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I have a table TRANS that contains the following records:
TRANS_ID TRANS_DT QTY
1 01-Aug-2020 5
1 01-Aug-2020 1
1 03-Aug-2020 2
2 02-Aug-2020 1
The expected output:
TRANS_ID TRANS_DT BEGBAL TOTAL END_BAL
1 01-Aug-2020 0 6 6
1 02-Aug-2020 6 0 6
1 03-Aug-2020 6 2 8
2 01-Aug-2020 0 0 0
2 02-Aug-2020 0 1 1
2 03-Aug-2020 1 0 1
Each trans_id starts with a beginning balance of 0 (01-Aug-2020). For succeeding days, the beginning balance is the ending balance of the previous day and so on.
I can create PL/SQL block to create the output. Is it possible to get the output in 1 SQL statement?
Thanks.
Try this following script using CTE-
Demo Here
WITH CTE
AS
(
SELECT DISTINCT A.TRANS_ID,B.TRANS_DT
FROM your_table A
CROSS JOIN (SELECT DISTINCT TRANS_DT FROM your_table) B
),
CTE2
AS
(
SELECT C.TRANS_ID,C.TRANS_DT,SUM(D.QTY) QTY
FROM CTE C
LEFT JOIN your_table D
ON C.TRANS_ID = D.TRANS_ID
AND C.TRANS_DT = D.TRANS_DT
GROUP BY C.TRANS_ID,C.TRANS_DT
ORDER BY C.TRANS_ID,C.TRANS_DT
)
SELECT F.TRANS_ID,F.TRANS_DT,
(
SELECT COALESCE (SUM(QTY), 0) FROM CTE2 E
WHERE E.TRANS_ID = F.TRANS_ID AND E.TRANS_DT < F.TRANS_DT
) BEGBAL,
(
SELECT COALESCE (SUM(QTY), 0) FROM CTE2 E
WHERE E.TRANS_ID = F.TRANS_ID AND E.TRANS_DT = F.TRANS_DT
) TOTAL ,
(
SELECT COALESCE (SUM(QTY), 0) FROM CTE2 E
WHERE E.TRANS_ID = F.TRANS_ID AND E.TRANS_DT <= F.TRANS_DT
) END_BAL
FROM CTE2 F
You can as well do like this (I would assume it's a bit faster): Demo
with
dt_between as (
select mindt + level - 1 as trans_dt
from (select min(trans_dt) as mindt, max(trans_dt) as maxdt from t)
connect by level <= maxdt - mindt + 1
),
dt_for_trans_id as (
select *
from dt_between, (select distinct trans_id from t)
),
qty_change as (
select distinct trans_id, trans_dt,
sum(qty) over (partition by trans_id, trans_dt) as total,
sum(qty) over (partition by trans_id order by trans_dt) as end_bal
from t
right outer join dt_for_trans_id using (trans_id, trans_dt)
)
select
trans_id,
to_char(trans_dt, 'DD-Mon-YYYY') as trans_dt,
nvl(lag(end_bal) over (partition by trans_id order by trans_dt), 0) as beg_bal,
nvl(total, 0) as total,
nvl(end_bal, 0) as end_bal
from qty_change q
order by trans_id, trans_dt
dt_between returns all the days between min(trans_dt) and max(trans_dt) in your data.
dt_for_trans_id returns all these days for each trans_id in your data.
qty_change finds difference for each day (which is TOTAL in your example) and cumulative sum over all the days (which is END_BAL in your example).
The main select takes END_BAL from previous day and calls it BEG_BAL, it also does some formatting of final output.
