2700)
,(7,100,cast('2022-01-03' as date),3000),(8,400,cast('2022-01-03' as date),1000),(9,600,cast('2022-01-03' as date),3000)
;
![MURDER DRONES Series Finale [TRAILER]](http://i.ytimg.com/vi/p6mhC5Iatns/mqdefault.jpg)
Select a.order_date,
Sum(Case when a.order_date = a.first_order_date then 1 else 0 end) as new_customer,
Sum(Case when a.order_date != a.first_order_date then 1 else 0 end) as repeat_customer
from(
Select customer_id, order_date, min(order_date) over(partition by customer_id) as first_order_date from customer_orders) a
group by a.order_date;
Good.
impressed
A very good solution, I feel mine is overcomplicating things
Nice one Satyam 😊
Nice one bro
Now i will never Forget CTE... Great teaching skill
Thanks, ankit for this brainstorming question, initially couldnt figure out the but the min(order_date) that you gave was the key. I accomplished this with subqueries:
select order_date,count(new_customer) as new,count(repeat_customer) as repeat from (
select order_date,
case when order_date=first_order_date then 'new_customer'
end as new_customer,
case when order_date!=first_order_date then 'repeat_customer'
end as repeat_customer
from (
select a.*,b.first_order_date from customer_orders a
join(
select customer_id,min(order_date) as first_order_date
from customer_orders
group by customer_id) b
on a.customer_id=b.customer_id)c)d
group by order_date
order by order_date asc;
Nice work Ankit, your way of solving the problem is simple but effective.
Hey Ankit , your channel is really addictive. Since yesterday I have picked more than 15 problems in a row (which indirectly means I watched 15 videos from your channel straight in a row). I am really enjoying it. People binge watch Netflix and here I am binge watching sql problem solving. Can't express in words, felt like I found the gem on the RUclips. It pumped adrenaline rush in my body when I am able to solve the problems without looking at the solution. At the end comparing my solution with your provided solution and that also is teaching me how to solve any problem in minimal joins and subqueries. Amazing....Amazing...Amazing....Thank you so much for all your hardwork and knowledge sharing.
Thats a big compliment for me. Keep rocking 😊
For me also same thing,I am not forcefully learning, by intrest I am coming and watching your videos....I can say I addicted to your channel. The positive of your channel is you will provide table with data so it make us to practice while watching your video
Same for me also
Exactly the same here too. I'm not able to sleep at night till i solve problems and i never see your solution till i solve it myself. Thanks for this good collection of questions
Exactly..Same here addicted to the explaination and over enthusiastic to solve. Initially to type single line query I was seeing and typing ,now I am watching whole video till end and then writing 5-6lines queries without seeing. @AnkitBansalYou got great teaching skills.
seriously no one in the entire youtube explained CTE like this. Made it so simple thank you ankit bro
🙏
Hi Ankit, 2 years I cracked the DA role with your help, now when I'm preparing again for a switch, this is my go to source material for SQL Prep, thanks for a splendid playlist.
Here is my solution:
SELECT order_date, sum(CASE WHEN rn2 = 1 THEN 1 ELSE 0 END) AS new, sum(CASE WHEN rn2 > 1 THEN 1 ELSE 0 END) AS repeat
FROM (SELECT *,
row_number() over(PARTITION BY customer_id ORDER BY order_date ASC, customer_id ASC) AS rn2 FROM customer_orders)
GROUP BY order_date ORDER BY order_date;
best step by step practice
Assignment query:
with cte as(
select order_date,order_amount, row_number() over(partition by customer_id order by
order_date asc) as rn from customer_orders)
select order_date, sum(case when rn=1 then 1 else 0 end) as new_customers,
sum(case when rn>1 then 1 else 0 end) as repeat_customers,
sum(case when rn=1 then order_amount else 0 end) as new_customers_order_amount,
sum(case when rn>1 then order_amount else 0 end) as repeat_customers_order_amount
from cte
group by order_date;
select * from customer_orders;
Hi Sir, Thank you for all your videos ..Really helpful for learning .
