YEAR MONTH BALANCE SSN
2016 1 3175 34/1043/03T
2016 1 2984 93/1194/07T
2016 1 2269 39/3149/00T
2015 12 3172 36/1011/03T
2015 12 2984 22/1224/07T
2015 12 2169 12/3143/00T
For example I have this table, but I have rows for each month of each year, and I have to choose the best ssn and balance of each month of each year. For example, here, I would like obtain this on my query:
YEAR MONTH BALANCE SSN
2016 1 3175 34/1043/03T
2015 12 3172 36/1011/03T
What can I do?
You can do this in several ways. A very Oracle'ish way is to use keep:
select year, month,
max(balance) as balance,
max(SSN) keep (dense_rank first order by balance desc) as ssn
from t
group by year, month;
Like most DBMSes Oracle supports ROW_NUMBER/RANK:
select *
from
(
select year, month, balance, SSN,
row_number()
over (partition by year, month
order by balance desc) as rn
from tab
) dt
where rn = 1
Related
In my data set, each customer has some orders on different dates.
For each customer each month, I want to check his/her last order in the previous month in which city.
For example, it is my data for one of the customers.
customer
year
month
day
order id
city id
1544
2022
2
6
413
9
1544
2022
2
17
39
10
1544
2022
3
5
115
21
1544
2022
5
29
2153
4
1544
2022
5
30
955
9
the result should be the same as this:
customer
year
month
city of last order of prev month(prevCity)
1544
2022
2
null or 9
1544
2022
3
10
1544
2022
5
21
(the first row of the above table is not my question now. )
I write my query using last_value the same as this:
select customer,
year,
month,
last_value(City) over (partition by customer, year, month order by created_at desc
ROWS BETWEEN CURRENT ROW AND UNBOUNDED FOLLOWING) as prevCity
from table1
but the result is false!
How can I correct this?
Using the window function lag() over() in concert with the WITH TIES clause
Select top 1 with ties
customer
,year
,month
,LastCityID = lag([city id],1) over (partition by customer order by year, month,day)
From YourTable
order by row_number() over (partition by customer,year,month order by year, month,day)
Or an Nudge More Perforamt
with cte as (
Select *
,LastCityID = lag([city id],1) over (partition by customer order by year, month,day)
,RN = row_number() over (partition by customer,year,month order by year, month,day)
From YourTable
)
Select customer
,year
,month
,LastCityID
From cte
Where RN =1
Results
customer year month LastCityID
1544 2022 2 NULL
1544 2022 3 10
1544 2022 5 21
I want to calculate the cumulative sum of monthly orders for each customer in my database.
For example, I have this data:
customer
year
month
no_orders
1544
2022
4
5
1544
2022
4
1
1544
2022
12
1
1544
2023
1
3
And the result should be the same as below:
customer
year
month
cumulative no_orders
1544
2022
4
0
1544
2022
12
6
1544
2023
1
7
I used lag() and in the next step, sum() over () but my result was false!
How can I solve this problem?
As #Larnu advises in the comments
Seems like you need to do several steps here. Aggregate (and group)
into months first, and then use a cumulative SUM but have the window
not include the current row.
Some SQL to implement this idea is below (DB FIDDLE)
SELECT customer,
year,
month,
cumulative_no_orders = ISNULL(SUM(SUM(no_orders))
OVER (
PARTITION BY customer
ORDER BY year, month
ROWS BETWEEN UNBOUNDED PRECEDING AND 1 PRECEDING)
, 0)
FROM YourTable
GROUP BY customer,
year,
month
It first does the aggregation
SELECT customer,
year,
month,
sum_no_orders = SUM(no_orders)
FROM YourTable
GROUP BY customer, year, month
to return the following
customer
year
month
sum_no_orders
1544
2022
4
6
1544
2022
12
1
1544
2023
1
3
and then calculates the running total of sum_no_orders from previous rows on top of that.
Can you try this,
SELECT
customer
, year
, month
, ISNULL(LAG(no_orders) OVER (PARTITION BY customer ORDER BY customer, year, month),0) Cum_Orders
FROM (
SELECT
DISTINCT customer, year, month
, SUM(no_orders) OVER (PARTITION BY customer ORDER BY customer, year, month) no_orders
FROM ABC
) a
I have this table in SQL Server:
Year Month Quantity
----------------------------
2015 January 10
2015 February 20
2015 March 30
2014 November 40
2014 August 50
How can I identify the different years and months adding two more columns that group the same years with a number and then different months in sequential way like the example
Year Month Quantity Group Subgroup
------------------------------------------------
2015 January 10 1 1
2015 February 20 1 2
2015 March 30 1 3
2014 November 40 2 1
2014 August 50 2 2
You can use DENSE_RANK to calculate the groups for you:
SELECT t1.*, DENSE_RANK() OVER (ORDER BY Year DESC) AS [Group],
DENSE_RANK() OVER (PARTITION BY Year ORDER BY DATEPART(month, Month + ' 01 2010')) AS [SubGroup]
FROM t1
ORDER BY 4, 5
See this fiddle.
To associate group and subgroup with a number you can do this:
WITH RankedTable AS (
SELECT year, month, quantity,
ROW_NUMBER() OVER (partition by year order by Month) AS rn
FROM yourtable)
SELECT year, month, quantity,
SUM (CASE WHEN rn = 1 THEN 1 ELSE 0 END) OVER (ORDER BY YEAR) as year_group,
rn AS subgroup
FROM RankedTable
Here ROW_NUMBER() OVER clause calculates rank of a month within a year.
And SUM() ... OVER calculates running SUM for the months with rank 1.
