I am running a query which I want to only show distinct customers from.
At the current time I am receiving records which have multiple records for example 3 records for Item A0003. I want to only return the last record in the sequence.
My code:
select OJCUNO AS Item,OJPRRF as code,OJFVDT as From Date, OJLVDT as To Date
from M3FDBPRD.OPRICH
WHERE
OJCUNO in ( Select max(OJCUNO) FROM OPRICH group by OJCUNO )
Data Sample:
Item Code From Date To Date
A0007 AD 20030301 20161231
A0008 AF 20030301 20161231
A0009 AL 20030301 20121229
A0009 AL 20030301 20121231
Expected Result:
Item Code From Date To Date
A0007 AD 20030301 20161231
A0008 AF 20030301 20161231
A0009 AL 20030301 20121231
Just use row_number():
select OJCUNO AS Item, OJPRRF as code ,OJFVDT as FromDate, OJLVDT as ToDate
from (select o.*,
row_number() over (partition by ojcuno order OJPRRF desc, OJLVDT desc) as seqnum
from M3FDBPRD.OPRICH o
) o
where seqnum = 1;
Your approach, using a correlated subquery would work if you used the right columns:
select OJCUNO AS Item,OJPRRF as code, OJFVDT as FromDate, OJLVDT as ToDate
from M3FDBPRD.OPRICH
where OJLVDT in ( Select max(OJLVDT) from OPRICH group by OJCUNO );
I'm just having a stab here because I don't have access to a DB2 database to test on, but:
SELECT *
FROM
(
select OJCUNO AS Item,
OJPRRF as code,
OJFVDT as From Date,
OJLVDT as To Date,
ROW_NUMBER() OVER (PARTITION BY OJCUNO, OJPRRF ORDER BY OJFVDT DESC, OJLVDT DESC) AS RNum
from M3FDBPRD.OPRICH
WHERE
OJCUNO in ( Select max(OJCUNO) FROM OPRICH group by OJCUNO )
) a
WHERE a.RNum = 1
Related
I have some prices for the month of January.
Date,Price
1,100
2,100
3,115
4,120
5,120
6,100
7,100
8,120
9,120
10,120
Now, the o/p I need is a non-overlapping date range for each price.
price,from,To
100,1,2
115,3,3
120,4,5
100,6,7
120,8,10
I need to do this using SQL only.
For now, if I simply group by and take min and max dates, I get the below, which is an overlapping range:
price,from,to
100,1,7
115,3,3
120,4,10
This is a gaps-and-islands problem. The simplest solution is the difference of row numbers:
select price, min(date), max(date)
from (select t.*,
row_number() over (order by date) as seqnum,
row_number() over (partition by price, order by date) as seqnum2
from t
) t
group by price, (seqnum - seqnum2)
order by min(date);
Why this works is a little hard to explain. But if you look at the results of the subquery, you will see how the adjacent rows are identified by the difference in the two values.
SELECT Lag.price,Lag.[date] AS [From], MIN(Lead.[date]-Lag.[date])+Lag.[date] AS [to]
FROM
(
SELECT [date],[Price]
FROM
(
SELECT [date],[Price],LAG(Price) OVER (ORDER BY DATE,Price) AS LagID FROM #table1 A
)B
WHERE CASE WHEN Price <> ISNULL(LagID,1) THEN 1 ELSE 0 END = 1
)Lag
JOIN
(
SELECT [date],[Price]
FROM
(
SELECT [date],Price,LEAD(Price) OVER (ORDER BY DATE,Price) AS LeadID FROM [#table1] A
)B
WHERE CASE WHEN Price <> ISNULL(LeadID,1) THEN 1 ELSE 0 END = 1
)Lead
ON Lag.[Price] = Lead.[Price]
WHERE Lead.[date]-Lag.[date] >= 0
GROUP BY Lag.[date],Lag.[price]
ORDER BY Lag.[date]
Another method using ROWS UNBOUNDED PRECEDING
SELECT price, MIN([date]) AS [from], [end_date] AS [To]
FROM
(
SELECT *, MIN([abc]) OVER (ORDER BY DATE DESC ROWS UNBOUNDED PRECEDING ) end_date
FROM
(
SELECT *, CASE WHEN price = next_price THEN NULL ELSE DATE END AS abc
FROM
(
SELECT a.* , b.[date] AS next_date, b.price AS next_price
FROM #table1 a
LEFT JOIN #table1 b
ON a.[date] = b.[date]-1
)AA
)BB
)CC
GROUP BY price, end_date
This is the question i was being asked at Apple onsite interview and it blew my mind. Data is like this:
orderdate,unit_of_phone_sale
20190806,3000
20190704,3789
20190627,789
20190503,666
20190402,765
I had to write a query to get the result for each month sale, we should have last 3 month sales values. Let me put the expected output here.