Here is my query
with cte as
(select customer_id,min(order_date) as first_visit_date
from customer_orders
group by customer_id)
select c.order_date,
sum(case when c.order_date = f.first_visit_date then 1 else 0 end) as first_visit_flag,
sum(case when c.order_date != f.first_visit_date then 1 else 0 end) as repeat_visit_flag,
sum(case when c.order_date = f.first_visit_date then order_amount else 0 end) as newCustAmount,
sum(case when c.order_date != f.first_visit_date then order_amount else 0 end) as repeatCustAmount
from customer_orders c
inner join cte f
on c.customer_id=f.customer_id
group by c.order_date
;
such a great explanation!!!
Wow Ankit, your videos on SQL are so good, informative and helpful. Thanks a lot for making them. Keep going.
Thank you 😊
This is a clear and concise explanation
Thank you so much Ankit Bansal. This is really helpful.
VERY Good Query
Very useful video
awesome explanation ...
MYSQL Query for the same:-
with cte as(
select order_date, row_number() over(partition by customer_id order by
order_date asc) as rn from customer_orders)
select order_date, sum(case when rn=1 then 1 else 0 end) as new_customers,
sum(case when rn>1 then 1 else 0 end) as repeat_customers from cte
group by order_date;
This is good. Thanks for posting 👏
I ended up being very close to your solution with a little difference.
with old_new_counter as (
SELECT *,row_number() over (partition by customer_id order by order_date) old_new_flag FROM customer_orders)
select order_date,
count(case when old_new_flag=1 then 'new_customer' end) count_new_customer,
count(case when old_new_flag>1 then 'old_customer' end) count_repeat_customer
from old_new_counter group by order_date
order by order_date;
Cheers
This query will not give expected output in a case where same user has more than 1 order the same date.
I tested using same records 2 times in a table .
Just my input..
You make it so easy, superb explanation
I was asked exactly the same question in my interview with dunnhumby and I failed to answer as I panicked and tried to give an answer hurriedly . Now after going through your video in steps , I completely understood the approach in how to deal with these questions. Looking forward to the rest of the playlist .
Cool
@@ankitbansal6 I was also asked this questions
For which job position ankit
@@sachinkapoor2424 DS
For how many years experience did you apply?
this what a real time problems...thanks and keep bring such
Sure. Thanks.
love your great content.
awesome problem, Thank you so much for posting.
MY only SQL guru.. Thank you guruji.. love you for ever.
Your videos are amazing. Keep up the good work!
Thank you 😊
Love the way of explanation with step by step. 😀
🙏
Best tutor
Thank you very much Sir, for this practical question and your step by step explanation.
You are most welcome
Nice totala
Hi Ankit,your channel is very helpful and the way you are explaining is just amazing. Here is my solution for this,WITH CTE AS (
SELECT ORDER_ID,CUSTOMER_ID,ORDER_DATE,ORDER_AMOUNT,
CASE WHEN PRIV IS NULL THEN 1 ELSE 0 END AS NEW_FLAG,
CASE WHEN PRIV IS NOT NULL THEN 1 ELSE 0 END AS OLD_FLAG FROM (
select *,
lag(ORDER_DATE) over(partition by CUSTOMER_ID order by ORDER_DATE) as PRIV from customer_orders) ORDER BY ORDER_ID)
SELECT ORDER_DATE,SUM(NEW_FLAG) AS NEW_CUSTOMER,SUM(OLD_FLAG) AS OLD_CUTOMER FROM CTE
GROUP BY ORDER_DATE;
Thank you for this video! Please come with problems like this. Thank you
Sure.
