I want to count the number of ids, but in a special way. For example, if the id is counted in April, I don't want to count that id again in May. so excluding id's that have been counted in previous months.
this is the query I am using.
select store, monthname(created_at), count(distinct customer_id) from a group by 1,2;
Hope you can help me.
Thanks in advance!
Just count the first time a customer is seen. One method is two levels of aggregation:
select store, date_trunc('month', min_created_at), count(*)
from (select store, customer_id, min(created_at) as min_created_at
from a
group by store, customer_id
) c
group by 1, 2;
Note: monthname() is not appropriate for defining a month, because it does not take the year into account. If you really do want to ignore the year, you can use monthname() but that seems unusual.
Hope this sql statement helps!
select store, monthname(created_at), count(distinct customer_id)
from a
where monthname(created_at)=MONTHNAME(current_date())
group by 1,2;
Related
My table consists of user_id, revenue, publish_month columns.
Right now I use group_by user_id and sum(revenue) to get revenue for all individual users.
Is there a single SQL query I can use to query for user revenue across a time period conditionally? If for a specific user, there is a row for this month, I want to query for this month, last month and the month before. If there is not yet a row for this month, I want to query for last month and the two months before.
Any advice with which approach to take would be helpful. If I should be using cases, if-elses with exists or if this is do-able with a single SQL query?
UPDATE---since I did a bad job of describing the question, I've come to include some example data and expected results
Where current month is not present for user 33
Where current month is present
Assuming publish_month is a DATE datatype, this should get the most recent three months of data per user...
SELECT
user_id, SUM(revenue) as s_revenue
FROM
(
SELECT
user_id, revenue, publish_month,
MAX(publish_month) OVER (PARTITION BY user_id) AS user_latest_publish_month
FROM
yourtableyoudidnotname
)
summarised
WHERE
publish_month >= DATEADD(month, -2, user_latest_publish_month)
GROUP BY
user_id
If you want to limit that to the most recent 3 months out of the last 4 calendar months, just add AND publish_month >= DATEADD(month, -3, DATE_TRUNC(month, GETDATE()))
The ambiguity here is why it is important to include a Minimal Reproducible Example
With input data and require results, we could test our code against your requirements
If you're using strings for the publish_month, you shouldn't be, and should fix that with utmost urgency.
You can use a windowing function to "number" the months. In this way the most recent one will have a value of 1, the prior 2, and the one before 3. Then you can only select the items with a number of 3 or less.
Here is how:
SELECT user_id, revienue, publish_month,
ROW_NUMBER() OVER(PARTITION BY user_id ORDER BY publish_month DESC) as RN
FROM yourtableyoudidnotname
now you just select the items with RN less than 3 and do your sum
SELECT user_id, SUM(revenue) as s_revenue
FROM (
SELECT user_id, revenue, publish_month,
ROW_NUMBER() OVER(PARTITION BY user_id ORDER BY publish_month DESC) as RN
FROM yourtableyoudidnotname
) X
WHERE RN <= 3
GROUP BY user_id
You could also do this without a sub query if you use the windowing function for SUM and a range, but I think this is easier to understand.
From the comment -- there could be an issue if you have months from more than one year. To solve this make the biggest number in the order by always the most recent. so instead of
ORDER BY publish_month DESC
you would have
ORDER BY (100*publish_year)+publish_month DESC
This means more recent years will always have a higher number so january of 2023 will be 202301 while december of 2022 will be 202212. Since january is a bigger number it will get a row number of 1 and december will get a row number of 2.
I have these tables
record: sid (string), cid (string), quarter (string), year (integer), grade(integer)
student: sid (string)
For every student who has taken at least one class, meaning a student is entered in the record table at least once, i need to get their GPA in the most recent quarter they were enrolled in. I need to display sid, quarter, year, and grade (gpa).
There are 3 quarters in a given calendar year, and it may be helpful to observe the order of the occurrence of quarters is in reverse alphabetical order ('W' > 'S' > 'F'). These stands for winter, spring, fall respectively. Fall being the latest quarter of the year.
this is what i came up with:
select sid, quarter, year, avg(grade) as gpa
from (select sid, min(quarter) as quarter, year, avg(grade) as grade
from (select *, max(year) as maxy
from record
group by sid)
group by sid)
group by sid;
this gives me the average grade for all quarters/years enrolled, and doesn't give me the latest quarter either.
I can only use functions such as NOT EXIST / EXIST, NOT IN/IN , group by, order by. I cannot use rank().
I was told that I should use NOT EXIST to get the latest quarter since the most recent quarter means for a specific quarter, there is no succeeding quarter.
any help would be greatly appreciated. thank you!
You want solution using not exists? Here you go.
Select t.*
From record t
Where not exists
(Select 1 from record tt
Where tt.year > t.year
And tt.quarter < t.quarter
And tt.sid = t.sid)
Above query will give you all the data of student for latest quarter, then you can use the aggregate function according to your requirement.
Use row_number():
select r.*
from (select r.*,
row_number() over (partition by sid
order by case quarter when 'W' then 1 when 'S' then 2 when 'F' then 3 else 4 end desc
) as seqnum
from records r
) r
where seqnum = 1;
Although this can be simplified a little bit by relying on the "alphabetical" ordering of quarter, I don't recommend that approach. Relying on alphabetic ordering for time periods is counterintuitive and will make the code harder to understand.
