I have 2 tables,
tblCustomer
CustomerID(PK), FirstName, Surname)
tblPurchases
PurchaseID(PK), PurchaseDate, Qty, CustomerID(FK).
I want to display all the customers who purchased products after five (5) days or more since their last purchase in the following manner.
FirstName diff in days since last purchase
Alex 7
Thanks!
Try with the below query.
SELECT FirstName, DATEDIFF(DAY, t.PurchaseDate, getdate()) as 'diff in days since last purchase'
FROM tblCustomer c
JOIN (SELECT CustomerID, MAX(PurchaseDate)PurchaseDate
FROM tblPurchases
GROUP BY CustomerID )t ON c.CustomerID=t.CustomerID
WHERE DATEDIFF(DAY, PurchaseDate, getdate())>5
SELECT FirstName,
DATEDIFF(DAY, t.PurchaseDate, getdate()) 'diff in days since last purchase'
FROM tblCustomer c
JOIN (SELECT CustomerID, MAX(PurchaseDate)PurchaseDate
FROM tblPurchases
GROUP BY CustomerID )t ON c.CustomerID=t.CustomerID
WHERE DATEDIFF(DAY, t.PurchaseDate, getdate())>5
Give a row number based on the last purchase date for each CustomerId by joining both the tables.
Then find the difference in number of days between current date and PurchaseDate by using DATEDIFF.And also give the number of days difference in the WHERE clause.
Query
;WITH CTE AS(
SELECT [rn] = ROW_NUMBER() OVER(
PARTITION BY t.[CustomerID]
ORDER BY t.[PurchaseDate] DESC
), t.[CustomerID], t.[FirstName], t.[PurchaseDate]
FROM (
SELECT t1.[CustomerID], t1.[FirstName], t2.[PurchaseDate]
FROM [tblCustomer] t1
JOIN [tblPurchases] t2
ON t1.[CustomerID] = t2.[CustomerID]
)t
)
SELECT [FirstName],
DATEDIFF(DAY, [PurchaseDate], GETDATE()) AS [diff in days since last purchase]
FROM CTE
WHERE [rn] = 1
AND DATEDIFF(DAY, [PurchaseDate], GETDATE()) > 5;
;WITH T AS
(
SELECT
*,
DATEDIFF(DAY, [PurchaseDate], GETDATE()) AS DiffInDays
FROM #tblPurchases
WHERE DATEDIFF(DAY, [PurchaseDate], GETDATE()) > 5
)
SELECT
C.FirstName,
MAX(DiffInDays) AS DiffInDays
FROM T
LEFT JOIN #tblCustomer C ON T.CustomerId=C.CustomerId
GROUP BY C.FirstName
Related
I have a table which has the following columns: DeskID *, ProductID *, Date *, Amount (where the columns marked with * make the primary key). The products in use vary over time, as represented in the image below.
Table format on the left, and a (hopefully) intuitive representation of the data on the right for one desk
The objective is to have the sum of the latest amounts of products by desk and date, including products which are no longer in use, over a date range.
e.g. using the data above the desired table is:
So on the 1st Jan, the sum is 1 of Product A
On the 2nd Jan, the sum is 2 of A and 5 of B, so 7
On the 4th Jan, the sum is 1 of A (out of use, so take the value from the 3rd), 5 of B, and 2 of C, so 8 in total
etc.
I have tried using a partition on the desk and product ordered by date to get the most recent value and turned the following code into a function (Function1 below) with #date Date parameter
select #date 'Date', t.DeskID, SUM(t.Amount) 'Sum' from (
select #date 'Date', t.DeskID, t.ProductID, t.Amount
, row_number() over (partition by t.DeskID, t.ProductID order by t.Date desc) as roworder
from Table1 t
where 1 = 1
and t.Date <= #date
) t
where t.roworder = 1
group by t.DeskID
And then using a utility calendar table and cross apply to get the required values over a time range, as below
select * from Calendar c
cross apply Function1(c.CalendarDate)
where c.CalendarDate >= '20190101' and c.CalendarDate <= '20191009'
This has the expected results, but is far too slow. Currently each desk uses around 50 products, and the products roll every month, so after just 5 years each desk has a history of ~3000 products, which causes the whole thing to grind to a halt. (Roughly 30 seconds for a range of a single month)
Is there a better approach?
