Best approach to display all the users who have more than 1 purchases in a month in SQL - sql

I have two tables in an Oracle Database, one of which is all the purchases done by all the customers over many years (purchase_logs). It has a unique purchase_id that is paired with a customer_id.The other table contains the user info of all the customers. Both have a common key of customer_id.
I want to display the user info of customers who have more than 1 unique item (NOT the item quantity) purchased in any month (i.e if A customer bought 4 unique items in february 2020 they would be valid as well as someone who bought 2 items in june). I was wondering what should my correct approach be and also how to correct execute that approach.
The two approaches that I can see are
Approach 1
Count the overall number of purchases done by all customers, filter the ones that are greater than 1 and then check if they any of them were done within a month.
Use this as a subquery in the where clause of the main query for retrieving the customer info for all the customer_id which match this condition.
This is what i've done so far,this retrieves the customer ids of all the customers who have more than 1 purchases in total. But I do not understand how to filter out all the purchases that did not occur in a single arbitrary month.
SELECT * FROM customer_details
WHERE customer_id IN (
SELECT cust_id from purchase_logs
group by cust_id
having count(*) >= 2);
Approach 2
Create a temporary table to Count the number of monthly purchases of a specific user_id then find the MAX() of the whole table and check if that MAX value is bigger than 1 or not. Then if it is provide it as true for the main query's where clause for the customer_info.
Approach 2 feels like the more logical option but I cannot seem to understand how to write the proper subquery for it as the command MAX(COUNT(customer_id)) from purchase_logs does not seem to be a valid query.
This is the DDL diagram.
This is the Sample Data of Purchase_logs
Customer_info
and Item_info
and the expected output for this sample data would be
It is certainly possible that there is a simpler approach that I am not seeing right now.
Would appreciate any suggestions and tips on this.

You need this query:
SELECT DISTINCT cust_id
FROM purchase_logs
GROUP BY cust_id, TO_CHAR(purchase_date, 'YYYY-MON')
HAVING COUNT(DISTINCT item_id) > 1;
to get all the cust_ids of the customers who have more than 1 unique item purchased in any month and you can use with the operator IN:
SELECT *
FROM customer_details
WHERE customer_id IN (
SELECT DISTINCT cust_id -- here DISTINCT may be removed as it does not make any difference when the result is used with IN
FROM purchase_logs
GROUP BY cust_id, TO_CHAR(purchase_date, 'YYYY-MON')
HAVING COUNT(DISTINCT item_id) > 1
);

One approach might be to try
with multiplepurchase as (
select customer_id,month(purchasedate),count(*) as order_count
from purchase_logs
group by customer_id,month(purchasedate)
having count(*)>=2)
select customer_id,username,usercategory
from mutiplepurchase a
left join userinfo b
on a.customer_id=b.customer_id

Expanding on #MT0 answer:
SELECT *
FROM customer_details CD
WHERE exists (
SELECT cust_id
FROM purchase_logs PL
where CD.customer_id = PL.customer_id
GROUP BY cust_id, item_id, to_char(purchase_date,'YYYYMM')
HAVING count(*) >= 2
);

I want to display the user info of customers who have more than 1 purchases in a single arbitrary month.
Just add a WHERE filter to your sub-query.
So assuming that you wanted the month of July 2021 and you had a purchase_date column (with a DATE or TIMESTAMP data type) in your purchase_logs table then you can use:
SELECT *
FROM customer_details
WHERE customer_id IN (
SELECT cust_id
FROM purchase_logs
WHERE DATE '2021-07-01' <= purchase_date
AND purchase_date < DATE '2021-08-01'
GROUP BY cust_id
HAVING count(*) >= 2
);
If you want the users where they have bought two-or-more items in any single calendar month then:
SELECT *
FROM customer_details c
WHERE EXISTS (
SELECT 1
FROM purchase_logs p
WHERE c.customer_id = p.cust_id
GROUP BY cust_id, TRUNC(purchase_date, 'MM')
HAVING count(*) >= 2
);

Related

How to get the customer info of the customer with the highest number of transactions

