Need a query to select "Running Total" as mentioned in the image
2017 year end total plus Every months new figure should add up to previous total.
https://i.stack.imgur.com/DL7p0.png "Example"
This following script will work for MSSQL and you can use the same logic for other databases as well-
WITH your_table(year,month,partersgrowth)
AS
(
SELECT '2019','jan', 100 UNION ALL
SELECT '2019','feb', 300 UNION ALL
SELECT '2019','mar', 400 UNION ALL
SELECT '2019','apr', 500 UNION ALL
SELECT '2018','Dec', 200
)
SELECT A.year,A.month,A.partersgrowth,
(
SELECT SUM(B.partersgrowth)
FROM your_table B
WHERE CAST(B.Year +'-'+B.month+'-01' AS DATE)
<= CAST(A.Year +'-'+A.month+'-01' AS DATE)
) Running_Total
FROM your_table A
ORDER BY CAST(A.Year +'-'+A.month+'-01' AS DATE)
using #mkRabbani solution.. you can simplify it like this:
;WITH your_table(year,month,partersgrowth)
AS
(
SELECT '2019','jan', 100 UNION ALL
SELECT '2019','feb', 300 UNION ALL
SELECT '2019','mar', 400 UNION ALL
SELECT '2019','apr', 500 UNION ALL
SELECT '2018','Dec', 200
)
select *, sum(partersgrowth) over ( order by [year],[month]) as running_total
from your_table
EDIT: As pointed out by comment below.. you want to order by a proper date in the sum part ( I would use order by year and then the month number rather than the month name)
If you use MSSQL you can run the following code
with cte as (
select t.*, lag(partersGrowth) over(partition by year order by month asc rows ROWS UNBOUNDED PRECEDING) prevTotal
from Table )
select years, month, partersGrowth, prevTotal || '+' partersGrowth as "Need Running Total"
from cte
Related
I'm working on outputting some data and I want to pull the daily average of some numbers.
As you can see, what I want to do is count the amount of rows received/results(think the row ID) and then divide it against the day value to make the daily average.(30/1) , (64/2) etc I've tried everything, but I keep running into a wall with this.
As it stands, I'm guessing to make this work a sub query of some sort is needed. I just don't know how to get the day(Row id 1,2,3,4 etc) to use for the division.
SELECT calendar_date, SUM(NY_dayscore * cAttendance)
FROM vw_Appointments
WHERE status = 'Confirmed'
Group by calendar_date
Attempted count with distinct, to no avail
SUM(NY_dayscore * cAttendance) ) / count(distinct calendar_date)
My original code is long and cba to post it all. So just attempting to post a small sample code to get guidance on the issue.
In SQL Server 2012+, you would use the cumulative average:
select calendar_date, sum(NY_dayscore * cAttendance),
avg(sum(NY_dayscore * cAttendance)) over (order by calendar_date) as running_average
from vw_appointments a
where status = 'Confirmed'
group by calendar_date
order by calendar_date;
In SQL Server 2008, this is more difficult:
with a as (
select calendar_date, sum(NY_dayscore * cAttendance) as showed
from vw_appointments a
where status = 'Confirmed'
group by calendar_date
)
select a.*, a2.running_average
from a outer apply
(select avg(showed) as running_average
from a a2
where a2.calendar_date <= a.calendar_date
) a2
order by calendar_date;
Is it ROW_NUMBER() that you are missing?
SELECT
calendar_date,
SUM(NY_dayscore * cAttendance) / (ROW_NUMBER() OVER (ORDER BY calendar_date ASC)) AS average
FROM vw_Appointments
WHERE status = 'Confirmed'
GROUP BY calendar_date
ORDER BY calendar_date
I think you need sum(showed) over (..)/row_number() over (..)