Hi Ankit
Thanks your videos are helping me to break down complex scnearios into smaller parts and then combine the whole query.....so i did the query in different way..Please do let me know if thats correct since the motive is only to find duplicate and new customers
with cte as (
select customer_id,count(1) as ranking from customer_orders group by customer_id)
select * , case when ranking>1 then 'Duplicate' else 'New' end as status_customer from cte;
customer_id ranking status_customer
100 3 Duplicate
200 1 New
300 1 New
400 2 Duplicate
500 1 New
600 1 New
What a Explanation mind blowing ❤️❤️❤️
Thank you so much 😀
Best. Than kyou.
Really good channel and informative videos.
Glad you like them!
This was smooth.
Subscribed on 2nd video in your channel, Great stuff.
Thank you 😊
Great explanation for both approach and solution
Glad you liked it
Fanstastic channel.
This is perfect!!
Thank you !!
You are awesome 💌
Hey Ankit,
Thank you for this problem questions. I specially like the assignment you give at the end. As I am a beginner for SQL its very my encouraging and confidence boosting for me.
Assignment sol:-
with first_visit_flag as
(SELECT customer_id, MIN(order_date) as first_visit_date
FROM customer_orders
GROUP BY customer_id),
repeat_visit_flag AS
(
SELECT co.order_date,
fv.first_visit_date,
CASE WHEN co.order_date=fv.first_visit_date THEN 1 ELSE 0 END AS first_visit,
CASE WHEN co.order_date!=fv.first_visit_date THEN 1 ELSE 0 END AS repeat_visit,
CASE WHEN co.order_date=fv.first_visit_date THEN SUM(order_amount) ELSE 0 END AS new_order,
CASE WHEN co.order_date!=fv.first_visit_date THEN SUM(order_amount) ELSE 0 END AS repeat_order
FROM customer_orders co
inner join first_visit_flag fv ON co.customer_id = fv.customer_id
GROUP BY co.order_date,fv.first_visit_date
)
SELECT order_date, SUM(first_visit) as new_customer,SUM(repeat_visit) as repeat_customer, SUM(new_order) AS new_order_amount, SUM(repeat_order) AS repeat_order_amount
FROM repeat_visit_flag
GROUP BY order_date;
Thanks, Ankit for this brainstorming question,
MY QUERY
SELECT order_date,
Count(CASE WHEN rnk = 1 THEN cnt END) AS "New Customer",
Count(CASE WHEN rnk > 1 THEN cnt END) AS "Old Customer"
FROM (
SELECT order_date, customer_id,
DENSE_RANK() OVER (PARTITION BY customer_id ORDER BY order_date) AS rnk,
COUNT(*) OVER (PARTITION BY customer_id, order_date) AS cnt
FROM customer_orders1
) A
GROUP BY order_date;
Nice explanation Brother
want more such videos
thank you Brother 🙂
Sure
sELECT order_date,
Sum(cASE WHEN ORDER_DATE = fIRST_dATE THEN 1 else 0 END) AS nEW,
Sum(cASE WHEN ORDER_DATE fIRST_dATE THEN 1 else 0 END )AS rEPEATCUS
FROM
(Select Customer_id,order_date,
Min(order_date) over (Partition by customer_id order by order_date) as First_Date
from customer_orders ) as a
group by order_date got the answer by this too thanks
Very good question and very well explained. Great video Ankit :)
Thank you 😊
I was asked the same question in curefit in 3rd round. There were 2 extra tables to refer but now I realize it could have been done using single table with min order date criteria. Glad I stumbled on your channel
Glad to know 🙏
You got this for which profile.
@@m04d10y1996 Product analytics
more help full
Hi Ankit, thanks for creating such videos.
Here is my approach:
with sequenced_order_table as(
select *, dense_rank() over(partition by customer_id order by order_date) as order_seq
from customer_orders)
SELECT order_date,
count(case when order_seq = 1 then customer_id end) as new_customer,
count(case when order_seq > 1 then customer_id end) as old_customer
FROM sequenced_order_table
group by 1
order by 1
Bro , did you check this in the db?
