Count distinct monthly running total [closed] - sql

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I am trying to find a SQL query to the question of finding the running unique count of distinct customers each month.
This is not the number of distinct customers each month, but rather what is the new incremental total number of unique customers as each month rolls by.
For example, in January I had 10 unique customers and in February I had another 10 customers, but 5 of these who transacted in February were repeat customers and had also transacted in January; so between January and February I really only have a running total of 15 unique customers (10 in January and 5 in February).
What would be the simplest solution to achieve the running unique customer count for each month of the year?
Example Output where (compared to January) in February there were an additional 5 unique customers and in March there were an additional 10 unique customers

This might do:
Select
month,
count(*) as custs,
(select
count(distinct cust_id)
from mytable b
where b.month<=a.month) as RunningUniqueCusts
From mytable a
group by month
Or for month & region
Select
month,
region,
count(*) as custs,
(select
count(distinct cust_id)
from mytable b
where b.month<=a.month
and b.region=a.region) as RunningUniqueCustsForRegion
From mytable a
group by month, region
Update 3-Mar-2022
The following would return the unique customer ids for each month where they didn't appear previously:
SELECT TM.MONTH_ID, TM.CUST_ID
FROM MYTABLE as TM
WHERE NOT EXISTS
(SELECT 1
FROM MYTABLE as PM
WHERE PM.CUST_ID = TM.CUST_ID
and PM.MONTH < TM.MONTH)
GROUP BY TM.MONTH_ID, TM.CUST_ID
ORDER BY TM.MONTH_ID, TM.CUST_ID

Related

attendance system CHECK-IN & CHECK-OUT values in SQL SERVER [closed]

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I have an SQL table where I have employee logs and the states of the log state 255 being (check-in) and state 1 being checked-out
I want to have the minimum check-in time and the max checkout time of each day date as a result.
This is the table that I have.
This is the Query
select [full name],
(SELECT MIN(time) AS CHECKIN FROM attendance WHERE state = 255 and fingerprint = id) checkin,
(SELECT MAX(time) AS CHECKIN FROM attendance WHERE state = 1 and fingerprint = id) checkout,
cast (date as date)
FROM employee full outer join attendance on fingerprint = id
group by id,fingerprint,[full name],date
The problem is that it selects the minimum and Max time but not by date
it basically searches the min and max times across all dates instead of each day as its own
notice how it gave the checkin (2:00) and checkout(8:00) for Jon considering the fact that dates are in diffrent years
Is the query good, or am I doing it wrong?
Your query is almost correct, the mistake is that the group by clause in your query groups by id and then fingerprint and groups further by [full name] and then by date. Instead, you need to group only by [full name] and date:
select [full name],
MIN(
CASE
WHEN state = 255 and fingerprint = id THEN time
ELSE 999999
END
) checkin,
MAX(
CASE
WHEN state = 1 and fingerprint = id THEN time
ELSE -1
END
) checkout,
cast (date as date)
FROM employee full outer join attendance on fingerprint = id
group by [full name],date

Find customers who ordered in the last week every day? [closed]

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Recently i was asked a question in the interview. Which is there is a customer table and an order table. How to get the customers who ordered in the last week everyday and how to find the customers who order in last three consecutive days?
Appreciate your help.
Thanks in advance.
To get customers who ordered every day for the past N days I would try something like:
select customer_id
from orders
where trunc(order_date) > trunc(sysdate) - :N
group by customer_id
having count(distinct trunc(order_date)) = :N
The idea is that the WHERE clause limits your data to orders made within the last N days, then we GROUP BY to form a group for each customer, and then in the HAVING clause we check each customer to see if COUNT(DISTINCT TRUNC(ORDER_DATE)) equals N.
One way is to convert the order table to a wide table with columns corresponding to each day of interest. Not sure if there are better ways.
A brute force method uses exists. For instance, for the last three days:
select c.*
from customers c
where exists (select 1
from orders o
where o.customer_id = c.customer_id and
trunc(o.order_date) = trunc(sysdate)
) and
exists (select 1
from orders o
where o.customer_id = c.customer_id and
trunc(o.order_date) = trunc(sysdate) - interval '1' day
) and
exists (select 1
from orders o
where o.customer_id = c.customer_id and
trunc(o.order_date) = trunc(sysdate) - interval '2' day
) ;
This is one method. There are others, that I'll let you think about.