Change your function to the following should be faster:
select #date 'Date', t.DeskID, SUM(t.Amount) 'Sum'
FROM (SELECT m.DeskID, m.ProductID, MAX(m.[Date) AS MaxDate
FROM Table1 m
where m.[Date] <= #date) d
INNER JOIN Table1 t
ON d.DeskID=t.DeskID
AND d.ProductID=t.ProductID
and t.[Date] = d.MaxDate
group by t.DeskID
The performance of TVF usually suffers. The following removes the TVF completely:
-- DROP TABLE Table1;
CREATE TABLE Table1 (DeskID int not null, ProductID nvarchar(32) not null, [Date] Date not null, Amount int not null, PRIMARY KEY ([Date],DeskID,ProductID));
INSERT Table1(DeskID,ProductID,[Date],Amount)
VALUES (1,'A','2019-01-01',1),(1,'A','2019-01-02',2),(1,'B','2019-01-02',5),(1,'A','2019-01-03',1)
,(1,'B','2019-01-03',4),(1,'C','2019-01-03',3),(1,'B','2019-01-04',5),(1,'C','2019-01-04',2),(1,'C','2019-01-05',2)
GO
DECLARE #StartDate date=N'2019-01-01';
DECLARE #EndDate date=N'2019-01-05';
;WITH cte_p
AS
(
SELECT DISTINCT DeskID,ProductID
FROM Table1
WHERE [Date] <= #EndDate
),
cte_a
AS
(
SELECT #StartDate AS [Date], p.DeskID, p.ProductID, ISNULL(a.Amount,0) AS Amount
FROM (
SELECT t.DeskID, t.ProductID
, MAX(t.Date) AS FirstDate
FROM Table1 t
WHERE t.Date <= #StartDate
GROUP BY t.DeskID, t.ProductID) f
INNER JOIN Table1 a
ON f.DeskID=a.DeskID
AND f.ProductID=a.ProductID
AND f.[FirstDate]=a.[Date]
RIGHT JOIN cte_p p
ON p.DeskID=a.DeskID
AND p.ProductID=a.ProductID
UNION ALL
SELECT DATEADD(DAY,1,a.[Date]) AS [Date], t.DeskID, t.ProductID, t.Amount
FROM Table1 t
INNER JOIN cte_a a
ON t.DeskID=a.DeskID
AND t.ProductID=a.ProductID
AND t.[Date] > a.[Date]
AND t.[Date] <= DATEADD(DAY,1,a.[Date])
WHERE a.[Date]<#EndDate
UNION ALL
SELECT DATEADD(DAY,1,a.[Date]) AS [Date], a.DeskID, a.ProductID, a.Amount
FROM cte_a a
WHERE NOT EXISTS(SELECT 1 FROM Table1 t
WHERE t.DeskID=a.DeskID
AND t.ProductID=a.ProductID
AND t.[Date] > a.[Date]
AND t.[Date] <= DATEADD(DAY,1,a.[Date]))
AND a.[Date]<#EndDate
)
SELECT [Date], DeskID, SUM(Amount)
FROM cte_a
GROUP BY [Date], DeskID;
I am trying to solve a problem using SQL query and need some expert's advice.
I have below transaction table.
-- UserID, ProductId, TransactionDate
-- 1 , 2 , 2014-01-01
-- 1 , 3 , 2014-01-05
-- 2 , 2 , 2014-01-02
-- 2 , 3 , 2014-05-07
.
.
.
What I am trying to achieve is to find all user who purchased more than one product WITHIN 30 DAYS .
My query so far is like
select UserID, COUNT(distinct ProductID)
from tableA
GROUP BY UserID HAVING COUNT(distinct ProductID) > 1
I am not sure where to apply "WITH IN 30 DAYS" logic in the query .
The outcome should be :
1, 2
2, 1
Thanks in advance for your help.