order_monnth,M-1_Sale, M-2_Sale, M-3_Sale
201908,3000,3789,789,666
201907,3789,789,666,765
201906,789,666,765,0
201905,666,765,0,0
201904,765,0,0
I could only got the month wise sale and and used case statement by hardcoding month which was wrong. I banged my head to write this sql, but i could not.
Can anyone help on this. It will be really helpful for me to prepare for sql interviews
Update: This is what i tried
with abc as(
select to_char(order_date,'YYYYMM') as yearmonth,to_char(order_date,'YYYY') as year,to_char(order_date,'MM') as moth, sum(unit_of_phone_sale) as unit_sale
from t1 group by to_char(order_date,'YYYYMM'),to_char(order_date,'YYYY'),to_char(order_date,'MM'))
select yearmonth, year, case when month=01 then unit_sale else 0 end as M1_Sale,
case when month=02 then unit_sale else 0 end as M2_Sale...
case when month=12 then unit_sale else 0 end as M12_Sale
from abc
You will first of all need to sum the month's data and then use the LAG function to get previous months' data as following:
SELECT
ORDER_MONTH,
LAG(UNIT_OF_PHONE_SALE, 1) OVER(
ORDER BY
ORDER_MONTH
) AS "M-1_Sale",
LAG(UNIT_OF_PHONE_SALE, 2) OVER(
ORDER BY
ORDER_MONTH
) AS "M-2_Sale",
LAG(UNIT_OF_PHONE_SALE, 3) OVER(
ORDER BY
ORDER_MONTH
) AS "M-3_Sale"
FROM
(
SELECT
TO_CHAR(ORDERDATE, 'YYYYMM') AS ORDER_MONTH,
SUM(UNIT_OF_PHONE_SALE) AS UNIT_OF_PHONE_SALE
FROM
DATAA
GROUP BY
TO_CHAR(ORDERDATE, 'YYYYMM')
)
ORDER BY
ORDER_MONTH DESC;
Output:
ORDER_ M-1_Sale M-2_Sale M-3_Sale
------ ---------- ---------- ----------
201908 3789 789 666
201907 789 666 765
201906 666 765
201905 765
201904
db<>fiddle demo
Cheers!!
-- Update --
For the requirement mentioned in the comments, Following query will work for it.
CTE AS (
SELECT
TRUNC(ORDERDATE, 'MONTH') AS ORDER_MONTH,
SUM(UNIT_OF_PHONE_SALE) AS UNIT_OF_PHONE_SALE
FROM
DATAA
GROUP BY
TRUNC(ORDERDATE, 'MONTH')
)
SELECT
TO_CHAR(C.ORDER_MONTH,'YYYYMM') as ORDER_MONTH,
NVL(C1.UNIT_OF_PHONE_SALE, 0) AS "M-1_Sale",
NVL(C2.UNIT_OF_PHONE_SALE, 0) AS "M-2_Sale",
NVL(C3.UNIT_OF_PHONE_SALE, 0) AS "M-3_Sale"
FROM
CTE C
LEFT JOIN CTE C1 ON ( C1.ORDER_MONTH = ADD_MONTHS(C.ORDER_MONTH, - 1) )
LEFT JOIN CTE C2 ON ( C2.ORDER_MONTH = ADD_MONTHS(C.ORDER_MONTH, - 2) )
LEFT JOIN CTE C3 ON ( C3.ORDER_MONTH = ADD_MONTHS(C.ORDER_MONTH, - 3) )
ORDER BY
C.ORDER_MONTH DESC
Output:
db<>fiddle demo of updated answer.
Cheers!!