Very nicely explained
Thank you so much 🙂
Good explanation bro
Thanks Juhair 😊
There is a small difference between the repeat customers (those purchased for the consecutive days) and old customers (no consecutive days condition).
For the repeat customers - purchasing for consecutive days
select order_date, sum(new_customer) as new_customer, sum(repeat_customer) as repeat_customer from
(select customer_id, order_date, case when order_date=first_visit then 1 else 0 end as new_customer,
case when date_diff(order_date, prev_day)=1 then 1 else 0 end as repeat_customer (
select customer_id, order_date, min(order_date) over (partition by customer_id) as first_visit , lag(order_date,1,0) over (partition by customer_id order by order_date) as prev_day from customer_orders)x)y group by order_date
Solved the question without looking into the solution in MySQL, I have used the concept of sum with case when after seeing it in your other video, it is very helpful and important concept:
with cte as
(
select *,
row_number() over(partition by customer_id) as rn
from customer_orders
order by order_date
)
select order_date,
sum(case when rn = 1 then 1 else 0 end) as new_customer_count,
sum(case when rn = 1 then 0 else 1 end) as repeat_customer_count
from cte
group by order_date;
Please let me know if there is some issue in this code
Nice
Thanks 🙏
Hey Ankit Thanks for providing this question my solution for this problem
with cte as (select order_id , customer_id , order_date ,
lag(customer_id)over(partition by customer_id order by order_date) as statements
from customer_orders)
select order_date , sum(case when statements is null then 1 else 0 end) as new_customer_count
, sum(case when statements is not null then 1 else 0 end) as old_customer_count
from cte
group by order_date order by order_date
What a beautiful question, make your brain to hit hard.
Thanks
thank you
hi ankith this is also working
with cte as (
SELECT *,min(order_date) over (partition by customer_id) as first_date
FROM customer_orders as a
)
select order_date,count(case when order_date first_date then customer_id end) as repeat,
count(case when order_date = first_date then customer_id end) as new,
count(customer_id) as total
from cte
group by order_date
Hey Ankit, Your videos are really awesome and informative.
Can you make some content or share resources regarding learn data analysis through python required for data analyst role
Will make. Thank you 😊
finished watching
Hi Ankit,
with cte as
(select *, row_number() over (partition by customer_id order by order_date) as order_flag
from customer_orders)
select order_date,
sum(case when order_flag=1 then 1 else 0 end) as new_customer_count,
sum(case when order_flag>1 then 1 else 0 end) as repeat_customer_count
from cte
group by order_date
Day 1 Example 2 done
Hi Ankit..Thanks for your efforts.. I have an alternate solution as well
WITH CTE AS(
select *,CASE WHEN(DENSE_RANK()OVER(PARTITION BY customer_id ORDER BY order_date)=1) THEN 'New'
ELSE 'Repeat' END AS IND_CUSTOMER
from customer_orders
)
SELECT order_date,count(CASE WHEN IND_CUSTOMER='New' THEN order_id END) AS no_new_customer,
count(CASE WHEN IND_CUSTOMER='Repeat' THEN order_id END) AS no_repeat_customer
FROM CTE
GROUP BY order_date
ORDER BY order_date
Thanks for posting. This is also good. 👍
Thanks Ankit, Here is my solution -
with cte as (
select *,
count(order_id) over(partition by customer_id
order by order_date
rows between unbounded preceding and current row) as cnt
from customer_orders
)
select order_date,
sum(case when cnt = 1 then 1 else 0 end) as new_cust_ind,
sum(case when cnt > 1 then 1 else 0 end) as repeat_cust_ind,
sum(case when cnt = 1 then order_amount else 0 end) as new_cust_amt,
sum(case when cnt > 1 then order_amount else 0 end) as repeat_cust_amt
from cte
group by order_date;
I am planning to complete all the SQL videos created by you in order to learn SQL. I will post a comment on each video and like it as a checklist for completed videos, starting from the beginning.