Edit: Within 30 Days
SQL Fiddle
SELECT
a.UserID,
COUNT(DISTINCT ProductID)
FROM TableA a
INNER JOIN (
SELECT UserID, TransactionDate = MAX(TransactionDate)
FROM TableA
GROUP BY UserID
) AS t
ON t.UserID = a.UserID
AND a.TransactionDate >= DATEADD(DAY, -30, t.TransactionDate)
AND a.TransactionDate <= t.TransactionDate
GROUP BY a.UserID
You can use GROUP BY YEAR(TransactionDate), MONTH(TransactionDate)
SELECT
UserID,
COUNT(DISTINCT ProductID)
FROM TableA
GROUP BY
UserID, YEAR(TransactionDate), MONTH(TransactionDate)
HAVING
COUNT(DISTINCT ProductID) > 1
Just add a where clause.
SELECT UserID, COUNT(DISTINCT ProductID) cnt
FROM tableA
WHERE TransactionDate >= CAST(DATEADD(DAY,-30,GETDATE()) AS DATE)
GROUP BY UserID
HAVING COUNT(DISTINCT ProductID) > 1
This works because the where clause is performed BEFORE the Group By and Having. So first it filters out all transactions over 30 days old and then returns only people who bought two distinct products.
Query Processing Order:
http://blog.sqlauthority.com/2009/04/06/sql-server-logical-query-processing-phases-order-of-statement-execution/
I have a date column Order_date and I am looking for ways to calculate the date difference between customer last order date and his recent previous ( previous form last) order_date ....
Example
Customer : 1, 2 , 1 , 1
Order_date: 01/02/2007, 02/01/2015, 06/02/2014, 04/02/2015
As you can see customer # 1 has three orders.
I want to know the date difference between his recent order date (04/02/2015) and his recent previous (06/02/2014).
For SQL Server 2012 & 2014 you could use LAG with a DATEDIFF to see the number of days between them.
For older versions, a CTE would probably be your best bet:
;WITH CTE AS
(
SELECT CustomerID,
Order_Date,
rn = ROW_NUMBER() OVER (PARTITION BY CustomerID ORDER BY Order_Date DESC)
)
SELECT c1.CustomerID,
DATEDIFF(d, c1.Order_Date, c2.Order_Date)
FROM CTE c1
INNER JOIN CTE c2 ON c2.rn = c1.rn + 1
In SQL Server 2012+, you can use lag() to get the difference between any two dates:
select t.*,
datediff(day, lag(order_date) over (partition by customer order by order_date),
order_date) as days_dff
from table t;
If you have an older version, you can do something similar with correlated subqueries or outer apply.
EDIT:
If you just want the difference between the two most recent dates, use conditional aggregation instead:
select customer,
datediff(day, max(case when seqnum = 2 then order_date end),
max(case when seqnum = 1 then order_date end)
) as MostRecentDiff
from (select t.*,
row_number() over (partition by customer order by order_date desc) as seqnum
from table t
) t
group by customer;
If you're using SQL Server 2008 or later, you can try CROSS APPLY.
SELECT [customers].[customer_id], DATEDIFF(DAY, MIN([recent_orders].[order_date]), MAX([recent_orders].[order_date])) AS [elapsed]
FROM [customers]
CROSS APPLY (
SELECT TOP 2 [order_date]
FROM [orders]
WHERE ([orders].[customer_id] = [customers].[customer_id])
) [recent_orders]
GROUP BY [customers].[customer_id]
SELECT DATEDIFF(DAY, Y.PrevLastOrderDate, Y.LastOrderDate) AS PreviousDays
FROM
(
SELECT X.LastOrderDate
, (SELECT MAX(OrderDate) FROM dbo.Orders SO WHERE SO.CustomerID=1 AND SO.OrderDate < X.LastOrderDate) AS PrevLastOrderDate
FROM
(
select MAX(OrderDate) AS LastOrderDate
FROM dbo.Orders O
WHERE O.CustomerID=1
)X
)Y
drop table #Invoices
create table #Invoices ( OrderId int , OrderDate datetime )
insert into #Invoices (OrderId , OrderDate )
select 101, '01/01/2001' UNION ALL Select 202, '02/02/2002' UNION ALL Select 303, '03/03/2003'
UNION ALL Select 808, '08/08/2008' UNION ALL Select 909, '09/09/2009'
;
WITH
