I'm trying to union two different results but I'm not sure how to remove the blank values. The paid_date and check_date are sometimes different. What do I need to do to modify the code? I've tried doing MAX but didn't seem to work. Thanks.
Results:
What I want:
WITH PAID AS
(
SELECT PIDATE
,SUM(PAY) AS PAID
,ROW_NUMBER() OVER(ORDER BY PAYDATE DESC) AS ROW_NUM
FROM CLAIM
WHERE PAY<>0
GROUP BY PAYDATE
)
,TOTAL_PAID AS
(
SELECT CONVERT(DATE,CONVERT(VARCHAR(10),PAYDATE)) AS PAID_DATE
,FORMAT(PAID,'C','EN-US') AS PAID
FROM PAID
WHERE ROW_NUM <= 10
)
,CHECKS AS
(
SELECT DISTINCT CHECK_DATE
,SUM(PAYMENT) AS PAYMENT
,ROW_NUMBER() OVER(ORDER BY CHECK_DATE DESC) AS ROW_NUM
FROM CR
GROUP BY CHECK_DATE
)
,TOTAL_CHECKS AS
(
SELECT CONVERT(DATE,CONVERT(VARCHAR(10),CHECK_DATE)) AS CHECK_DATE
,FORMAT(PAYMENT,'C','EN-US') AS PAYMENT
FROM CHECKS
WHERE ROW_NUM <= 10
)
,FINAL AS
(
SELECT PAID_DATE
,PAID
,'' AS CHECK_DATE
,'' AS PAYMENT
FROM TOTAL_PAID
UNION ALL
SELECT '' AS PAID_DATE
,'' AS PAID
,CHECK_DATE
,PAYMENT
FROM TOTAL_CHECKS
)
SELECT *
FROM FINAL
In this scenario, you must not use union. you must use join instead.
Try join two sequence on paid_date = check_date and select proper field from each set.
Related
Can someone please help me to find the average time between first and second purchase on a product level.
This is what I have written -
Select A.CustomerId,A.ProductId , A.OrderSequence, (Case WHEN OrderSequence = 1 THEN OrderDate END) AS First_Order_Date,
MAX(Case WHEN OrderSequence = 2 THEN OrderDate END) AS Second_Order_Date
From
(
Select t.CustomerId, t.ProductId, t.OrderDate,
Dense_RANK() OVER (PARTITION BY t.CustomerId, t.ProductId ORDER BY OrderDate Asc) as OrderSequence
From Transactions t (NOLOCK)
Where t.SiteKey = 01
Group by t.CustomerId, t.ProductId, t.OrderDate)
A
Where A.OrderSequence IN (1,2)
Group By A.Customer_Id, A.ProductId, A.OrderSequence, A.OrderDate
Sample Data:
It looks like row-numbering and LEAD should do the trick for you here.
Don't use NOLOCK unless you really know what you're doing
It's unclear if you want the results to be partitioned by CustomerId also. If not, you can remove it everywhere in the query
SELECT
A.CustomerId,
A.ProductId,
AVG(DATEDIFF(day, OrderDate, NextOrderDate))
FROM
(
SELECT
t.CustomerId,
t.ProductId,
t.OrderDate,
ROW_NUMBER() OVER (PARTITION BY t.CustomerId, t.ProductId ORDER BY OrderDate) AS rn,
LEAD(OrderDate) OVER (PARTITION BY t.CustomerId, t.ProductId ORDER BY OrderDate) AS NextOrderDate
FROM Transactions t
WHERE t.SiteKey = '01'
) t
WHERE t.rn = 1
GROUP BY
t.Customer_Id,
t.ProductId;
I’m struggling a bit here. The data is fabricated, but the query concept is very real.
I need to select the Customer, Current Amount, Previous Amount, Sequence and Date
WHERE DATE < 1190105
AND the DATE/SEQ is the maximum date/seq prior to that date point grouping by customer.
I’ve spent quite a few days trying all sorts of things using HAVING, nested select to try and obtain the max-date/amount and min-date/amount by customer and can’t quite get my head around it. I’m sure it should be quite easy, but any help you can offer would be really appreciated.
Thanks
**SEQ DATE CUSTOMER AMOUNT**
1 1181225 Bob 400
2 1181226 Fred 300
3 1190101 Bob 100
4 1190104 Fred 500
5 1190104 George 200
6 1190105 Bob 150
7 1190106 Bob 200
8 1190110 Fred 160
9 1190110 Bob 300
10 1190112 Fred 400
Opt 1 use row number and lag functions
SELECT
ROW_NUMBER() OVER (Partition By CustomerID Order By [Date]) as Sec,
[Date],
Customer,
Amount as CurrentAmount,
Lead(Amount) OVER (Partition By CustomerID, Order By [Date]) as PreviousAmount
FROM
YourTable
WHERE
[DATE] < 1190105
Opt use outer apply
SELECT
ROW_NUMBER() OVER (Partition By Customer Order By [Date]) as Sec,
[Date],
Customer,
Amount as CurrentAmount,
Prev.Amount as PreviousAmount
FROM
YourTable T
OUTER APPLY (
SELECT TOP 1 Amount FROM YourTable
WHERE Customer = T.Customer AND [Date] < T.[Date]
ORDER BY [DATE] DESC
) Prev
WHERE
DATE < 1190105
Opt 3 use a correlated subquery
SELECT
ROW_NUMBER() OVER (Partition By Customer Order By [Date]) as Sec,
[Date],
Customer,
Amount as CurrentAmount,
(
SELECT TOP 1 Amount FROM YourTable
WHERE Customer = T.Customer AND [Date] < T.[Date]
ORDER BY [DATE] DESC
) as PreviousAmount
FROM YourTable
WHERE
DATE < 1190105
First restrict the rows with the date filter, then search for the max by customer.