Today exactly same question was asked to Me for Cummins.
Here is my query
With cte_1 as (
Select *, rank() over(partition by customer_id order by order_date) as ranked from customer_orders
),
cte_2 as (
Select order_date, case when ranked = 1 then 'new' else 'repeat' end as new_or_repeat from cte_1
)
Select order_date, sum(case when new_or_repeat = 'new' then 1 else 0 end) as new_customer,sum(case when new_or_repeat = 'repeat' then 1 else 0 end) as repeat_customer from cte_2
group by order_date;
Thank you for your efforts
Day 2 of 47
Thanks for the video series
with cte1 as
(select order_date, case when customer_id= rep_cs then 1 else 0 end as rep_flag,
case when customer_id rep_cs then 1 else 0 end as new_flag from
(select order_date,customer_id,lag(customer_id,3,0) over(order by order_id )as rep_cs
from customer_orders) e1)
select order_date,sum(new_flag) as new_customer, sum(rep_flag) as rep_customer
from cte1
group by order_date
using lag function with cte
Thanks Ankit for great explanatory video>
Here is solution of assignment given in video
WITH first_visit AS (
SELECT customer_id, min(order_date) AS first_visit_date
FROM customer_orders
GROUP BY customer_id)
SELECT co.order_date,
SUM(CASE WHEN co.order_date = fv.first_visit_date THEN 1 ELSE 0 END) AS first_visit_customer,
SUM(CASE WHEN co.order_date != fv.first_visit_date THEN 1 ELSE 0 END) AS repeat_visit_customer,
SUM(CASE WHEN co.order_date = fv.first_visit_date THEN order_amount ELSE 0 END) AS first_visit_customer_order,
SUM(CASE WHEN co.order_date != fv.first_visit_date THEN order_amount ELSE 0 END) AS repeat_visit_customer_order
FROM customer_orders co
INNER JOIN first_visit fv ON co.customer_id = fv.customer_id
GROUP BY co.order_date
Hey Ankit, I have used a different approach:
with new_table as(
select order_date,count(customer_id) as new_customer from customer_orders a
where 0=( select count(*) from customer_orders b where a.order_date>b.order_date and a.customer_id=b.customer_id) group by order_date),
repeat_table as (select order_date,count(customer_id) as old_customer from customer_orders a
where ( select count(*) from customer_orders b where a.order_date>b.order_date and a.customer_id=b.customer_id)>0 group by order_date)
select case when a.order_date is null then b.order_date else a.order_date end as date ,new_customer,old_customer from
new_table a full outer join repeat_table b on a.order_date=b.order_date;
Thanks Ankit for your guidance. Please have a look below query
select sum(case when tc> 1 then 1 else 0 end )as repeat_customer,
sum(case when tc= 1 then 1 else 0 end )as new_customer from (select customer_id, count( customer_id) as tc
from customer_orders
group by customer_id) a ;
Can you pls share consolidated list of Easy and Medium SQLs if possible I'm xls or pdf format?
Great content as always.
Here is my attempt to the homework
with first_order_table as
(select customer_id, min(order_date) as first_order_date
from customer_orders
group by customer_id)
select co.order_date,
sum(case when co.order_date = fot.first_order_date then co.order_amount else 0 end) as order_amount_by_new_customer,
sum(case when co.order_date fot.first_order_date then co.order_amount else 0 end) as order_amount_by_repeat_customer
from customer_orders co
join first_order_table fot on fot.customer_id = co.customer_id
group by co.order_date
order by 1
very helpful. Can you please create a playlist for python qiestions asked in Data Engineering interview ?
Ok
@@ankitbansal6 Ankit please bring python questions too... as same as sql. You have got great teaching skills.
