Company_Name Amount Cumulative Total
---------------------------------------------
Company 6 100 100
Company 6 200 300
Company 6 150 450
Company 7 700 700
Company 7 1100 1800
Company 7 500 2300
How can I do cumulative sum group by company as shown in this example?
First, you need a column that specifies the ordering, because SQL tables represent unordered sets. Let me assume you have such a column.
Then the function is sum() as a window function:
select t.*,
sum(amount) over (partition by company order by <ordering col>)
from t;
Note: This does not return 0 for the "first" row for each company, so it really is a cumulative sum. For your logic, you need an additional conditional:
select t.*,
(case when row_number() over (partition by company order by <ordering col>) = 1
then 0
else sum(amount) over (partition by company order by <ordering col>)
end)
from t;
Related
I have a table that contains Transactions of Customers.
I should Find Customers That had have at least 2 transaction with amount>20000 in Three consecutive days each month.
For example , Today is 2022/03/12 , I should Gather Data Of Transactions From 2022/02/13 To 2022/03/12, Then check These Data and See If a Customer had at least 2 Transaction With Amount>=20000 in Three consecutive days.
For Example, Consider Below Table:
Id
CustomerId
Transactiondate
Amount
1
1
2022-01-01
50000
2
2
2022_02_01
20000
3
3
2022_03_05
30000
4
3
2022_03_07
40000
5
2
2022_03_07
20000
6
4
2022_03_07
30000
7
4
2022_03_07
30000
The Out Put Should be : CustomerId =3 and CustomerId=4
I write query that Find Customer For Special day , but i don't know how to find these customers in one month with out using loop.
the query for special day is:
With cte (select customerid, amount, TransactionDate,Dateadd(day,-2,TransactionDate) as PrevDate
From Transaction
Where TransactionDate=2022-03-12)
Select CustomerId,Count(*)
From Cte
Where
TransactionDate>=Prevdate and TransactionDate<=TransactionDate
And Amount>=20000
Group By CustomerId
Having count(*)>=2
Hi there are many options how to achieve this.
I think that easies (from perfomance maybe not) is using LAG function:
WITH lagged_days AS (
SELECT
ISNULL(LAG(Transactiondate) OVER(PARTITION BY CustomerID ORDER BY id),
LEAD(Transactiondate) OVER(PARTITION BY CustomerID ORDER BY id)) lagged_dt
,*
FROM Transaction
), valid_cust_base as (
SELECT
*
FROM lagged_days
WHERE DATEPART(MONTH, lagged) = DATEPART(MONTH, Transactiondate)
AND datediff(day, Transactiondate, lagged_dt) <= 3
AND Amount >= 20000
)
SELECT
CustomerID
FROM valid_cust_base
GROUP BY CustomerID
HAVING COUNT(*) >= 2
First I have created lagged TransactionDate over customer (I assume that id is incremental). Then I have Selected only transactions within one month, with amount >= 20000 and where date difference between transaction is less then 4 days. Then just select customers who had more than 1 transaction.
In LAG First value is always missing per Customer missing, but you still need to be able say: 1st and 2nd transaction are within 3 days. Thats why I am replacing first NULL value with LEAD. It doesn't matter if you use:
ISNULL(LAG(Transactiondate) OVER(PARTITION BY CustomerID ORDER BY id),
LEAD(Transactiondate) OVER(PARTITION BY CustomerID ORDER BY id)) lagged_dt
OR
ISNULL(LEAD(Transactiondate) OVER(PARTITION BY CustomerID ORDER BY id),
LAG(Transactiondate) OVER(PARTITION BY CustomerID ORDER BY id)) lagged_dt
The main goal is to have for each transaction closest TransactionDate.
I am using SQL Server to solve the following question:
My table T1 has the following data:
Date Id Name Rent Number
01/01/2019 1 A 100 10
01/02/2019 1 A 200 30
01/03/2019 1 A 300 40
.
.
.
12/31/2019 1 A 150 25
The data exists for different combinations of Id and Name. I am trying to find average Rent and Number for 7 days:
Final Output
Date Id Name Rent Number
01/01/2019 - 01/07/2019 1 A Avg(rent for 7 days) Avg(Number for 7 days)
01/08/2019 - 01/14/2019 1 A Avg(rent for 7 days) Avg(Number for 7 days)
The final data should be grouped by Id and Name
My code:
SELECT min(date), Id, Name,
AVG(Rent) as Rent,
AVG(Number) Number,
AVG(AVG(Rent)) OVER (ORDER BY min(date) ROWS BETWEEN 6 PRECEDING AND CURRENT ROW) as AvgRent,
AVG(AVG(Number)) OVER (ORDER BY min(date) ROWS BETWEEN 6 PRECEDING AND CURRENT ROW) as AvgNumber
FROM T1
WHERE date >= '2019-01-01'
AND date < '2019-12-31'
GROUP BY Id, Name
My output has only row.
You need to group the dates by week which you can do by finding the difference between '01/01/2019' and your Date column using the DATEDIFF function and then divide it by 7. Since both the dividend and the divisor are integers the quotient will be an integer as well with the effect of group your dates into weeks.