SQL Fiddle
I'm looking to query the database to find highest revenue month for all the customers in the system. I have got the query working to pull customers monthly revenue from all the years for which the data is present. But I'm struggling to figure out how to get highest revenue month-year from this data.
The database is SQL Server 2008 R2.
The columns are: Customer name, Year, Month, and Revenue.
I even tried using Row_Number() and tried partitioning by customer name/year and ordering by revenue. But it didn't work. Maybe I'm making some mistake there.
Here's how I tried to build the base query.
Select Customer, Year(orderdatetime) as Year, Month(orderdatetime) as Month, SUM(Revenue)
From Orders
Group By Customer, Year(orderdatetime), Month(orderdatetime)
This is how I tried to use Row_Number()
WITH Max_Revenue AS
(
Select Customer, Year(orderdatetime) as Year, Month(orderdatetime) as Month, SUM(Revenue), RowNumber = ROW_NUMBER() OVER(PARTITION By Year Order By Revenue DESC)
From Orders
Group By Customer, Year(orderdatetime), Month(orderdatetime)
)
Select Max_Revenue.Customer, Max_Revenue.Year, Max_Revenue.Month, Max_Revenue.Revenue
From Max_Revenue
Where Max_Revenue.RowNumber = 1
Order By Max_Revenue.Customer asc
The data I get back is like:
Customer Month Year Revenue
ABC 2 2012 100
ABC 3 2013 150
ABC 5 2012 200
XYZ 4 2011 500
XYZ 6 2012 650
XYZ 7 2012 800
What I want as the output is
Customer Month Year Revenue
ABC 5 2012 200
XYZ 7 2012 800
So every customer's best month and respective year in terms of revenue.
SELECT Customer,
Year,
Revenue,
Month
FROM (
SELECT Customer,
Year,
ROW_NUMBER() OVER(PARTITION By Customer Order By Revenue DESC) as rank,
Revenue,
Month
FROM (
Select Customer,
Year(orderdatetime) as Year,
Month(orderdatetime) as Month,
SUM(Revenue) as Revenue
From Orders
Group By
Customer,
Year(orderdatetime),
Month(orderdatetime)
) BS
GROUP BY Customer,
Year,
Month) BS2
WHERE BS2.rank = 1
OR change = ROW_NUMBER() OVER(PARTITION By Year Order By Revenue DESC to
= ROW_NUMBER() OVER(PARTITION By Customer Order By Revenue DESC
my working table, Table name: sales
Here Is MY TABLE, [sl_no is primary key] table structure:
CREATE TABLE SALES
( SL_NO NUMBER PRIMARY KEY, REGION VARCHAR2(10) NOT NULL,
MONTH VARCHAR2(20) NOT NULL, YEAR NUMBER NOT NULL,
SALES_AMOUNT NUMBER NOT NULL )
and here is table data:
SQL> select * from sales;
SL_NO REGION MONTH YEAR SALES_AMOUNT
---------- ---------- -------------------- ---------- ------------
1 east december 2011 750000
2 east august 2011 800000
3 west january 2012 640000
5 east march 2012 1200000
6 west february 2011 580000
4 west april 2011 555000
6 rows selected.
I have tried this query to view total sales amount of those[2011,2012] year;
SELECT year, SUM(sales_amount) FROM sales GROUP BY year;
YEAR SUM(SALES_AMOUNT)
---------- -----------------
2011 2685000
2012 1840000
MY GOAL:> I want to find out the year of maximum sales amount.
I tried this,and work perfectly...but when i want to display that year also, it gives an Error.
SQL> select max(sum(sales_amount)) from sales group by year;
MAX(SUM(SALES_AMOUNT))
----------------------
2685000
SQL> select year, max(sum(sales_amount)) from sales group by year;
select year, max(sum(sales_amount)) from sales group by year
*
ERROR at line 1:
ORA-00937: not a single-group group function
Extra addition: if multiple rows have same value means....when sales amount of both year[2011,2012] remain same, Then....
plZ help me to Solve this problem.
This should work.
with yr_agg as (
select year, sum(sales_amount) as total
from sales
group by year
)
select year, total as max_total
from yr_agg
where total = (select max(total)
from yr_agg);
I think the simplest way is to order the results and take the first row:
select year, sales_amount
from (SELECT year, SUM(sales_amount) as sales_amount
FROM sales
GROUP BY year
order by sum(sales_amount) desc
) t
where rownum = 1;
EDIT:
If you need to display all the matching rows (which isn't mentioned in the question), I would suggest using the dense_rank() analytic function:
select year, sales_amount
from (SELECT year, SUM(sales_amount) as sales_amount,
dense_rank(over order by SUM(sales_amount) desc) as seqnum
FROM sales
GROUP BY year
order by sum(sales_amount) desc
) t
where seqnum = 1;
Or, you might like the max() version instead:
select year, sales_amount
from (SELECT year, SUM(sales_amount) as sales_amount,
max(sum(sales_amount)) over () as maxsa
FROM sales
GROUP BY year
order by sum(sales_amount) desc
) t
where sales_amount = maxsa;
Following select should do what you need (untested, do not have Oracle at home):
select year, total
from (
select year, sum(sales_amount) total
from sales
group by year
)
where total = (select max(total_amount)
from (
select year, sum(sales_amount) total_amount
from sales
group by year
))
Take in account, though, that it might give you different years in each execution if two of them have exactly the same total amount. You might want to include some more conditions to avoid this.
Here is my Query where multiple row can select
SELECT year,MAX(total_sale) as max_total
FROM
(SELECT year,SUM(sales_amount) AS total_sale FROM sales GROUP BY year)
GROUP BY
year HAVING MAX(total_sale) =
(SELECT MAX(total_sale) FROM (SELECT SUM(sales_amount) AS total_sale FROM sales GROUP BY year));