I need to select only the lines where MIN date is 7 days ago.
I have Customers with invoices, the invoices have due dates. I can select MIN due date of these invoices per Customer. I am now stuck as I cannot select only the Customers who are exactly 7 days overdue with their oldest invoice.
This is what I have:
select
customerID,
MIN(dueDate) as min_due_date
from invoices
where (invoiceStatus NOT IN ('PaidPosted','Cancelled'))
and (entity = 'HQ')
group by (customerID)
I tried adding:
and min_due_date = dateadd(day, -7, SYSDATE())
This does not work.
What am I doing wrong?
Thanks.
Use a having clause to filter the minimum date:
select customerID, min(dueDate) as min_due_date
from invoices
where invoiceStatus not in ('PaidPosted', 'Cancelled') and
entity = 'HQ'
group by customerID
having min(duedate) = current_date - interval '7 day';
Note that date functions are highly database specific. The above is standard SQL, but the exact syntax depends on the database you are using.
Thank you, Gordon, you put me on the right track!
This is what eventually did the trick:
select customerID, min(dueDate) as min_due_date
from invoices
where invoiceStatus not in ('PaidPosted', 'Cancelled') and
entity = 'HQ'
having MIN(dueDate) = trunc(sysdate) -7
group by customerID
Related
So I am trying to find count of customers who placed order both in this month and previous month. I have to find this from the beginning of last year. I came up with a query which obviously doesn't work. Can I get some help with this please?
Query:
SELECT DATE_TRUNC('month', month_column), COUNT(DISTINCT(customer_id))
FROM table
WHERE month_column >= '2021-01-01' AND customer_id IN (
SELECT customer_id
FROM table
WHERE month_column = month_column - INTERVAL '1 month')
GROUP BY 1
NOTE: month_column has only month number i.e., '2021-01-01', '2021-02-01' etc.
I am using postgresql.
This is my first stack overflow question. So, if I didn't abide by any rules, I apologize.
To make this trivial, you can use 2 queries. Get customerID's from this month (insert into temp table 1) and customerID's from last month (insert into temp table 2). Lastly just inner join both tables on customerID
something like the below
SELECT customer_id
INTO #thisMonth
FROM customer
WHERE month_column > DATEADD(month, 0, GETDATE())
SELECT customer_id
INTO #prevMonth
FROM customer
WHERE month_column > DATEADD(month, -1, GETDATE())
SELECT COUNT(customer_id)
FROM #thisMonth tm
INNERJOIN #prevMonth pm ON tm.customer_id = pm.customerID
I want to know the trick to find the list of customers who are transacting for consecutive 3 months ,that could be any 3 consecutive months with any number of occurrence.
example: suppose there is customer who transact in January then keep transacting till march then he stopped transacting.I want the list of these customer from my database .
I am working on AWS Athena.
One method uses aggregation and window functions:
select customer_id, yyyymm_2
from (select date_trunc(month, transactdate) as yyyymm, customer_id,
lag(date_trunc(month, transactdate), 2) over (partition by customer_id order by date_trunc(month, transactdate)) as prev_yyyymm_2
from t
where transactdate >= '2017-01-01' and
transactadte < '2019-01-01'
)
where prev_dt_2 = yyyymm - interval '2' month;
This aggregates transactions by month and looks at the transaction date two rows earlier. The outer filter checks that that date is exactly 2 months earlier.
I am trying to create a report to show me the last date a customer filed a ticket.
Customers can file dozens of tickets. I want to know when the last ticket was filed and show how many days it's been since they have done so.
The fields I have are:
Customer,
Ticket_id,
Date_Closed
All from the Same table "Tickets"
I'm thinking I want to do a ranking of tickets by min date? I tried this query to grab something but it's giving me all the tickets from the customer. (I'm using SQL in a product called Domo)
select * from (select *, rank() over (partition by "Ticket_id"
order by "Date_Closed" desc) as date_order
from tickets ) zd
where date_order = 1
This should be simple enough,
SELECT customer,
MAX (date_closed) last_date,
ROUND((SYSDATE - MAX (date_closed)),0) days_since_last_ticket_logged
FROM emp
GROUP BY customer
select Customer, datediff(day, date_closed, current_date) as days_since_last_tkt
from
(select *, rank() over (partition by Customer order by "Date_Closed" desc) as date_order
from tickets) zd
join tickets t on zd.date_closed = t.date_closed
where zd.date_order = 1
Or you can simply do
select customer, datediff(day, max(Date_closed), current_date) as days_since_last_tkt
from tickets
group by customer
To select other fields
select t.*
from tickets t
join (select customer, max(Date_closed) as mxdate,
datediff(day, max(Date_closed), current_date) as days_since_last_tkt
from tickets
group by customer) tt
on t.customer = tt.customer and tt.mxdate = t.date_closed
I would do this with a simple sub-query to select the last closed date for the customer. Then compare this to today with datediff() to get the number of days since last closed.
Select
LastTicket.Customer,
LastTicket.LastClosedDate,
DateDiff(day,LastTicket.LastClosedDate,getdate()) as DaysSinceLastClosed
From
(select
tickets.customer
max(tickets.dateClosed) as LastClosedDate
from tickets
Group By tickets.Customer) as LastTicket
Based on the responses this is what I did:
select "Customer",
Max("date_closed") "last_date,
round(datediff(DAY, CURRENT_DATE, max("date_closed")), 0) as "Closed_date"
from tickets
group by "Customer"
ORDER BY "Customer"
I want to create a one time reporting job that generate day wise orders placed and and revenue earned for last 5 days. please help me.
Query 1:
SELECT order_date,
COUNT(web_order_number) AS num_orders,
SUM(order_total) AS daily_total
FROM purchase_orders_owner
WHERE order_date>=date_sub(CURRENT_DATE, INTERVAL 5 DAY)
GROUP BY order_date;
Query 2:
SELECT order_date ,
COUNT(web_order_number) AS num_orders ,
SUM(order_total) AS daily_total
FROM purchase_orders_owner
GROUP BY DATE(order_date)
In Oracle, you would do this as:
SELECT trunc(order_date) as order_date,
COUNT(web_order_number) AS num_orders,
SUM(order_total) AS daily_total
FROM purchase_orders_owner
WHERE order_date >= trunc(sysdate - 5)
GROUP BY trunc(order_date);
Your queries use MySQL functions that Oracle does not support.
I am essentially doing the following query (edited):
Select count(orders)
From Orders_Table
Where Order_Open_Date<=##/##/####
and Order_Close_Date>=##/##/####
Where the ##/##/##### is the same date. So in essence the number of 'open' orders for any given day. However I am wanting this same count for every single day for a year and don't want to write a separate query for each day for the whole year. I'm sorry this is probably really simple but I am new to SQL and I guess I don't know how to search for an answer to this question since my searches have come up with nothing. Thanks for any help you can offer.
why not
select Order_Date, count(orders) from Orders_Table group by Order_Date
and for last year
select Order_Date, count(orders) from Orders_Table where Order_Date > DATE_SUB(CURDATE(), INTERVAL 1 YEAR) group by Order_Date;
SELECT CONVERT(VARCHAR, Order_Date, 110), count(orders)
FROM Orders_Table
WHERE Order_Date = BETWEEN #A AND #B
GROUP BY CONVERT(VARCHAR, Order_Date, 110)
If you want to have every day of the year, including those with no orders, you will need to generate a temporary table or similar containing every date in the range and left/right join it to the Orders_Table data. This depends upon which RDBMS you're using. In SQL Server I have done this using a user defined function which returns a table variable.