Make videos to crack DE for Amazon
What if the same customer visits the website twice or thrice and orders each time? In that case, he should be a repeat customer. However, according to your solution, he won't be counted as a repeated customer as his min(order_date) = order_date.
What do you think?
However, your tutorials have been really helpful to me. Really appreciate your effort.
Hey so what do you think the solution for this ?
Could you please help in this ?
then I think we need timestamp as well not only the date part.
Well, he will be both new AND repeat customer as per current problem explanation. So in this case new problem explanation will be needed.
I have implemented with this logic.
with cte as (
select *,ROW_NUMBER() over(partition by customer_id order by order_date asc) as rnk from customer_orders),
cte_not_1 as (select order_date,count(*) as cnt from cte where rnk 1 group by order_date)
select * from (
select order_date,count(*) as new_customer_count from cte t1
where t1.rnk = 1
group by order_date) t1 left join cte_not_1 t2
on t1.order_date = t2.order_date
Hi Ankit if we need to add Customer_id also in the select list means, what we should do?
I used below query,
with cte as(
select *,min(order_date) over(partition by customer_id) as first_visit from customer_orders
)
select order_date,
sum(
case when order_date=first_visit then 1 else 0 end
) as first_time,
sum(
case when order_datefirst_visit then 1 else 0 end
) as sec_time
from cte
group by order_date
Thank you for posting.
@ankit bansal
can we also write this query in this way Ankit? Same output
select order_date,
count(case when rn = 1 then 'new' end) newcust,
count(case when rn != 1 then 'repeated' end) repcust from
(select
customer_id,
order_date,
row_number() over(partition by customer_id order by order_date asc ) 'rn'
from newrepeat) newt
group by order_date
order by order_date asc
Hi Ankit could you suggest some resources for data modeling
Hi Ankit can you also upload the video of assignment too ?
can i also use window function
hi there may i know why i am getting syntax error following the same in pg admin and mysql
Which one approach is considered optimized the join one or without join ?
Hi Ankit,
Please have a look on below answer using windows function.
with ft as (select customer_id, order_date,
Dense_rank() over(Partition by customer_id order by order_date) as R
from customer_orders)
select order_date,sum(new_customer) as new_customer,
sum(existing_customer) as existing_customer
from
(select *,
case when R > 1 then 1 else 0 end as existing_customer,
case when R = 1 then 1 else 0 end as new_customer
from ft) t1
group by order_date
order by order_date;
I solved this using Window functions and cte,
My solution:
with cte as (select customer_id,row_number() over(partition by customer_id order by customer_id) num,order_date
from customer_orders group by customer_id,order_date)
select order_date, sum(case when num=1 then 1 else 0 end ) as new_cust,sum(case when num>1 then 1 else 0 end) repeat_cust
from cte group by order_date;
I have tried a bit different approach
with counter as
(
select *,
count(customer_id) over(partition by customer_id order by order_date) as ct
from customer_orders
)
select order_date,
sum(case when ct=1 then 1 else 0 end) as New_customer_counter,
sum(case when ct!=1 then 1 else 0 end) as old_customer_counter,
sum(case when ct=1 then order_amount else 0 end ) as New_customer_revenue,
SUM(case when ct!=1 then order_amount else 0 end ) as Old_customer_revenue
from counter
select order_date,
sum(case when rank1=1 then 1 else 0 end) as new_customer,
sum(case when rank1>1 then 1 else 0 end) as repeated_customer
from (select customer_id,order_date,row_number() over(partition by customer_id order by order_date)
rank1 from customer_orders) a
group by order_date
first time i tried in my own.
Thanks Ankit. Can you please share github link for code. That would be really helpful.
Thankyou for another great question!