MyCTE /* http://technet.microsoft.com/en-us/library/ms175972.aspx */
( OrderId,OrderDate,ROWID) AS
(
SELECT
OrderId,OrderDate
, ROW_NUMBER() OVER ( ORDER BY OrderDate ) as ROWID
FROM
#Invoices inv
)
SELECT
OrderId,OrderDate
,(Select Max(OrderDate) from MyCTE innerAlias where innerAlias.ROWID = (outerAlias.ROWID-1) ) as PreviousOrderDate
,
[MyDiff] =
CASE
WHEN (Select Max(OrderDate) from MyCTE innerAlias where innerAlias.ROWID = (outerAlias.ROWID-1) ) iS NULL then 0
ELSE DATEDIFF (mm, OrderDate , (Select Max(OrderDate) from MyCTE innerAlias where innerAlias.ROWID = (outerAlias.ROWID-1) ) )
END
, ROWIDMINUSONE = (ROWID-1)
, ROWID as ROWID_SHOWN_FOR_KICKS , OrderDate as OrderDateASecondTimeForConvenience
FROM
MyCTE outerAlias
ORDER BY outerAlias.OrderDate Desc , OrderId
service table:
claimid, customerid, serv-start-date, service-end-date, charge
1, A1, 1-1-14 , 1-5-14 , $200
2, A1, 1-6-14 , 1-8-14 , $300
3, A1, 2-1-14 , 2-1-14 , $100
4, A2, 2-1-14 , 2-1-14 , $100
5, A2, 2-3-14 , 2-5-14 , $100
6, A2, 2-6-14 , 2-8-14 , $100
Problem:
Basically to see the maximum total consecutive days Service start date and end date.
for customer A1 it would be 8 days (1-5 plus 6-8) and customer A2 it would be 5 6 days (3-5 plus 6-8) ... (claimid is unique PK).
Dates are in m-d-yy notation.
This gets a little messy since you could possibly have customers without multiple records. This uses a common-table-expressions, along with the max aggregate and union all to determine your results:
with cte as (
select s.customerid,
s.servicestartdate,
s2.serviceenddate,
datediff(day,s.servicestartdate,s2.serviceenddate)+1 daysdiff
from service s
join service s2 on s.customerid = s2.customerid
and s2.servicestartdate in (s.serviceenddate, dateadd(day,1,s.serviceenddate))
)
select customerid, max(daysdiff) daysdiff
from cte
group by customerid
union all
select customerid, max(datediff(day, servicestartdate, serviceenddate))
from service s
where not exists (
select 1
from cte
where s.customerid = cte.customerid
)
group by customerid
SQL Fiddle Demo
The second query in the union statement is what determines those service records without multiple records with consecutive days.
Here ya go, I think it's the simplest way:
SELECT customerid, sum(datediff([serv-end-date],[serv-start-date]))
FROM [service]
GROUP BY customerid
You will have to decide if same day start/end records count as 1. If they do, then add one to the datediff function, e.g. sum(datediff([serv-end-date],[serv-start-date]) + 1)
If you don't want to count same day services but DO want to count start/end dates inclusively when you sum them up, you will need to add a function that does the +1 only when start and end dates are different. Let me know if you want ideas on how to do that.
The only way I could think to solve the issue described by Jonathan Leffler (in a comment on another answer) was to use a temp table to merge contiguous date ranges. This would be best accomplished in an SP - but failing that the following batch may produce the output you are looking for:-
select *, datediff(day,servicestartdate,serviceenddate)+1 as numberofdays
into #t
from service
while ##rowcount>0 begin
update t1 set
t1.serviceenddate=t2.serviceenddate,
t1.numberofdays=datediff(day,t1.servicestartdate,t2.serviceenddate)+1
from #t t1
join #t t2 on t2.customerid=t1.customerid
and t2.servicestartdate=dateadd(day,1,t1.serviceenddate)
end
select
customerid,
max(numberofdays) as maxconsecutivedays
from #t
group by customerid
The update to the temp table needs to be in a loop because the date range could (I assume) be spread over any number of records (1->n). Interesting problem.