How to retrieve the minimum date having 3 columns (day), (month) and (year) with RSQLite? [closed]

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I am working with SQL in R. I want to get the minimum date by choosing the minimum value of the column 'day;, the minimum value of the column 'month' and the minimum value of the column 'year'.
I have tried this by the following code:
dbGetQuery(db, "SELECT day, month, year
FROM surveys
WHERE year = (SELECT MIN(year) FROM surveys);")
But my output is not one value, how can I get one value in my output and not a list of values?
Right now your query returns rows on the minimum year, not minimum date. Consider generating a date column by concatenating the date parts to identify minimum:
sql = "WITH sub AS (
SELECT day, month, year
, DATE(year || '-' ||
CASE
WHEN length(month)=1
THEN '0' || month
ELSE month
END || '-' ||
CASE
WHEN length(day)=1
THEN '0' || day
ELSE day
END) AS [date]
FROM surveys
)
SELECT DISTINCT day, month, year, [date]
FROM sub
WHERE [date] = (SELECT MIN([date]) FROM sub)"
dbGetQuery(db, sql)
Online Demo
Using the test data shown we order it by year, month and day and select the first row of the sorted table.
library(sqldf)
surveys <- data.frame(year = 2001:2005, month = 5:1, day = 1:5)
sqldf("select day, month, year from surveys order by year, month, day limit 1")
## day month year
## 1 1 5 2001

how to get recent 6 dates (Max dates) [closed]

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I need recent 6 dates count
My code is
select DUE_DATE, count(*)
from DATA
group by DUE_DATE
03/24/2018 10
03/17/2018 20
03/10/2018 15
03/03/2018 23
02/24/2018 42
02/17/2018 32
02/10/2018 15
02/03/2018 17
01/27/2018 23
select DUE_DATE, count(*) from DATA group by DUE_DATE order by DUE_DATE desc limit 6
The simplest case would be if every date is presented and/or you're only interested in 6 most recent dates regardless of whether they're presented in your table or not:
select DUE_DATE, count(*)
from DATA
where DUE_DATE >= trunc(sysdate) - 5
group by DUE_DATE
order by DUE_DATE desc
(Note: dates that are not presented in your table won't be shown).
On the other hand, if you need 6 most recent dates from the subgroup of dates you have in your table, then you'll first need a subquery to fetch you those dates and than use count on those dates only:
select DUE_DATE, count(*)
from DATA
where DUE_DATE in (select distinct DUE_DATE
from DATA
order by DUE_DATE desc limit 6)
group by DUE_DATE
I hope I helped!
Are you looking for fetch first?
select DUE_DATE, count(*)
from DATA
group by DUE_DATE
order by DUE_DATE desc
fetch first 6 rows only;

Top 10 based on last month showing 6 previous months

I want to show a graph with income from different parties over the last 6 months, but based on the top income of 10 people only based on the last month.
So this can change each month as the top 10 people can change when they deposit more money, so the graph will show these 10 people's deposits of the last 6 months, based on the last month deposit only.
I already used a LAG function and a RANK() OVER PARTITION function.
I don't understand why you'll need rank or lag functions.
You can simply use an IN statement:
SELECT * FROM YourTable t
WHERE t.depositDate between StartRangeDate and EndRangeDate
AND t.ID in(select ID from(SELECT s.id,sum(s.depositAmount) as total
from YourTable s
where s.date between ThisMonthStart and ThisMonthEnd
group by s.id)
order by total
limit 10)
You can play with the first select to select what ever you want/add a group by and sum them or I don't know.