SQL DATEDIFF in an sql query - sql-server

Question

I have two tables Customers and Purchases:
Customers table:
+------------+-----------+----------+
| CustomerID | FirstName | Surname |
+------------+-----------+----------+
| 101 | Jeff | Smith |
| 102 | Alex | Jones |
| 103 | Pam | Clark |
| 104 | Zola | Lona |
| 105 | Simphele | Ndima |
| 106 | Andre | Williams |
| 107 | Wayne | Shelton |
| 108 | Bob | Banard |
| 109 | Ken | Davidson |
| 110 | Sally | Ivan |
+------------+-----------+----------+
Purchases table:
+------------+--------------+------------+-----------+
| PurchaseId | PurchaseDate | CustomerID | ProductID |
+------------+--------------+------------+-----------+
| 1 | 2012-08-15 | 105 | a510 |
| 2 | 2012-08-15 | 102 | a510 |
| 3 | 2012-08-15 | 103 | a506 |
| 4 | 2012-08-16 | 105 | a510 |
| 5 | 2012-08-17 | 106 | a507 |
| 6 | 2012-08-17 | 107 | a509 |
| 7 | 2012-08-18 | 108 | a502 |
| 8 | 2012-08-19 | 108 | a510 |
| 9 | 2012-08-19 | 109 | a502 |
| 10 | 2012-08-20 | 110 | a503 |
| 11 | 2012-08-21 | 101 | a510 |
| 12 | 2012-08-22 | 102 | a507 |
+------------+--------------+------------+-----------+
My question (which I have been struggling with for the last 2 days): create a query that will display all the customers who purchased products after five days or more, since their last purchase.
Desired outputs:
+-----------+------------------+
| Firstname | Daysdifference |
+-----------+------------------+
| Alex | 7 |
+-----------+------------------+

Answer 1

select c.FirstName, t.dif as Daysdifference from customer c
inner join
(
select p1.CustomerID,
datediff(day,p1.PurchaseDate,p2.PurchaseDate) as dif
from purchases p1
inner join purchases p2
on p1.CustomerID=p2.CustomerID
where datediff(day,p1.PurchaseDate,p2.PurchaseDate)>=5
) t
on t.CustomerID= c.CustomerID

Answer 2

Here you go:
DECLARE #Customers TABLE (CustomerID INT, FirstName VARCHAR(30), Surname VARCHAR(30));
DECLARE #Purchases TABLE (PurchaseId INT, PurchaseDate DATE, CustomerID INT, ProductID VARCHAR(10) );
/**/
INSERT INTO #Customers VALUES
(101,'Jeff ' , 'Smith '),
(102,'Alex ' , 'Jones '),
(103,'Pam ' , 'Clark '),
(104,'Zola ' , 'Lona '),
(105,'Simphele' , 'Ndima '),
(106,'Andre ' , 'Williams'),
(107,'Wayne ' , 'Shelton '),
(108,'Bob ' , 'Banard '),
(109,'Ken ' , 'Davidson'),
(110,'Sally ' , 'Ivan ');
INSERT INTO #Purchases VALUES
(1, '2012-08-15' ,105, 'a510'),
(2, '2012-08-15' ,102, 'a510'),
(3, '2012-08-15' ,103, 'a506'),
(4, '2012-08-16' ,105, 'a510'),
(5, '2012-08-17' ,106, 'a507'),
(6, '2012-08-17' ,107, 'a509'),
(7, '2012-08-18' ,108, 'a502'),
(8, '2012-08-19' ,108, 'a510'),
(9, '2012-08-19' ,109, 'a502'),
(10,'2012-08-20' ,110, 'a503'),
(11,'2012-08-21' ,101, 'a510'),
(12,'2012-08-22' ,102, 'a507');
--
WITH CTE AS (
SELECT Pur1.CustomerID, DATEDIFF(DAY, Pur1.PurchaseDate, Pur2.PurchaseDate) Daysdifference
FROM #Purchases Pur1 INNER JOIN #Purchases Pur2 ON Pur1.CustomerID = Pur2.CustomerID
)
SELECT Cus.FirstName, CTE.Daysdifference
FROM #Customers Cus INNER JOIN CTE ON Cus.CustomerID = CTE.CustomerID
WHERE CTE.Daysdifference >= 5;
Result:
+-----------+------------------+
| Firstname | Daysdifference |
+-----------+------------------+
| Alex | 7 |
+-----------+------------------+
Demo

Answer 3

You can solve it like this:
Create a ranking based on date desc and partitioned by customer id
Next check date diff between consecutive ranks to find those customers
Query below
; with cte as
(
select
*,
row_number() over(partition by CustomerID order by PurchaseDate desc) r
from
Purchases
)
select
Name= c.FirstName,
Daysdifference =datediff(d,c1.PurchaseDate, c2.PurchaseDate)
from
Customers c join
cte c1
on c.customerid=c1.customerid
join cte c2
on c1.CustomerID=c2.CustomerId
and c1.r-1=c2.r
and datediff(d,c1.PurchaseDate, c2.PurchaseDate) >=5
See working demo

Answer 4

Since SQL Server 2012 and the addition of the LAG & LEAD functions, there is no reason at all to do a self join for something like this...
Note... Ranking function can be extremely efficient compared to other methods BUT they do need the help of a proper index to perform their best (note the additional POC index in the test script).
CREATE TABLE #Customers (
CustomerID INT PRIMARY KEY,
FirstName VARCHAR(30),
Surname VARCHAR(30)
);
CREATE TABLE #Purchases (
PurchaseId INT PRIMARY KEY,
PurchaseDate DATE,
CustomerID INT,
ProductID VARCHAR(10)
);
INSERT INTO #Customers VALUES
(101,'Jeff ' , 'Smith '),
(102,'Alex ' , 'Jones '),
(103,'Pam ' , 'Clark '),
(104,'Zola ' , 'Lona '),
(105,'Simphele' , 'Ndima '),
(106,'Andre ' , 'Williams'),
(107,'Wayne ' , 'Shelton '),
(108,'Bob ' , 'Banard '),
(109,'Ken ' , 'Davidson'),
(110,'Sally ' , 'Ivan ');
INSERT INTO #Purchases VALUES
(1, '2012-08-15' ,105, 'a510'),
(2, '2012-08-15' ,102, 'a510'),
(3, '2012-08-15' ,103, 'a506'),
(4, '2012-08-16' ,105, 'a510'),
(5, '2012-08-17' ,106, 'a507'),
(6, '2012-08-17' ,107, 'a509'),
(7, '2012-08-18' ,108, 'a502'),
(8, '2012-08-19' ,108, 'a510'),
(9, '2012-08-19' ,109, 'a502'),
(10,'2012-08-20' ,110, 'a503'),
(11,'2012-08-21' ,101, 'a510'),
(12,'2012-08-22' ,102, 'a507');
-- add POC index...
CREATE NONCLUSTERED INDEX ix_POC ON #Purchases (CustomerID, PurchaseDate);
--===========================================================
SELECT
c.FirstName,
p2.Daysdifference
FROM
#Customers c
JOIN (
SELECT
p.CustomerID,
Daysdifference = DATEDIFF(DAY, p.PurchaseDate, LEAD(p.PurchaseDate, 1) OVER (PARTITION BY p.CustomerID ORDER BY p.PurchaseDate))
FROM
#Purchases p
) p2
ON c.CustomerID = p2.CustomerID
WHERE
p2.Daysdifference >= 5;
Results...
FirstName Daysdifference
------------------------------ --------------
Alex 7