I've made updates to the code so that the temp table ends up with an extra column that holds the number of days in the date range on each record. This allows the following:-
select x.customerid, x.maxconsecutivedays, max(x.serviceenddate) as serviceenddate
from (
select t1.customerid, t1.maxconsecutivedays, t2.serviceenddate
from (
select
customerid,
max(numberofdays) as maxconsecutivedays
from #t
group by customerid
) t1
join #t t2 on t2.customerid=t1.customerid and t2.numberofdays=t1.maxconsecutivedays
) x
group by x.customerid, x.maxconsecutivedays
To identify the longest block of consecutive days (or the latest/longest if there is a tie) for each customer. This would allow you to subsequently dive back into the temp table to pull out the rows related to that block - by searching on the customerid and the serviceenddate (not maxconsecutivedays). Not sure this fits with your use case - but it may help.
WITH chain_builder AS
(
SELECT ROW_NUMBER() OVER(ORDER BY s.customerid, s.CLAIMID) as chain_ID,
s.customerid,
s.serv-start-date, s.service-end-date, s.CLAIMID, 1 as chain_count
FROM services s
WHERE s.serv-start-date <> ALL
(
SELECT DATEADD(d, 1, s2.service-end-date)
FROM services s2
)
UNION ALL
SELECT chain_ID, s.customerid, s.serv-start-date, s.service-end-date,
s.CLAIMID, chain_count + 1
FROM services s
JOIN chain_builder as c
ON s.customerid = c.customerid AND
s.serv-start-date = DATEADD(d, 1, c.service-end-date)
),
chains AS
(
SELECT chain_ID, customerid, serv-start-date, service-end-date,
CLAIMID, chain_count
FROM chain_builder
),
diff AS
(
SELECT c.chain_ID, c.customerid, c.serv-start-date, c.service-end-date,
c.CLAIMID, c.chain_count,
datediff(day,c.serv-start-date,c.service-end-date)+1 daysdiff
FROM chains c
),
diff_sum AS
(
SELECT chain_ID, customerid, serv-start-date, service-end-date,
CLAIMID, chain_count,
SUM(daysdiff) OVER (PARTITION BY chain_ID) as total_diff
FROM diff
),
diff_comp AS
(
SELECT chain_ID, customerid,
MAX(total_diff) OVER (PARTITION BY customerid) as total_diff
FROM diff_sum
)
SELECT DISTINCT ds.CLAIMID, ds.customerid, ds.serv-start-date,
ds.service-end-date, ds.total_diff as total_days, ds.chain_count
FROM diff_sum ds
JOIN diff_comp dc
ON ds.chain_ID = dc.chain_ID AND ds.customerid = dc.customerid
AND ds.total_diff = dc.total_diff
ORDER BY customerid, chain_count
OPTION (maxrecursion 0)
I have hospital patient admission data in Microsoft SQL Server r2 that looks something like this:
PatientID, AdmitDate, DischargeDate
Jones. 1-jan-13 01:37. 1-jan-13 17:45
Smith 1-jan-13 02:12. 2-jan-13 02:14
Brooks. 4-jan-13 13:54. 5-jan-13 06:14
I would like count the number of patients in the hospital day by day and hour by hour (ie at
1-jan-13 00:00. 0
1-jan-13 01:00. 0
1-jan-13 02:00. 1
1-jan-13 03:00. 2
And I need to include the hours when there are no patients admitted in the result.
I can't create tables so making a reference table listing all the hours and days is out, though.
Any suggestions?
To solve this problem, you need a list of date-hours. The following gets this from the admit date cross joined to a table with 24 hours. The table of 24 hours is calculating from information_schema.columns -- a trick for getting small sequences of numbers in SQL Server.
The rest is just a join between this table and the hours. This version counts the patients at the hour, so someone admitted and discharged in the same hour, for instance is not counted. And in general someone is not counted until the next hour after they are admitted:
with dh as (
select DATEADD(hour, seqnum - 1, thedatehour ) as DateHour
from (select distinct cast(cast(AdmitDate as DATE) as datetime) as thedatehour
from Admission a
) a cross join
(select ROW_NUMBER() over (order by (select NULL)) as seqnum
from INFORMATION_SCHEMA.COLUMNS
) hours
where hours <= 24
)
select dh.DateHour, COUNT(*) as NumPatients
from dh join
Admissions a
on dh.DateHour between a.AdmitDate and a.DischargeDate
group by dh.DateHour
order by 1
This also assumes that there are admissions on every day. That seems like a reasonable assumption. If not, a calendar table would be a big help.