First of all, you need to generate dates, then you need to aggregate your values by TRANS_DT, and then left join your aggregated data to dates. The easiest way to get required sums is to use analitic window functions:
with dates(dt) as ( -- generating dates between min(TRANS_DT) and max(TRANS_DT) from TRANS
select min(trans_dt) from trans
union all
select dt+1 from dates
where dt+1<=(select max(trans_dt) from trans)
)
,trans_agg as ( -- aggregating QTY in TRANS
select TRANS_ID,TRANS_DT,sum(QTY) as QTY
from trans
group by TRANS_ID,TRANS_DT
)
select -- using left join partition by to get data on daily basis for each trans_id:
dt,
trans_id,
nvl(sum(qty) over(partition by trans_id order by dates.dt range between unbounded preceding and 1 preceding),0) as BEGBAL,
nvl(qty,0) as TOTAL,
nvl(sum(qty) over(partition by trans_id order by dates.dt),0) as END_BAL
from dates
left join trans_agg tr
partition by (trans_id)
on tr.trans_dt=dates.dt;
Full example with sample data:
alter session set nls_date_format='dd-mon-yyyy';
with trans(TRANS_ID,TRANS_DT,QTY) as (
select 1,to_date('01-Aug-2020'), 5 from dual union all
select 1,to_date('01-Aug-2020'), 1 from dual union all
select 1,to_date('03-Aug-2020'), 2 from dual union all
select 2,to_date('02-Aug-2020'), 1 from dual
)
,dates(dt) as ( -- generating dates between min(TRANS_DT) and max(TRANS_DT) from TRANS
select min(trans_dt) from trans
union all
select dt+1 from dates
where dt+1<=(select max(trans_dt) from trans)
)
,trans_agg as ( -- aggregating QTY in TRANS
select TRANS_ID,TRANS_DT,sum(QTY) as QTY
from trans
group by TRANS_ID,TRANS_DT
)
select
dt,
trans_id,
nvl(sum(qty) over(partition by trans_id order by dates.dt range between unbounded preceding and 1 preceding),0) as BEGBAL,
nvl(qty,0) as TOTAL,
nvl(sum(qty) over(partition by trans_id order by dates.dt),0) as END_BAL
from dates
left join trans_agg tr
partition by (trans_id)
on tr.trans_dt=dates.dt;
You can use a recursive query to generate the overall date range, cross join it with the list of distinct tran_id, then bring the table with a left join. The last step is aggregation and window functions:
with all_dates (trans_dt, max_dt) as (
select min(trans_dt), max(trans_dt) from trans group by trans_id
union all
select trans_dt + interval '1' day, max_dt from all_dates where trans_dt < max_dt
)
select
i.trans_id,
d.trans_dt,
coalesce(sum(sum(t.qty)) over(partition by i.trans_id order by d.trans_dt), 0) - coalesce(sum(t.qty), 0) begbal,
coalesce(sum(t.qty), 0) total,
coalesce(sum(sum(t.qty)) over(partition by i.trans_id order by d.trans_dt), 0) endbal
from all_dates d
cross join (select distinct trans_id from trans) i
left join trans t on t.trans_id = i.trans_id and t.trans_dt = d.trans_dt
group by i.trans_id, d.trans_dt
order by i.trans_id, d.trans_dt
I have the following problem.
Part of a task is to determine the visitor(s) with the most money spent between 2000 and 2020.
It just looks like this.
SELECT UserEMail FROM Visitor
JOIN Ticket ON Visitor.UserEMail = Ticket.VisitorUserEMail
where Ticket.Date> date('2000-01-01') AND Ticket.Date < date ('2020-12-31')
Group by Ticket.VisitorUserEMail
order by SUM(Price) DESC;
Is it possible to output more than one person if both have spent the same amount?
Use rank():
SELECT VisitorUserEMail
FROM (SELECT VisitorUserEMail, SUM(PRICE) as sum_price,
RANK() OVER (ORDER BY SUM(Price) DESC) as seqnum
FROM Ticket t
WHERE t.Date >= date('2000-01-01') AND Ticket.Date <= date('2021-01-01')
GROUP BY t.VisitorUserEMail
) t
WHERE seqnum = 1;
Note: You don't need the JOIN, assuming that ticket buyers are actually visitors. If that assumption is not true, then use the JOIN.
Use a CTE that returns all the total prices for each email and with NOT EXISTS select the rows with the top total price:
WITH cte AS (
SELECT VisitorUserEMail, SUM(Price) SumPrice
FROM Ticket
WHERE Date >= '2000-01-01' AND Date <= '2020-12-31'
GROUP BY VisitorUserEMail
)
SELECT c.VisitorUserEMail
FROM cte c
WHERE NOT EXISTS (
SELECT 1 FROM cte
WHERE SumPrice > c.SumPrice
)
or:
WITH cte AS (
SELECT VisitorUserEMail, SUM(Price) SumPrice
FROM Ticket
WHERE Date >= '2000-01-01' AND Date <= '2020-12-31'
GROUP BY VisitorUserEMail
)
SELECT VisitorUserEMail
FROM cte
WHERE SumPrice = (SELECT MAX(SumPrice) FROM cte)
Note that you don't need the function date() because the result of date('2000-01-01') is '2000-01-01'.
Also I think that the conditions in the WHERE clause should include the =, right?
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 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