WITH Table1(date, showed) AS
(
SELECT '2019-01-02', 30 UNION ALL
SELECT '2019-01-03', 34 UNION ALL
SELECT '2019-01-03', 41 UNION ALL
SELECT '2019-01-04', 48
)
SELECT date,
sum(showed) over (order by date) /
row_number() over (order by date)
as daily_average
FROM Table1
GROUP BY showed, date;
date daily_average
2019-01-02 30
2019-01-03 52
2019-01-03 35
2019-01-04 38
Demo
I have a transaction table where I have to find the first and second date of transaction of every customer. Finding first date is very simple where I can use MIN() func to find the first date but the second and in particular finding the difference between the two is getting very challenging and somehow I am not able to find out any feasible way:
select a.customer_id, a.transaction_date, a.Row_Count2
from ( select
transaction_date as transaction_date,
reference_no as customer_id,
row_number() over (partition by reference_no
ORDER BY reference_no, transaction_date) AS Row_Count2
from transaction_detail
) a
where a.Row_Count2 < 3
ORDER BY a.customer_id, a.transaction_date, a.Row_Count2
Gives me this :
What I want is , following columns:
||CustomerID|| ||FirstDateofPurchase|| ||SecondDateofPuchase|| ||Diff. b/w Second & First Date ||
You can use window functions LEAD/LAG to return results you are looking for
First try to find all the leading dates by reference number using LEAD, generate row number for each row using your original logic. You can then do difference on dates for row number value 1 row from the result set.
Ex (I'm not excluding same day transactions and treating them as separate and generating row number based on result set from your query above, you can easily change the sql below to consider these as one and remove them so that you get next date as second date):
declare #tbl table(reference_no int, transaction_date datetime)
insert into #tbl
select 1000, '2018-07-11'
UNION ALL
select 1001, '2018-07-12'
UNION ALL
select 1001, '2018-07-12'
UNIOn ALL
select 1001, '2018-07-13'
UNIOn ALL
select 1002, '2018-07-11'
UNIOn ALL
select 1002, '2018-07-15'
select customer_id, transaction_date as firstdate,
transaction_date_next seconddate,
datediff(day, transaction_date, transaction_date_next) diff_in_days
from
(
select reference_no as customer_id, transaction_date,
lead(transaction_date) over (partition by reference_no
order by transaction_date) transaction_date_next,
row_number() over (partition by reference_no ORDER BY transaction_date) AS Row_Count
from #tbl
) src
where Row_Count = 1
You can do this with CROSS APPLY.
SELECT td.customer_id, MIN(ca.transaction_date), MAX(ca.transaction_date),
DATEDIFF(day, MIN(ca.transaction_date), MAX(ca.transaction_date))
FROM transaction_detail td
CROSS APPLY (SELECT TOP 2 *
FROM transaction_detail
WHERE customer_id = td.customer_id
ORDER BY transaction_date) ca
GROUP BY td.customer_id
I am loading data into a table. I don't have any info on how frequent or when the source data is loaded, all I know is I need data from the source to run my script.
Here's the issue, if I run max(date) I get the latest date from the source, but I don't know if the data is still loading. I've ran into cases where I've only gotten a percentage of the data. Thus, I need the next business day after max date.
I want to know is there a way to get the second latest date in the system. I know I can get max(date) - 1, but that give me literally the day after. I don't need the literal day after.
Example, if I run the script on Tuesday, max(date) will be Monday, but since weekend are not in the source system, I need to get Friday instead of Monday.
DATE
---------
2017-04-29
2017-04-25
2017-04-21
2017-04-19
2017-04-18
2017-04-15
2017-04-10
max(date) = 2017-04-29
how do I get 2017-04-25?
Depending on your version of SQL Server, you can use a windowing function like row_number:
select [Date]
from
(
select [Date],
rn = row_number() over(order by [Date] desc)
from #yourtable
) d
where rn = 2
Here is a demo.
Should you have multiple of the same date, you can perform a distinct first:
;with cte as
(
select distinct [date]
from #yourtable
)
select [date]
from
(
select [date],
rn = row_number() over(order by [date] desc)
from cte
) x
where rn = 2;
You can use row_number and get second as below
select * from ( select *, Rown= row_number() over (order by date desc) from yourtable ) a
where a.RowN = 2
More recent SQL Server versions support FETCH FIRST:
select date
from tablename
order by date desc
offset 1 fetch first 1 row only
OFFSET 1 means skip one row. (The 2017-04-29 row.)
;With cte([DATE])
AS
(
SELECT '2017-04-29' union all
SELECT '2017-04-25' union all
SELECT '2017-04-21' union all
SELECT '2017-04-19' union all
SELECT '2017-04-18' union all
SELECT '2017-04-15' union all
SELECT '2017-04-10'
)
SELECT [DATE] FROM
(
SELECT *,ROW_NUMBER()OVER(ORDER BY Seq)-1 As Rno FROM
(
SELECT *,MAX([DATE])OVER(ORDER BY (SELECT NULL))Seq FROM cte
)dt
)Final
WHERE Final.Rno=1
OutPut
DATE
-----
2017-04-25
You can also use FIRST_VALUE with a dynamic date something like DATEADD(DD, -1, GETDATE()). The example below has the date hard coded.