Here is one (ugly) way:
;WITH DayHours AS
(
SELECT 0 DayHour
UNION ALL
SELECT DayHour+1
FROM DayHours
WHERE DayHour+1 <= 23
)
SELECT B.AdmitDate, A.DayHour, COUNT(DISTINCT PatientID) Patients
FROM DayHours A
CROSS JOIN (SELECT DISTINCT CONVERT(DATE,AdmitDate) AdmitDate
FROM YourTable) B
LEFT JOIN YourTable C
ON B.AdmitDate = CONVERT(DATE,C.AdmitDate)
AND A.DayHour = DATEPART(HOUR,C.AdmitDate)
GROUP BY B.AdmitDate, A.DayHour
This is a bit messy and includes a temp table with the test data you provided but
CREATE TABLE #HospitalPatientData (PatientId NVARCHAR(MAX), AdmitDate DATETIME, DischargeDate DATETIME)
INSERT INTO #HospitalPatientData
SELECT 'Jones.', '1-jan-13 01:37:00.000', '1-jan-13 17:45:00.000' UNION
SELECT 'Smith', '1-jan-13 02:12:00.000', '2-jan-13 02:14:00.000' UNION
SELECT 'Brooks.', '4-jan-13 13:54:00.000', '5-jan-13 06:14:00.000'
;WITH DayHours AS
(
SELECT 0 DayHour
UNION ALL
SELECT DayHour+1
FROM DayHours
WHERE DayHour+1 <= 23
),
HospitalPatientData AS
(
SELECT CONVERT(nvarchar(max),AdmitDate,103) as AdmitDate ,DATEPART(hour,(AdmitDate)) as AdmitHour, COUNT(PatientID) as CountOfPatients
FROM #HospitalPatientData
GROUP BY CONVERT(nvarchar(max),AdmitDate,103), DATEPART(hour,(AdmitDate))
),
Results AS
(
SELECT MAX(h.AdmitDate) as Date, d.DayHour
FROM HospitalPatientData h
INNER JOIN DayHours d ON d.DayHour=d.DayHour
GROUP BY AdmitDate, CountOfPatients, DayHour
)
SELECT r.*, COUNT(h.PatientId) as CountOfPatients
FROM Results r
LEFT JOIN #HospitalPatientData h ON CONVERT(nvarchar(max),AdmitDate,103)=r.Date AND DATEPART(HOUR,h.AdmitDate)=r.DayHour
GROUP BY r.Date, r.DayHour
ORDER BY r.Date, r.DayHour
DROP TABLE #HospitalPatientData
This may get you started:
BEGIN TRAN
DECLARE #pt TABLE
(
PatientID VARCHAR(10)
, AdmitDate DATETIME
, DischargeDate DATETIME
)
INSERT INTO #pt
( PatientID, AdmitDate, DischargeDate )
VALUES ( 'Jones', '1-jan-13 01:37', '1-jan-13 17:45' ),
( 'Smith', '1-jan-13 02:12', '2-jan-13 02:14' )
, ( 'Brooks', '4-jan-13 13:54', '5-jan-13 06:14' )
DECLARE #StartDate DATETIME = '20130101'
, #FutureDays INT = 7
;
WITH dy
AS ( SELECT TOP (#FutureDays)
ROW_NUMBER() OVER ( ORDER BY name ) dy
FROM sys.columns c
) ,
hr
AS ( SELECT TOP 24
ROW_NUMBER() OVER ( ORDER BY name ) hr
FROM sys.columns c
)
SELECT refDate, COUNT(p.PatientID) AS PtCount
FROM ( SELECT DATEADD(HOUR, hr.hr - 1,
DATEADD(DAY, dy.dy - 1, #StartDate)) AS refDate
FROM dy
CROSS JOIN hr
) ref
LEFT JOIN #pt p ON ref.refDate BETWEEN p.AdmitDate AND p.DischargeDate
GROUP BY refDate
ORDER BY refDate
ROLLBACK