I have a list of dates in a YYYYMM format and I am trying to rank them in a Year over Year format that would look like the following:
MonthDisplay YearMonth Rank MonthNumber YearNumber
Aug-2013 201308 1 8 2013
Aug-2012 201208 2 8 2012
Jul-2013 201307 3 7 2013
Jul-2012 201207 4 7 2012
I have been able to get it close by using the following Rank and get the results below:
RANK() OVER(PARTITION BY 1 ORDER BY MonthNumber DESC, YearNumber DESC)
Month YearMonth Rank
Dec-2012 201212 1
Dec-2011 201112 2
Nov-2012 201211 114
Nov-2011 201111 115
Oct-2012 201210 227
Oct-2011 201110 228
However, this starts with Dec-2012 instead of the Aug-2013 (current month). I can't figure out how to get it to start with the current month. I am sure it something super easy and I am just missing it. Thanks!
select
T.YearMonth,
rank() over (order by R.rnk asc, D.YearNumber desc) as [Rank],
D.MonthNumber, D.YearNumber
from Table1 as T
outer apply (
select
month(getdate()) as CurMonthNumber,
cast(right(T.YearMonth, 2) as int) as MonthNumber,
cast(left(T.YearMonth, 4) as int) as YearNumber
) as D
outer apply (
select
case
when D.MonthNumber <= D.CurMonthNumber then
D.CurMonthNumber - D.MonthNumber
else
12 + D.CurMonthNumber - D.MonthNumber
end as rnk
) as R
sql fiddle example
Related
I need to find the last three distinct loaddates for each month in various tables for reporting purposes. Example: If I have data from 2021 February to today: I need the three loaddates of Feb 2021, March 2021 and so on till. Dec 2022
So far, I'm able to create the below query in SQL Server which gives me the result for a particular month that I pass in the where condition.
SELECT ROW_NUMBER() OVER (ORDER BY loaddate desc) AS myrank, loaddate
FROM <tablename>
where year(loaddate) = 2022 and month(loaddate) = 6
group by loaddate
It gives me:
myrank loaddate
1 2022-08-29 00:00:00.000
2 2022-08-25 00:00:00.000
3 2022-08-18 00:00:00.000
4 2022-08-17 00:00:00.000
5 2022-08-11 00:00:00.000
From this I can easily select the top three dates with the below query:
SELECT myrank, loaddate
FROM
(
SELECT ROW_NUMBER() OVER (ORDER BY loaddate desc) AS myrank, loaddate
FROM <tablename>
where year(loaddate) = 2022 and month(loaddate) = 6
group by loaddate
) as daterank
WHERE daterank.myrank <= 3
which outputs:
rank loaddate
1 2022-08-29 00:00:00.000
2 2022-08-25 00:00:00.000
3 2022-08-18 00:00:00.000
But this is only for one month. I'm manually passing the month number in the where condition. How to make this ranking query give me the the last 3 distinct loaddates for each month of data that exists in the table?
And also, how to do I run such a generic query on list of 400+ tables instead of changing the tablename manually for each table in the list?
You just add the PARTITION BY clause to ROW_NUMBER() and partition by month (and year since your data might cross a year boundary).
WITH cte AS (
SELECT *
, ROW_NUMBER() OVER (PARTITION BY DATEPART(year, loaddate), DATEPART(month, loaddate) ORDER BY loaddate desc) AS myrank
FROM #MyTable
)
SELECT *
FROM cte
WHERE myrank <= 3
ORDER BY loaddate;
Note: The CTE is doing the same thing as your sub-query - don't let that confuse you - I just prefer it for neatness.
If I understand your request I think this would help you:
SELECT myrank, loaddate, monthofyear
FROM (
SELECT
ROW_NUMBER() OVER (PARTITION BY month(loaddate) ORDER BY loaddate DESC) AS myrank
, loaddate, month(loaddate) as monthofyear
FROM Db15.dbo.mytable
GROUP BY loaddate
) AS daterank
WHERE daterank.myrank <= 3
I have following query
My #dates table has following records:
month year saledate
9 2020 2020-09-01
10 2020 2020-10-01
11 2020 2020-11-01
with monthlysalesdata as(
select month(salesdate) as salemonth, year(salesdate) as saleyear,salesrepid, salespercentage
from salesrecords r
join #dates d on d.saledate = r.salesdate
group by salesrepid, salesdate),
averagefor3months as(
select 0 as salemonth, 0 as saleyear, salesrepid, salespercentage
from monthlysalesdata
group by salesrepid)
finallist as(
select * from monthlysalesdata
union
select * from averagefor3months
This query returns following records which gives duplicate for a averagefor3months result set when there is null record in the first monthlyresultdata. how to achieve average for 3 months as one record instead of having duplicates?