SELECT DISTINCT
FIRST_VALUE([date]) OVER(ORDER BY [date] DESC) AS FirstDate
FROM CTE
WHERE [date] < '2017-04-25'
Another way
DECLARE #T TABLE ([DATE] DATE)
INSERT INTO #T VALUES
('2017-04-29'),
('2017-04-25'),
('2017-04-21'),
('2017-04-19'),
('2017-04-18'),
('2017-04-15'),
('2017-04-10');
SELECT
MAX([DATE]) AS [DATE]
FROM #T
WHERE DATENAME(DW,[DATE]) NOT IN ('Saturday','Sunday')
Another way of doing it, just for example sake...
SELECT MIN(A.date)
FROM
(
SELECT TOP 2 DISTINCT date
FROM YourTable AS C
ORDER BY date DESC
) AS A
I have an sql table like that:
Id Date Price
1 21.09.09 25
2 31.08.09 16
1 23.09.09 21
2 03.09.09 12
So what I need is to get min and max date for each id and dif in days between them. It is kind of easy. Using SQLlite syntax:
SELECT id,
min(date),
max(date),
julianday(max(date)) - julianday(min(date)) as dif
from table group by id
Then the tricky one: how can I receive the price per day during this difference period. I mean something like this:
ID Date PricePerDay
1 21.09.09 25
1 22.09.09 0
1 23.09.09 21
2 31.08.09 16
2 01.09.09 0
2 02.09.09 0
2 03.09.09 12
I create a cte as you mentioned with calendar but dont know how to get the desired result:
WITH RECURSIVE
cnt(x) AS (
SELECT 0
UNION ALL
SELECT x+1 FROM cnt
LIMIT (SELECT ((julianday('2015-12-31') - julianday('2015-01-01')) + 1)))
SELECT date(julianday('2015-01-01'), '+' || x || ' days') as date FROM cnt
p.s. If it will be in sqllite syntax-would be awesome!
You can use a recursive CTE to calculate all the days between the min date and max date. The rest is just a left join and some logic:
with recursive cte as (
select t.id, min(date) as thedate, max(date) as maxdate
from t
group by id
union all
select cte.id, date(thedate, '+1 day') as thedate, cte.maxdate
from cte
where cte.thedate < cte.maxdate
)
select cte.id, cte.date,
coalesce(t.price, 0) as PricePerDay
from cte left join
t
on cte.id = t.id and cte.thedate = t.date;
One method is using a tally table.
To build a list of dates and join that with the table.
The date stamps in the DD.MM.YY format are first changed to the YYYY-MM-DD date format.
To make it possible to actually use them as a date in the SQL.
At the final select they are formatted back to the DD.MM.YY format.
First some test data:
create table testtable (Id int, [Date] varchar(8), Price int);
insert into testtable (Id,[Date],Price) values (1,'21.09.09',25);
insert into testtable (Id,[Date],Price) values (1,'23.09.09',21);
insert into testtable (Id,[Date],Price) values (2,'31.08.09',16);
insert into testtable (Id,[Date],Price) values (2,'03.09.09',12);
The SQL:
with Digits as (
select 0 as n
union all select 1
union all select 2
union all select 3
union all select 4
union all select 5
union all select 6
union all select 7
union all select 8
union all select 9
),
t as (
select Id,
('20'||substr([Date],7,2)||'-'||substr([Date],4,2)||'-'||substr([Date],1,2)) as [Date],
Price
from testtable
),
Dates as (
select Id, date(MinDate,'+'||(d2.n*10+d1.n)||' days') as [Date]
from (
select Id, min([Date]) as MinDate, max([Date]) as MaxDate
from t
group by Id
) q
join Digits d1
join Digits d2
where date(MinDate,'+'||(d2.n*10+d1.n)||' days') <= MaxDate
)
select d.Id,
(substr(d.[Date],9,2)||'.'||substr(d.[Date],6,2)||'.'||substr(d.[Date],3,2)) as [Date],
coalesce(t.Price,0) as Price
from Dates d
left join t on (d.Id = t.Id and d.[Date] = t.[Date])
order by d.Id, d.[Date];
The recursive SQL below was totally inspired by the excellent answer from Gordon Linoff.
And a recursive SQL is probably more performant for this anyway.
(He should get the 15 points for the accepted answer).