Using GROUP BY:
DECLARE #FilterDate INT = 1190105
;WITH MaxDateByCustomer AS
(
SELECT
T.CUSTOMER,
MaxSEQ = MAX(T.SEQ)
FROM
YourTable AS T
WHERE
T.Date < #FilterDate
GROUP BY
T.CUSTOMER
)
SELECT
T.*
FROM
YourTable AS T
INNER JOIN MaxDateByCustomer AS M ON
T.CUSTOMER = M.CUSTOMER AND
T.SEQ = M.MaxSEQ
Using ROW_NUMBER window function:
DECLARE #FilterDate INT = 1190105
;WITH DateRankingByCustomer AS
(
SELECT
T.*,
DateRanking = ROW_NUMBER() OVER (PARTITION BY T.CUSTOMER ORDER BY T.SEQ DESC)
FROM
YourTable AS T
WHERE
T.Date < #FilterDate
)
SELECT
D.*
FROM
DateRankingByCustomer AS D
WHERE
D.DateRanking = 1
Hi I am trying to create a windowed query in SQL that shows me the days since last order for each customer.
It now shows me the days in between each order.
What do I need to change in my query to have it only show the days since the last and the previous order per customer? Now it shows it for every order the customer made.
Query:
SELECT klantnr,besteldatum,
DATEDIFF(DAY,LAG(besteldatum) OVER(PARTITION BY klantnr ORDER BY besteldatum),besteldatum) AS DaysSinceLastOrder
FROM bestelling
GROUP BY klantnr,besteldatum;
You can use row_number() to order the rows by besteldatum for each klantnr, and return the latest two using a derived table (subquery) or common table expression.
derived table version:
select klantnr, besteldatum, DaysSinceLastOrder
from (
select klantnr, besteldatum
, DaysSinceLastOrder = datediff(day,lag(besteldatum) over (partition by klantnr order by besteldatum),besteldatum)
, rn = row_number() over (partition by klantnr order by besteldatum desc)
from bestelling
group by klantnr, besteldatum
) t
where rn = 1
common table expression version:
;with cte as (
select klantnr, besteldatum
, DaysSinceLastOrder = datediff(day,lag(besteldatum) over (partition by klantnr order by besteldatum),besteldatum)
, rn = row_number() over (partition by klantnr order by besteldatum desc)
from bestelling
group by klantnr, besteldatum
)
select klantnr, besteldatum, DaysSinceLastOrder
from cte
where rn = 1
If you want one row per customer, rn = 1 is the proper filter. If you want n number of latest rows, use rn < n+1.
I have this table:
ID COLOR TYPE DATE
-------------------------------
1 blue A 2012.02.05
2 white V 2010.10.23
3 white V 2014.03.05
4 black S 2013.02.14
I'd like to select only the ID, but in case of 2nd and 3rd rows I want to select the 3rd row because of its latest DATE value.
I have tried this query but it gives back all the two rows:
SELECT
ID, MAX(DATE) OVER(PARTITION BY COLOR, TYPE)
FROM
TABLE
WHERE
...
How can I select just one column value while I group the rows by other columns, please?
;WITH CTE AS
(
SELECT * , ROW_NUMBER() OVER (PARTITION BY COLOR,[TYPE] ORDER BY [DATE] DESC) rn
FROM TableName
)
SELECT ID
,COLOR
,[TYPE]
,[DATE]
FROM CTE
WHERE rn = 1
OR
SELECT ID
,COLOR
,[TYPE]
,[DATE]
FROM
(
SELECT * , ROW_NUMBER() OVER (PARTITION BY COLOR,[TYPE] ORDER BY [DATE] DESC) rn
FROM TableName
) A
WHERE rn = 1
I have the following query
;WITH tmp AS
(
SELECT *, ROW_NUMBER()
OVER
(PARTITION BY to_tel, duration, call_date
ORDER BY rates_start DESC) as rn
FROM ##TempTable
)
SELECT *
FROM tmp
WHERE rn = 1
ORDER BY customer_id, to_code, duration
But I would like to modify it where it doesn't give me the maximum rates_start, but the maximum rates_start before a certain date. Is there any way I can do this?
You can add WHERE inside the cte part. I'm not sure if you still want to partition by call_date in this case (I removed it). Change the PARTITION BY part if needed.
;WITH tmp AS
(
SELECT *, ROW_NUMBER()
OVER
(PARTITION BY to_tel, duration
ORDER BY rates_start DESC) as rn
FROM ##TempTable
WHERE call_date < #somedate
)
SELECT *
FROM tmp
WHERE rn = 1
ORDER BY customer_id, to_code, duration