salesrepid salemonth saleyear percentage
232 0 0 null -------------this is the duplicate record
232 0 0 90
232 9 2020 80
232 10 2020 null
232 11 2020 100
My first cte has this result:
salerepid month year percentage
---------------------------------------------
232 9 2020 80
232 10 2020 null
232 11 2020 100
My second cte has this result:
salerepid month year percentage
---------------------------------------------
232 0 0 null
232 0 0 90
How to avoid the duplicate record in my second cte,
I suspect that you want a summary row per sales rep based on some aggregation. Your question is not clear on what is needed for the aggregation, but something like this:
with ym as (
select r.salesrepid, d.year, d.month, sum(<something>) as whatever
from salesrecords r join
#dates d
on d.saledate = r.salesdate
group by r.salesrepid, d.year, d.month
)
select ym.*
from ym
union all
select salesrepid, null, null, avg(whatever)
from hm
group by salesrepid;
I updated to selected the group by from the table directly instead of the previous cte and got my results. Thank you all for helping
with ym as (
select r.salesrepid, d.year, d.month, sum(<something>) as whatever
from salesrecords r join
#dates d
on d.saledate = r.salesdate
group by r.salesrepid, d.year, d.month
),
threemonthsaverage as(
select r.salesrepid, r.year, r.month, sum(something) as whatever
from salesrecords as r
group by salesrepid)
select ym *
union
select threemonthsaverage*
This is SQL Query
SELECT
ROW_NUMBER() OVER (ORDER BY (SELECT 1)) [Sno],
_Date,
SUM(Payment) Payment
FROM
DailyPaymentSummary
GROUP BY
_Date
ORDER BY
_Date
This returns output like this
Sno _Date Payment
---------------------------
1 2017-02-02 46745.80
2 2017-02-03 100101.03
3 2017-02-06 140436.17
4 2017-02-07 159251.87
5 2017-02-08 258807.51
6 2017-02-09 510986.79
7 2017-02-10 557399.09
8 2017-02-13 751405.89
9 2017-02-14 900914.45
How can I get the additional column like below
Sno _Date Payment Diff
--------------------------------------
1 02/02/2017 46745.80 46745.80
2 02/03/2017 100101.03 53355.23
3 02/06/2017 140436.17 40335.14
4 02/07/2017 159251.87 18815.70
5 02/08/2017 258807.51 99555.64
6 02/09/2017 510986.79 252179.28
7 02/10/2017 557399.09 46412.30
8 02/13/2017 751405.89 194006.80
9 02/14/2017 900914.45 149508.56
I have tried the following query but not able to solve the error
WITH cte AS
(
SELECT
ROW_NUMBER() OVER (ORDER BY (SELECT 1)) [Sno],
_Date,
SUM(Payment) Payment
FROM
DailyPaymentSummary
GROUP BY
_Date
ORDER BY
_Date
)
SELECT
t.Payment,
t.Payment - COALESCE(tprev.col, 0) AS diff
FROM
DailyPaymentSummary t
LEFT OUTER JOIN
t tprev ON t.seqnum = tprev.seqnum + 1;
Can anyone help me?
Use a order by with column(s) to get consistent results.
Use lag function to get data from previous row and do the subtraction like this:
with t
as (
select ROW_NUMBER() over (order by _date) [Sno],
_Date,
sum(Payment) Payment
from DailyPaymentSummary
group by _date
)
select *,
Payment - lag(Payment, 1, 0) over (order by [Sno]) diff
from t;
You can use lag() to get previous row values
coalesce(lag(sum_payment_col) OVER (ORDER BY (SELECT 1)),0)
I havea table as table_A . table_A includes these columns
-CountryName
-Min_Date
-Max_Date
-Number
I want to duplicate data with seperating by months. For example
Argentina | 2015-01-04 | 2015-04-07 | 100
England | 2015-02-08 | 2015-03-11 | 90
I want to see a table as this (Monthly seperated)
Argentina | 01-2015 | 27 //(days to end of the min_date's month)
Argentina | 02-2015 | 29 //(days full month)
Argentina | 03-2015 | 31 //(days full month)
Argentina | 04-2015 | 7 //(days from start of the max_date's month)
England | 02-2015 | 21 //(days)
England | 03-2015 | 11 //(days)
I tried too much thing to made this for each records. But now my brain is so confusing and my project is delaying.