The difference in this version is that the datestamps are first formatted to YYYY-MM-DD.
with t as (
select Id,
('20'||substr([Date],7,2)||'-'||substr([Date],4,2)||'-'||substr([Date],1,2)) as [Date],
Price
from testtable
),
cte as (
select Id, min([Date]) as [Date], max([Date]) as MaxDate from t
group by Id
union all
select Id, date([Date], '+1 day'), MaxDate from cte
where [Date] < MaxDate
)
select cte.Id,
(substr(cte.[Date],9,2)||'.'||substr(cte.[Date],6,2)||'.'||substr(cte.[Date],3,2)) as [Date],
coalesce(t.Price, 0) as PricePerDay
from cte
left join t
on (cte.Id = t.Id and cte.[Date] = t.[Date])
order by cte.Id, cte.[Date];
I have a table with is simply a list of dates and user IDs (not aggregated).
We define a metric called active users for a given date by counting the distinct number of IDs that appear in the previous 45 days.
I am trying to run a query in BigQuery that, for each day, returns the day plus the number of active users for that day (count distinct user from 45 days ago until today).
I have experimented with window functions, but can't figure out how to define a range based on the date values in a column. Instead, I believe the following query would work in a database like MySQL, but does not in BigQuery.
SELECT
day,
(SELECT
COUNT(DISTINCT visid)
FROM daily_users
WHERE day BETWEEN DATE_ADD(t.day, -45, "DAY") AND t.day
) AS active_users
FROM daily_users AS t
GROUP BY 1
This doesn't work in BigQuery: "Subselect not allowed in SELECT clause."
How to do this in BigQuery?
BigQuery documentation claims that count(distinct) works as a window function. However, that doesn't help you, because you are not looking for a traditional window frame.
One method would adds a record for each date after a visit:
select theday, count(distinct visid)
from (select date_add(u.day, n.n, "day") as theday, u.visid
from daily_users u cross join
(select 1 as n union all select 2 union all . . .
select 45
) n
) u
group by theday;
Note: there may be simpler ways to generate a series of 45 integers in BigQuery.
Below should work with BigQuery
#legacySQL
SELECT day, active_users FROM (
SELECT
day,
COUNT(DISTINCT id)
OVER (ORDER BY ts RANGE BETWEEN 45*24*3600 PRECEDING AND CURRENT ROW) AS active_users
FROM (
SELECT day, id, TIMESTAMP_TO_SEC(TIMESTAMP(day)) AS ts
FROM daily_users
)
) GROUP BY 1, 2 ORDER BY 1
Above assumes that day field is represented as '2016-01-10' format.
If it is not a case , you should adjust TIMESTAMP_TO_SEC(TIMESTAMP(day)) in most inner select
Also please take a look at COUNT(DISTINC) specifics in BigQuery
Update for BigQuery Standard SQL
#standardSQL
SELECT
day,
(SELECT COUNT(DISTINCT id) FROM UNNEST(active_users) id) AS active_users
FROM (
SELECT
day,
ARRAY_AGG(id)
OVER (ORDER BY ts RANGE BETWEEN 3888000 PRECEDING AND CURRENT ROW) AS active_users
FROM (
SELECT day, id, UNIX_DATE(PARSE_DATE('%Y-%m-%d', day)) * 24 * 3600 AS ts
FROM daily_users
)
)
GROUP BY 1, 2
ORDER BY 1
You can test / play with it using below dummy sample
#standardSQL
WITH daily_users AS (
SELECT 1 AS id, '2016-01-10' AS day UNION ALL
SELECT 2 AS id, '2016-01-10' AS day UNION ALL
SELECT 1 AS id, '2016-01-11' AS day UNION ALL
SELECT 3 AS id, '2016-01-11' AS day UNION ALL
SELECT 1 AS id, '2016-01-12' AS day UNION ALL
SELECT 1 AS id, '2016-01-12' AS day UNION ALL
SELECT 1 AS id, '2016-01-12' AS day UNION ALL
SELECT 1 AS id, '2016-01-13' AS day
)
SELECT
day,
(SELECT COUNT(DISTINCT id) FROM UNNEST(active_users) id) AS active_users
FROM (
SELECT
day,
ARRAY_AGG(id)
OVER (ORDER BY ts RANGE BETWEEN 86400 PRECEDING AND CURRENT ROW) AS active_users
FROM (
SELECT day, id, UNIX_DATE(PARSE_DATE('%Y-%m-%d', day)) * 24 * 3600 AS ts
FROM daily_users
)
)
GROUP BY 1, 2
ORDER BY 1