SELECT MIN(Date) AS [Start Date]
, MAX(Date) AS [End Date]
, Id
, Name
, AVG(Rent) AS [Avg Rent]
, AVG(Number) AS [Avg Number]
FROM T1
GROUP BY DATEDIFF(DAY, '01/01/2019', Date) / 7;
This is the input table:
Customer_ID Date Amount
1 4/11/2014 20
1 4/13/2014 10
1 4/14/2014 30
1 4/18/2014 25
2 5/15/2014 15
2 6/21/2014 25
2 6/22/2014 35
2 6/23/2014 10
There is information pertaining to multiple customers and I want to get a rolling sum across a 3 day window for each customer.
The solution should be as below:
Customer_ID Date Amount Rolling_3_Day_Sum
1 4/11/2014 20 20
1 4/13/2014 10 30
1 4/14/2014 30 40
1 4/18/2014 25 25
2 5/15/2014 15 15
2 6/21/2014 25 25
2 6/22/2014 35 60
2 6/23/2014 10 70
The biggest issue is that I don't have transactions for each day because of which the partition by row number doesn't work.
The closest example I found on SO was:
SQL Query for 7 Day Rolling Average in SQL Server
but even in that case there were transactions made everyday which accomodated the rownumber() based solutions
The rownumber query is as follows:
select customer_id, Date, Amount,
Rolling_3_day_sum = CASE WHEN ROW_NUMBER() OVER (partition by customer_id ORDER BY Date) > 2
THEN SUM(Amount) OVER (partition by customer_id ORDER BY Date ROWS BETWEEN 2 PRECEDING AND CURRENT ROW)
END
from #tmp_taml9
order by customer_id
I was wondering if there is way to replace "BETWEEN 2 PRECEDING AND CURRENT ROW" by "BETWEEN [DATE - 2] and [DATE]"
One option would be to use a calendar table (or something similar) to get the complete range of dates and left join your table with that and use the row_number based solution.
Another option that might work (not sure about performance) would be to use an apply query like this:
select customer_id, Date, Amount, coalesce(Rolling_3_day_sum, Amount) Rolling_3_day_sum
from #tmp_taml9 t1
cross apply (
select sum(amount) Rolling_3_day_sum
from #tmp_taml9
where Customer_ID = t1.Customer_ID
and datediff(day, date, t1.date) <= 3
and t1.Date >= date
) o
order by customer_id;
I suspect performance might not be great though.
I am trying to find the difference of expenses of previous and current month in sql.
I have a table like this
Date Amount Category
2/18/2015 100 Salary
2/12/2015 150 Rent
2/21/2015 200 Allowances
1/4/2015 200 Salary
1/17/2015 50 Rent
1/20/2015 100 Allowances
Now I want a result like this
Category CurrentMonthAmount PreviousMonthAmount Difference
Salary 100 200 100
Rent 150 50 100
Allowances 200 100 100
Try using conditional Aggregate
;WITH cte
AS (SELECT Category,
Max(CASE WHEN Month([date]) = Month(Getdate()) and year([date]) =year(getdate()) THEN amount END) CurrentMonthAmount,
Max(CASE WHEN Month([date]) = Month(Getdate()) - 1 and year([date]) =year(getdate()) THEN amount END) PreviousMonthAmount
FROM Yourtable
GROUP BY Category)
SELECT Category,
CurrentMonthAmount,
PreviousMonthAmount,
[Difference]=Abs(CurrentMonthAmount - PreviousMonthAmount)
FROM cte
SQLFIDDLE DEMO
Are you trying to do the computations within your SQL only or via some scripts like PHP.? Furthermore, can you state if you are choosing specific records for this operation (specific rows i mean). Give some more clarification
I want to find the rows which are similar to each other, and replace them with a new row. My table looks like this:
OrderID | Price | Minimum Number | Maximum Number | Volume
1 45 2 10 250
2 46 2 10 250
3 60 2 10 250
"Similar" in this context means that the rows that have same Maximum Number, Minimum Number, and Volume. Prices can be different, but the difference can be at most 2.
In this example, orders with OrderID of 1 and 2 are similar, but 3 is not (since even if it has same Minimum Number, Maximum Number, and Volume, its price is not within 2 units from orders 1 and 2).
Then, I want orders 1 and 2 be replaced by a new order, let's say OrderID 4, which has same Minimum Number and Maximum Number. Its Volume hass to be sum of volumes of the orders it is replacing. Its price can be the Price of any of the previous orders that will be deleted in the output table (45 or 46 in this example). So, the output for the example above would be:
OrderID | Price | Minimum Number | Maximum Number | Volume
4 45 2 10 500
3 60 2 10 250
Here is a way to do this in SQL Server 2012 or Oracle. The idea is to use lag() to find where groups should begin and end and then aggregate.
select min(id) as id, min(price) as price, MinimumNumber, MaximumNumber, sum(Volume)
from (select t.*,
sum(case when prev_price < price - 2 then 1 else 0 end) over
(partition by MinimumNumber, MaximumNumber, Volume order by price) as grp
from (select t.*,
lag(price) over (partition by MinimumNumber, MaximumNumber, Volume
order by price
) as prev_price
from table t
) t
) t
group by grp, price, MinimumNumber, MaximumNumber;
The only issue is the setting of the id. I'm not sure what the exact rule is for that.