How can i get count of customers per day by unique and repeat customer for specific date?

Question

I am trying to get a result from my order table to get list of counts of customers who 1st time ordered and repeat orders. Something like below.

Date               


        

Answer 1

this is my answer but not sure is still can improve.

SELECT userID, COUNT(*) AS repeat_order_cnt FROM
(SELECT DATE(OrderDate) AS order_DT, userID, COUNT(*) AS no_of_order FROM order
AND YEAR(orderDate) = '2015'
AND MONTH(orderDate) = '01'
GROUP BY order_DT,userID) AS order2
GROUP BY userID
HAVING COUNT(*) > 1

Answer 2

A slightly complicated query but this should do:

First Time Users: Just Group by customerID to get the min orderdate and then group by on that date to get the number of new users on a particular day. Query would look like this:

select date(mdate) as day, COUNT(*) from (select customerid, min(orderdate) as mDate from orders GROUP BY CustomerID)q1 GROUP BY day;

Repeat Users: First filter out all such orderno which were placed as first orders and then do a group by orderdate to get repeat. Query would be :

select date(orderdate) day, COUNT(*) from (select * from orders where orderno not in (select orders.orderno  from orders JOIN (select customerid, min(orderdate) as mdate from orders GROUP BY CustomerID)as order2 ON (orders.customerid = order2.customerid) and (orders.orderdate = order2.mdate))) as q1 GROUP BY day;

You can do a join on day for both these queries to get combined results in a way you mentioned. Let me know if doesn't work

EDIT: This would be the complete query: Here I am doing a UNION on both left and right outer joins since it might happen that you come across where there are no new requests or no repeated requests. This would take care of both the scenarios.

select q2.*, q3.repeated from (select date(mdate) as day, COUNT(*) as first from (select customerid, min(orderdate) as mDate from orders GROUP BY CustomerID)q1 GROUP BY day) as q2 LEFT OUTER JOIN (select date(orderdate) day, COUNT(*) as repeated from (select * from orders where orderno not in (select orders.orderno  from orders JOIN (select customerid, min(orderdate) as mdate from orders GROUP BY CustomerID)as order2 ON (orders.customerid = order2.customerid) and (orders.orderdate = order2.mdate))) as q1 GROUP BY day) as q3 on q2.day = q3.day UNION select q2.*, q3.repeated from (select date(mdate) as day, COUNT(*) as first from (select customerid, min(orderdate) as mDate from orders GROUP BY CustomerID)q1 GROUP BY day) as q2 RIGHT OUTER JOIN (select date(orderdate) day, COUNT(*) as repeated from (select * from orders where orderno not in (select orders.orderno  from orders JOIN (select customerid, min(orderdate) as mdate from orders GROUP BY CustomerID)as order2 ON (orders.customerid = order2.customerid) and (orders.orderdate = order2.mdate))) as q1 GROUP BY day) as q3 on q2.day = q3.day

Answer 3

You can get the total orders per day by grouping on OrderDate:

SELECT OrderDate, COUNT(OrderNumber) AS total FROM orders GROUP BY OrderDate

And you can get the no. of first orders per day from the following query :

SELECT OrderDate, COUNT(q1.CustomerID) AS first FROM (SELECT CustomerID, min(OrderDate) AS OrderDate FROM orders GROUP BY CustomerID)q1 GROUP BY q1.OrderDate

Now join these two on OrderDate to get the distribution of first and repeated orders :

SELECT a.OrderDate, a.first, (b.total - a.first) AS repeated FROM
(SELECT OrderDate, COUNT(q1.CustomerID) AS first FROM (SELECT CustomerID, min(OrderDate) AS OrderDate FROM orders GROUP BY CustomerID)q1 GROUP BY q1.OrderDate)a
JOIN
(SELECT OrderDate, COUNT(OrderNumber) AS total FROM orders GROUP BY OrderDate)b
on(a.OrderDate = b.OrderDate)