Does anybody know how can i solve this. I tried to duplicate each rows with datediff count but it is not working
WITH cte AS (
SELECT CountryName, ISNULL(DATEDIFF(M,Min_Date ,Max_Date )+1,1) as count FROM table_A
UNION ALL
SELECT CountryName, count-1 FROM cte WHERE count>1
)
SELECT CountryName,count FROM cte
-Generate all the dates between min and max dates for each country.
-Then get the month start and month end dates for each country,year,month.
-Finally get the date differences of the month start and month end.
WITH cte AS (
SELECT Country, min_date dt,min_date,max_date FROM t
UNION ALL
SELECT Country, dateadd(dd,1,dt),min_date,max_date FROM cte WHERE dt < max_date
)
,monthends as (
SELECT country,year(dt) yr,month(dt) mth,max(dt) monthend,min(dt) monthstart
FROM cte
GROUP BY country,year(dt),month(dt))
select country
,cast(mth as varchar(2))+'-'+cast(yr as varchar(4)) yr_month
,datediff(dd,monthstart,monthend)+1 days_diff
from monthends
Sample Demo
EDIT: Another option would be to generate all the dates once (the example shown here generates 51 years of dates from 2000 to 2050) and then joining it to the table to get the days by month.
WITH cte AS (
SELECT cast('2000-01-01' as date) dt,cast('2050-12-31' as date) maxdt
UNION ALL
SELECT dateadd(dd,1,dt),maxdt FROM cte WHERE dt < maxdt
)
SELECT country,year(dt) yr,month(dt) mth, datediff(dd,min(dt),max(dt))+1 days_diff
FROM cte c
JOIN t on c.dt BETWEEN t.min_date and t.max_date
GROUP BY country,year(dt),month(dt)
OPTION (MAXRECURSION 0)
I think you have the right idea. But you need to construct the months:
WITH cte AS (
SELECT CountryName, Min_Date as dte, Min_Date, Max_Date
FROM table_A
UNION ALL
SELECT CountryName, DATEADD(month, 1, dte), Min_Date, Max_Date
FROM cte
WHERE dte < Max_date
)
SELECT CountryName, dte
FROM cte;
Getting the number of days in the month is a bit more complicated. That requires some thought.
Oh, I forgot about EOMONTH():
select countryName, dte,
(case when dte = min_date
then datediff(day, min_date, eomonth(dte)) + 1
when dte = max_date
then day(dte)
else day(eomonth(dte))
end) as days
from cte;
Using a Calendar Table makes this stuff pretty easy. RexTester: http://rextester.com/EBTIMG23993
begin
create table #enderaric (
CountryName varchar(16)
, Min_Date date
, Max_Date date
, Number int
)
insert into #enderaric values
('Argentina' ,'2015-01-04' ,'2015-04-07' ,'100')
, ('England' ,'2015-02-08' ,'2015-03-11' ,'90')
end;
-- select * from #enderaric
--*/"
declare #FromDate date;
declare #ThruDate date;
set #FromDate = '2015-01-01';
set #ThruDate = '2015-12-31';
with x as (
select top (cast(sqrt(datediff(day, #FromDate, #ThruDate)) as int) + 1)
[number]
from [master]..spt_values v
)
/* Date Range CTE */
,cal as (
select top (1+datediff(day, #FromDate, #ThruDate))
DateValue = convert(date,dateadd(day,
row_number() over (order by x.number)-1,#FromDate)
)
from x cross join x as y
order by DateValue
)
select
e.CountryName
, YearMonth = convert(char(7),left(convert(varchar(10),DateValue),7))
, [Days]=count(c.DateValue)
from #enderaric as e
inner join cal c on c.DateValue >= e.min_date
and c.DateValue <= e.max_date
group by
e.CountryName
, e.Min_Date
, e.Max_Date
, e.Number
, convert(char(7),left(convert(varchar(10),DateValue),7))
results in:
CountryName YearMonth Days
---------------- --------- -----------
Argentina 2015-01 28
Argentina 2015-02 28
Argentina 2015-03 31
Argentina 2015-04 7
England 2015-02 21
England 2015-03 11
More about calendar tables:
Aaron Bertrand - Generate a set or sequence without loops
generate-a-set-1
generate-a-set-2
generate-a-set-3
David Stein - Creating a Date Table/Dimension on SQL 2008
Michael Valentine Jones - F_TABLE_DATE
I have the following table:
ID GROUPID oDate oValue
1 A 2014-06-01 100
2 A 2014-06-02 200
3 A 2014-06-03 300
4 A 2014-06-04 400
5 A 2014-06-05 500
FF. until the end of the month
30 A 2014-06-30 600
I have 3 kinds of GROUPID, and each group will create one record per day.