I think LEAD function can help here -
SELECT TO_CHAR(orderdate, 'YYYYMM') "DATE"
,unit_of_phone_sale M_1_Sale
,LEAD(unit_of_phone_sale,1,0) OVER(ORDER BY TO_CHAR(orderdate, 'YYYYMM') DESC) M_2_Sale
,LEAD(unit_of_phone_sale,2,0) OVER(ORDER BY TO_CHAR(orderdate, 'YYYYMM') DESC) M_3_Sale
,LEAD(unit_of_phone_sale,3,0) OVER(ORDER BY TO_CHAR(orderdate, 'YYYYMM') DESC) M_4_Sale
FROM table_sales
Here is the DB Fiddle
You can use this query:
select a.order_month, a.unit_of_phone_sale,
LEAD(unit_of_phone_sale, 1, 0) OVER (ORDER BY rownum) AS M_1,
LEAD(unit_of_phone_sale, 2, 0) OVER (ORDER BY rownum) AS M_2,
LEAD(unit_of_phone_sale, 3, 0) OVER (ORDER BY rownum) AS M_3
from (
select TO_CHAR(orderdate, 'YYYYMM') order_month,
unit_of_phone_sale,
rownum
from Y
order by order_month desc) a
I have Four columns like
Date Customer InvoiceNo StockBalance
11/29/2017 A IN000414 5000
11/30/2017 B IN000415 4000
12/27/2017 A IN000416 3500
12/30/2017 B IN000417 2000
I want to get Stockbalance of every end of month, I need the output as
11/30/2017 B IN000415 4000
12/30/2017 B IN000417 2000
how could i get this could anybody guide to me?
You can use row_number() function :
select t.*
from (select *, row_number() over (partition by year(date), month(date) order by date desc) seq
from table
) t
where seq = 1;
EDIT : You want apply :
select t.*
from table t cross apply
( select top (1) t1.*
from table t1
where t1.Customer = t.Customer and
EOMONTH(t1.Dat) = t.Dat
order by t1.Dat desc
) t1;
Use row_number(), but be sure you include the year and month in the calculation:
select t.*
from (select t.*,
row_number() over (partition by year(date), month(date)
order by date desc
) as seqnum
from t
) t
where seqnum = 1;
I do have a table license_Usage where which works like a log of the usage of licenses in a day
ID User license date
1 1 A 22/2/2015
2 1 A 23/2/2015
3 1 B 22/2/2015
4 2 A 22/2/2015
Where I want to Count how many licenses per user in a day, the result shoul look like:
QuantityOfLicenses User date
2 1 22/2/2015
1 2 22/2/2015
For that I did the following query :
select count(license) as [Quantity of licenses],[user],[date]
From license_Usage
where date = '22/2/2015'
Group by [date], [user]
which works, but know I want to know which user have used the most number of licenses, for that I did the following query:
select MAX(result.[Quantity of licenses])
From (
select count(license) as [Quantity of licenses],[user],[date]
From license_Usage
Group by [date], [user]
) as result
And it returns the max value of 2, but when I want to know which user have used 2 licenses,I try this query with no success :
select result.user, MAX(result.[Quantity of licenses])
From (
select count(license) as [Quantity of licenses],[user],[date]
From license_Usage
Group by [date], [user]
) as result
Group by result.user
You can use something like this:
select top 1 *
From (
select count(license) as Quantity,[user],[date]
From license_Usage
Group by [date], [user]
) as result
order by Quantity desc
If you need to have a fetch that fetches all the rows that have max in case there's several, then you'll have to use rank() window function
Use RANK to rank the users by the number of licenses per day.
SELECT
LicPerDay.*,
RANK() OVER (PARTITION BY [date] ORDER BY Qty DESC) AS User_Rank
FROM (
SELECT
COUNT(license) AS Qty,
User,
[date]
FROM license_usage
GROUP BY User, [date]
) LicPerDay
Any user with User_Rank = 1 will have the most licenses for that day.
If you only want the top user for each day, wrap the query above as a subquery and filter on User_Rank = 1:
SELECT * FROM (
SELECT
LicPerDay.*,
RANK() OVER (PARTITION BY [date] ORDER BY Qty) AS User_Rank
FROM (
SELECT
COUNT(license) AS Qty,
User,
[date]
FROM license_usage
GROUP BY User, [date]
) LicPerDay
) LicPerDayRanks
WHERE User_Rank = 1
Use a Windowed Aggregate Function, RANK, to get the highest count:
SELECT * FROM (
SELECT
User,
[date]
COUNT(license) AS Qty,
-- rank by descending number for each day ??
--RANK() OVER (PARTITION BY [date] ORDER BY COUNT(license) DESC) AS rnk
-- rank by descending number
RANK() OVER (ORDER BY COUNT(license) DESC) AS rnk
FROM license_usage
GROUP BY User, [date]
) dt
WHERE rnk = 1
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