I have an oracle sql database consisting of three tables and I was wondering,
What is the most efficient subquery that can be written to retrieve the information of the customer stored in the table customer_info who has performed the highest amount of purchases in total.(The purchase data is in the table purchase_logs). i.e the number of transactions one customer has performed NOT the quantity of the items purchased.
i.e my aim is to retrieve the customer details of the customer witht he highest amount of purchases done.
I have 3 tables one for the customer_info, one as the purchase_logs and the last one being the item_info.
My current Approach
SELECT * FROM customer_info
WHERE customer_id = (SELECT cust_id
FROM purchase_logs
GROUP BY cust_id
ORDER BY COUNT(*)
DESC LIMIT 1);
This doesn't seem to give me any results at all unfortunately.
This is my Database Schema along with the Sample Data of purchase_logs, customer_info, item_info and the Expected Output
I would really appreciate any help in understanding what the proper approach to solving this problem would be.
There is no limit 1 in Oracle SQL, use row limiting clause instead (fetch first in the example below):
SELECT *
FROM
(SELECT cust_id, count(*) cnt
FROM purchase_logs
GROUP BY cust_id
ORDER BY cnt desc
fetch first 1 row only with ties
) vc
join customer_info
on customer_id = vc.cust_id;

Show duplicate rows(all columns of that row) where all columns are duplicate except one column

In below table, I need to select duplicate records where all columns are duplicate except Customer Type and Price for a particular week.
For e.g
Week Customer Product Customer Type Price
1 Alex Cycle Consumer 100
1 Alex Cycle Reseller 101
2 John Motor Consumer 200
3 John Motor Consumer 200
3 John Motor Reseller 201
I am using below query but this query doesn't show me both costumer type, it just shows me consumer count(*) for a combination.
select Week, Customer, product, count(distinct Customer Type)
from table
group by Week, Customer, product
having count(distinct Customer Type) > 1
I would like to see below result, that shows me duplicate values and not just the count(*) of duplicate row. I am trying to see customers assigned to multiple customer types in a particular week for a product and at the same time show me all columns. It doesn't matter if the price is different.
Week Customer Product Customer Type Price
1 Alex Cycle Consumer 100
1 Alex Cycle Reseller 101
3 John Motor Consumer 200
3 John Motor Reseller 201
Thanks
Shaki
WITH CustomerDistribution_CTE (WeekC ,CustomerC, ProductC)
AS
(
select Week, Customer, product
from Your_Table_Name group by Week, Customer,
product having count(distinct CustomerType) > 1
)
SELECT Y.*
FROM CustomerDistribution_CTE C
inner join Your_Table_Name Y on C.WeekC =Y.Week
and C.CustomerC =Y.Customer and C.productC =Y.product
Note :Please replace "Your_Table_Name" with exact table name and Try.
One way to achieve this, using generic SQL, is to use a "derived table" like this:
select x.*
from tablex x
inner join (
select Week, Customer, Product
from tablex
group by Week, Customer, Product
having count(*) > 1
) d on x.Week = d.Week and x.Customer = d.Customer and x.Product = d.Product
You can do that by using DISTINCT like
select DISTINCT Customer,Product,Customer_Type,Price from Your_Table_Name
will look for DISTINCT combination.
Note: This query if of SQL Server
From the expected result that you have pasted, it looks like you are not concerned about the week.
If you have a ID (incremental PK), it would be much simpler like below
select * from table where ID not in
(select max(ID) from table group by Customer, Product, CustomerType having count(*) > 1 )
This is tested on MySQL. Do you have a ID column?
In case you don't have a ID column, try the below:
select max(week) week, Customer, Product, CustomerType, max(price) from device group by Customer, Product, CustomerType;
I have not verified this one.
This will return your expected result set:
select *
from table
-- Teradata syntax to filter the result of an OLAP-function
-- (similar to HAVING after GROUP BY)
qualify
count(*)
over (partition by Week, Customer, product) > 1
For other DBMSes you will need to nest your query:
select *
from
(
select ...,
count(*)
over (partition by Week, Customer, product) as cnt
from table
) as dt
where cnt > 1
Edit:
After re-reading your description above Select might be not exactly what you want, because it will also return rows with a single type. Then switch to:
select *
from table
-- Teradata syntax to filter the result of an OLAP-function
-- (similar to HAVING after GROUP BY)
qualify -- at least two different types:
min(Customer_Type) over (partition by Week, Customer, product)
<> max(Customer_Type) over (partition by Week, Customer, product)

SQL grouping with multiple rows

There's a table that I use that lists invoice detail. So for instance let's say a customer checks out with 2 items, there are 2 rows for each item.
Right now my SQL Query looks like this:
Select date
,order_id
,count(distinct(item_name))
from Table_1
group by 1,2
Rather than grouping it by order_id. Is there anyway to modify this query to find the number of Orders that have X amount of items on a specific date. So on 1/1/1990 5 orders have 3 items, 6 orders have 2 items, etc.
Thanks for the help!
If I'm understanding your question correctly, you could use a subquery grouping by the item count:
select t.date, t.itemCount, t.count(order_id)
from (
Select date
,order_id
,count(distinct(item_name)) AS itemCount
from Table_1
group by 1,2
) AS t
group by date, itemCount