I want to calculate the total of oValue from the 2nd day of each month until the end of the month. So the total of June would be from 2/Jun/2014 until 30/Jun/2014. If July, then the total would be from 2/Jul/2014 until 31/Jul/2014.
The output will be like this (sample):
GROUPID MONTH YEAR tot_oValue
A 6 2014 2000
A 7 2014 3000
B 6 2014 1500
B 7 2014 5000
Does anyone know how to solve this with sql syntax?
Thank you.
You can use a correlated subquery to get this:
SELECT T.ID,
T.GroupID,
t.oDate,
T.oValue,
ct.TotalToEndOfMonth
FROM T
OUTER APPLY
( SELECT TotalToEndOfMonth = SUM(oValue)
FROM T AS T2
WHERE T2.GroupID = T.GroupID
AND T2.oDate >= T.oDate
AND T2.oDate < DATEADD(MONTH, DATEDIFF(MONTH, 0, T.oDate) + 1, 0)
) AS ct;
For your example data this gives:
ID GROUPID ODATE OVALUE TOTALTOENDOFMONTH
1 A 2014-06-01 100 2100
2 A 2014-06-02 200 2000
3 A 2014-06-03 300 1800
4 A 2014-06-04 400 1500
5 A 2014-06-05 500 1100
30 A 2014-06-30 600 600
Example on SQL Fiddle
For future reference if you ever upgrade, in SQL Server 2012 (and later) this becomes even easier with windowed aggregate functions that allow ordering:
SELECT T.*,
TotalToEndOfMonth = SUM(oValue)
OVER (PARTITION BY GroupID,
DATEPART(YEAR, oDate),
DATEPART(MONTH, oDate)
ORDER BY oDate DESC)
FROM T
ORDER BY oDate;
Example on SQL Fiddle
EDIT
If you only want this for the 2nd of each month, but still need all the fields then you can just filter the results of the first query I posted:
SELECT T.ID,
T.GroupID,
t.oDate,
T.oValue,
ct.TotalToEndOfMonth
FROM T
OUTER APPLY
( SELECT TotalToEndOfMonth = SUM(oValue)
FROM T AS T2
WHERE T2.GroupID = T.GroupID
AND T2.oDate >= T.oDate
AND T2.oDate < DATEADD(MONTH, DATEDIFF(MONTH, 0, T.oDate) + 1, 0)
) AS ct
WHERE DATEPART(DAY, T.oDate) = 2;
Example on SQL Fiddle
If you are only concerned with the total then you can use:
SELECT T.GroupID,
[Month] = DATEPART(MONTH, oDate),
[Year] = DATEPART(YEAR, oDate),
tot_oValue = SUM(T.oValue)
FROM T
WHERE DATEPART(DAY, T.oDate) >= 2
GROUP BY T.GroupID, DATEPART(MONTH, oDate), DATEPART(YEAR, oDate);
Example on SQL Fiddle
Not sure whether you have data for different years
Select YEAR(oDate),MONTH(oDate),SUM(Value)
From #Temp
Where DAY(oDate)>1
Group By YEAR(oDate),MONTH(oDate)
If you want grouped per GROUPID, year and month this should do it:
SELECT
GROUPID,
[MONTH] = MONTH(oDate),
[YEAR] = YEAR(oDate),
tot_oValue = SUM(ovalue)
FROM your_table
WHERE DAY(odate) > 1
GROUP BY GROUPID, YEAR(oDate), MONTH(oDate)
ORDER BY GROUPID, YEAR(oDate), MONTH(oDate)
This query produces required output:
SELECT GROUPID, MONTH(oDate) AS "Month", YEAR(oDate) AS "Year", SUM(oValue) AS tot_oValue
FROM table_name
WHERE DAY(oDate) > 1
GROUP BY GROUPID, YEAR(oDate), MONTH(oDate)
ORDER BY GROUPID, YEAR(oDate), MONTH(oDate)