My solution to this question:
SELECT new.order_date,
COUNT(CASE WHEN previous.customer_id IS NULL THEN 1 END) AS new_cust,
COUNT(DISTINCT(CASE WHEN previous.customer_id IS NOT NULL THEN previous.customer_id END)) AS repeat_cust,
SUM(CASE WHEN previous.customer_id IS NULL THEN new.order_amount END) AS amount_by_new_cust,
SUM(DISTINCT(CASE WHEN previous.customer_id IS NOT NULL THEN new.order_amount ELSE 0 END)) AS amount_by_repeat_cust
FROM customer_orders new
LEFT JOIN customer_orders previous
ON
previous.customer_id = new.customer_id
AND previous.order_date < new.order_date
GROUP BY new.order_date;
Hi everyone,
HW Task: add two columns of first_visit_order_amount, last_first_visit_order_amount
Solution:
sum(case when fv.first_visit_date = co.order_date then co.order_amount else 0 end) as first_visit_Order_amt_flag
, sum(case when fv.first_visit_date != co.order_date then co.order_amount else 0 end) as repeat_visit_order_amt_flag
add this two columns in Ankit's solution.
Hi Ankit,
Nice job.
I have a query - lets say like given scenario of new and repeat customer, if we do have few more scenarios like New, retained, Unretained, Reactivated and Lapsed customers. I/p table has millions of rows. Then, instead of self joining main table again and again, can we apply required filters and take data in CTE and then use that as a main table will help to optimize performance?
Yes . If with the same filters you need to use table multiple times then its a good idea to create cte and use that. You can also explore option to create temp table and use that.
@@ankitbansal6 i had to create a view so, was not had option to create temp table
@@supriyakolate2575 Can you share the data for the possible scenario, if you have and approach to do it?
My solution was:
with firsttable(customerid,firstorderdate) as
(
select customer_id,min(order_date) as first_order_date from customer_orders
group by customer_id
)
select order_date,sum(case when (customerid = customer_id and firstorderdate = order_date) then 1 else 0 end) as newcustomer,
sum(case when (customerid = customer_id and firstorderdate != order_date) then 1 else 0 end) as oldcustomer
from customer_orders,firsttable
group by order_date
with cte as (
select *,
row_number () over (partition by customer_id order by order_date ) as rn
from customer_orders),
cte1 as
(select order_date , count(1) as new_cust from cte
where rn=1 group by order_date),
cte2 as (
select order_date , count(1) as rep_cust from cte
where rn>1 group by order_date)
select a.order_date,new_cust, case when rep_cust is null then 0 else rep_cust end as rep_cust from cte1 a left join cte2 b on a.order_date=b.order_date
Hi Ankit, I just first saw the question and tried to do it myself. After I was done and I came back to the video, even though the result was same the methods were different. I am not sure on how to validate my answer on huge data though. Can you please help me to know if my method lacks anything?
WITH CTE AS (
SELECT ROW_NUMBER() OVER (PARTITION BY customer_id ORDER BY order_date ASC ) AS RN ,customer_id, order_date
FROM customer_orders
)
SELECT order_date
,SUM( CASE WHEN RN=1 THEN 1 else 0 END )AS NEW_COUNT
,SUM(CASE WHEN RN1 THEN 1 else 0 END ) AS REPEAT_COUNT
FROM CTE
GROUP BY order_date
ORDER BY order_date
Thanks in advance!
Looks good. Thanks for posting 👏
Hi Aniket,
Here is my solution :)
with cte1 as(
select *, ROW_NUMBER() OVER(PARTITION BY customer_id order by order_date) as row_num
from customer_orders
)
select order_date, sum(case when row_num = 1 then 1 else 0 end) as new_cus,
sum(case when row_num > 1 then 1 else 0 end) as rep_cus,
sum(case when row_num =1 then order_amount else 0 end) as newCus_order_amnt,
sum(case when row_num >1 then order_amount else 0 end) as repCus_order_amnt
from cte1
group by order_date
Please make video on thumb rules that are must to be followed
Sure.
if i want to know Repeat Customer vs New Customer Last year vs current year pls advise