MS-Access: HAVING clause not returning any records

I have a Select query to extract Customer Names and Purchase Dates from a table. My goal is to select only those names and dates for customers who have ordered on more than one distinct date. My code is as follows:
SELECT Customer, PurchDate
FROM (SELECT DISTINCT PurchDate, Customer
FROM (SELECT CDate(FORMAT(DateAdd("h",-7,Mid([purchase-date],1,10)+""+Mid([purchase-date],12,8)), "Short Date")) AS PurchDate,
[buyer-name] AS Customer
FROM RawImport
WHERE sku ALIKE "%RE%"))
GROUP BY Customer, PurchDate
HAVING COUNT(PurchDate)>1
ORDER BY PurchDate
This returns no results, even though there are many customers with more than one Purchase Date. The inner two Selects work perfectly and return a set of distinct dates for each customer, so I believe there is some problem in my GROUP/HAVING/ORDER clauses.
Thanks in advance for any help!
You are doing in the inner select
SELECT DISTINCT PurchDate, Customer
and in the outter select
GROUP BY Customer, PurchDate
That mean all are
having count(*) = 1
I cant give you the exact sintaxis in access but you need something like this
I will use YourTable as a replacement of your inner derivated table to make it easy to read
SELECT DISTINCT Customer, PurchDate
FROM YourTable
WHERE Customer IN (
SELECT Customer
FROM (SELECT DISTINCT Customer, PurchDate
FROM YourTable)
GROUP BY Customer
HAVING COUNT(*) > 1
)
inner select will give you which customer order on more than one day.
outside select will bring you those customer on all those days.
.
Maybe you can try something simple to get the list of customer who brought in more than one day like this
SELECT [buyer-name]
FROM RawImport
WHERE sku ALIKE "%RE%"
GROUP BY [buyer-name]
HAVING Format(MAX(purchase-date,"DD/MM/YYYY")) <>
Format(MIN(purchase-date,"DD/MM/YYYY"))

select multiple records based on order by

i have a table with a bunch of customer IDs. in a customer table is also these IDs but each id can be on multiple records for the same customer. i want to select the most recently used record which i can get by doing order by <my_field> desc
say i have 100 customer IDs in this table and in the customers table there is 120 records with these IDs (some are duplicates). how can i apply my order by condition to only get the most recent matching records?
dbms is sql server 2000.
table is basically like this:
loc_nbr and cust_nbr are primary keys
a customer shops at location 1. they get assigned loc_nbr = 1 and cust_nbr = 1
then a customer_id of 1.
they shop again but this time at location 2. so they get assigned loc_nbr = 2 and cust_Nbr = 1. then the same customer_id of 1 based on their other attributes like name and address.
because they shopped at location 2 AFTER location 1, it will have a more recent rec_alt_ts value, which is the record i would want to retrieve.
You want to use the ROW_NUMBER() function with a Common Table Expression (CTE).
Here's a basic example. You should be able to use a similar query with your data.
;WITH TheLatest AS
(
SELECT *, ROW_NUMBER() OVER (PARTITION BY group-by-fields ORDER BY sorting-fields) AS ItemCount
FROM TheTable
)
SELECT *
FROM TheLatest
WHERE ItemCount = 1
UPDATE: I just noticed that this was tagged with sql-server-2000. This will only work on SQL Server 2005 and later.
Since you didn't give real table and field names, this is just psuedo code for a solution.
select *
from customer_table t2
inner join location_table t1
on t1.some_key = t2.some_key
where t1.LocationKey = (select top 1 (LocationKey) as LatestLocationKey from location_table where cust_id = t1.cust_id order by some_field)
Use an aggregate function in the query to group by customer IDs:
SELECT cust_Nbr, MAX(rec_alt_ts) AS most_recent_transaction, other_fields
FROM tableName
GROUP BY cust_Nbr, other_fields
ORDER BY cust_Nbr DESC;
This assumes that rec_alt_ts increases every time, thus the max entry for that cust_Nbr would be the most recent entry.
By using time and date we can take out the recent detail for the customer.
use the column from where you take out the date and the time for the customer.
eg:
SQL> select ename , to_date(hiredate,'dd-mm-yyyy hh24:mi:ss') from emp order by to_date(hiredate,'dd-mm-yyyy hh24:mi:ss');