If the quarter were stored as a number then it would be appropriate for use in order by.
**Is there a way to count how many strings in a specific column are seen for
Since the value in the column 2 gets repeated sometimes due to the fact that some clients make several transactions in different times (the client can make a transaction in the 1st month then later in the next year).
Is there a way for me to count how many IDs are completely new per month through a group by (never seen before)?
Please let me know if you need more context.
Thanks!
A simple way is two levels of aggregation. The inner level gets the first date for each customer. The outer summarizes by year and month:
select year(min_date), month(min_date), count(*) as num_firsts
from (select customerid, min(date) as min_date
from t
group by customerid
) c
group by year(min_date), month(min_date)
order by year(min_date), month(min_date);
Note that date/time functions depends on the database you are using, so the syntax for getting the year/month from the date may differ in your database.
You can do the following which will assign a rank to each of the transactions which are unique for that particular customer_id (rank 1 therefore will mean that it is the first order for that customer_id)
The above is included in an inline view and the inline view is then queried to give you the month and the count of the customer id for that month ONLY if their rank = 1.
I have tested on Oracle and works as expected.
SELECT DISTINCT
EXTRACT(MONTH FROM date_of_transaction) AS month,
COUNT(customer_id)
FROM
(
SELECT
date_of_transaction,
customer_id,
RANK() OVER(PARTITION BY customer_id
ORDER BY
date_of_transaction ASC
) AS rank
FROM
table_1
)
WHERE
rank = 1
GROUP BY
EXTRACT(MONTH FROM date_of_transaction)
ORDER BY
EXTRACT(MONTH FROM date_of_transaction) ASC;
Firstly you should generate associate every ID with year and month which are completely new then count, while grouping by year and month:
SELECT count(*) as new_customers, extract(year from t1.date) as year,
extract(month from t1.date) as month FROM table t1
WHERE not exists (SELECT 1 FROM table t2 WHERE t1.id==t2.id AND t2.date<t1.date)
GROUP BY year, month;
Your results will contain, new customer count, year and month
my table name is CustomerDetails and it has the following columns:
customer_id, login_id, session_id, login_date
i am trying to write a query that calculates the average number of customers login in per day.
i tried this:
select avg(session_id)
from CustomerDetails
where exists (select count(session_id) from CustomerDetails as 'no_of_entries')
.
but then i realized it was going straight to the column and just calculating the average of that column but that's not what i want to do. can someone help me?
thanks
The first thing you need to do is get logins per day:
SELECT login_date, COUNT(*) AS loginsPerDay
FROM CustomerDetails
GROUP BY login_date
Then you can use that to get average logins per day:
SELECT AVG(loginsPerDay)
FROM (
SELECT login_date, COUNT(*) AS loginsPerDay
FROM CustomerDetails
GROUP BY login_date
)
If your login_date is a DATE type you're all set. If it has a time component then you'll need to truncate it to date only:
SELECT AVG(loginsPerDay)
FROM (
SELECT CAST(login_date AS DATE), COUNT(*)
FROM CustomerDetails
GROUP BY CAST(login_date AS DATE)
)
i am trying to write a query that calculates the average number of customers login in per day.
Count the number of customers. Divide by the number of days. I think that is:
select count(*) * 1.0 / count(distinct cast(login_date as date))
from customerdetails;
I understand that you want do count the number of visitors per day, not the number of visits. So if a customer logged twice on the same day, you want to count him only once.
If so, you can use distinct and two levels of aggregation, like so:
select avg(cnt_visitors) avg_cnt_vistors_per_day
from (
select count(distinct customer_id) cnt_visitors
from customer_details
group by cast(login_date as date)
) t
The inner query computes the count of distinct customers for each day, he outer query gives you the overall average.
I have a select that group by customers spending of the past two months by customer id and date. What I need to do is to associate for each row the total amount spent by that customer in the whole first week of the two month time period (of course it would be a repetition for each row of one customer, but for some reason that's ok ). do you know how to do that without using a sub query as a column?
I was thinking using some combination of OVER PARTITION, but could not figure out how...
Thanks a lot in advance.
Raffaele
Query:
select customer_id, date, sum(sales)
from transaction_table
group by customer_id, date
If it's a specific first week (e.g. you always want the first week of the year, and your data set normally includes January and February spending), you could use sum(case...):
select distinct customer_id, date, sum(sales) over (partition by customer_ID, date)
, sum(case when date between '1/1/15' and '1/7/15' then Sales end)
over (partition by customer_id) as FirstWeekSales
from transaction_table
In response to the comments below; I'm not sure if this is what you're looking for, since it involves a subquery, but here's my best shot:
select distinct a.customer_id, date
, sum(sales) over (partition by a.customer_ID, date)
, sum(case when date between mindate and dateadd(DD, 7, mindate)
then Sales end)
over (partition by a.customer_id) as FirstWeekSales
from transaction_table a
left join
(select customer_ID, min(date) as mindate
from transaction_table group by customer_ID) b
on a.customer_ID = b.customer_ID