This question already has answers here:
Calculate a Running Total in SQL Server
(15 answers)
Closed 3 years ago.
I have the following table:
________________________
date | amount
________________________
01-01-2019 | 10
01-01-2019 | 10
01-01-2019 | 10
01-01-2019 | 10
02-01-2019 | 5
02-01-2019 | 5
02-01-2019 | 5
02-01-2019 | 5
03-01-2019 | 20
03-01-2019 | 20
These are mutation values by date. I would like my query to return the summed amount by date. So for 02-01-2019 I need 40 ( 4 times 10) + 20 ( 4 times 5). For 03-01-2019 I would need ( 4 times 10) + 20 ( 4 times 5) + 40 ( 2 times 20) and so on. Is this possible in one query? How do I achieve this?
My current query to get the individual mutations:
Select s.date,
Sum(s.amount) As Sum_amount
From dbo.Financieel As s
Group By s.date
You can try below -
DEMO
select dateval,
SUM(amt) OVER(ORDER BY dateval ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW) as amt
from
(
SELECT
dateval,
SUM(amount) amt
FROM t2 group by dateval
)A
OUTPUT:
dateval amt
01/01/2019 00:00:00 40
01/02/2019 00:00:00 60
01/03/2019 00:00:00 100
Try this below script to get your desired output-
SELECT A.date,
(SELECT SUM(amount) FROM <your_table> WHERE Date <= A.Date) C_Total
FROM <your_table> A
GROUP BY date
ORDER BY date
Output is-
date C_Total
01-01-2019 40
02-01-2019 60
03-01-2019 100
I suggest to use a window function, like this:
select date, sum(amount) over( order by date)
from table
Related
I have a table of EMPLOYEES that contains information about the DATE and WORKTIME per that day. Fx:
ID | DATE | WORKTIME |
----------------------------------------
1 | 1-Sep-2014 | 4 |
2 | 2-Sep-2014 | 6 |
1 | 3-Sep-2014 | 5.5 |
1 | 4-Sep-2014 | 7 |
2 | 4-Sep-2014 | 4 |
1 | 9-Sep-2014 | 8 |
and so on.
Question: How can I create a query that would allow me to calculate amount of time worked per week (HOURS_PERWEEK). I understand that I need a summation of WORKTIME together with grouping considering both, ID and week, but so far my trials as well as googling didnt yield any results. Any ideas on this? Thank you in advance!
edit:
Got a solution of
select id, sum (worktime), trunc(date, 'IW') week
from employees
group by id, TRUNC(date, 'IW');
But will need somehow to connect that particular output with DATE table by updating a newly created column such as WEEKLY_TIME. Any hints on that?
You can find the start of the ISO week, which will always be a Monday, using TRUNC("DATE", 'IW').
So if, in the query, you GROUP BY the id and the start of the week TRUNC("DATE", 'IW') then you can SELECT the id and aggregate to find the SUM the WORKTIME column for each id.
Since this appears to be a homework question and you haven't attempted a query, I'll leave it at this to point you in the correct direction and you can complete the query.
Update
Now I need to create another column (lets call it WEEKLY_TIME) and populate it with values from the current output, so that Sep 1,3,4 (for ID=1) would all contain value 16.5, specifying that on that day (that is within the certain week) that person worked 16.5 in total. And for ID=2 it would then be a value of 10 for both Sep 2 and 4.
For this, if I understand correctly, you appear to not want to use aggregation functions and want to use the analytic version of the function:
select id,
"DATE",
trunc("DATE", 'IW') week,
worktime,
sum (worktime) OVER (PARTITION BY id, trunc("DATE", 'IW'))
AS weekly_time
from employees;
Which, for the sample data:
CREATE TABLE employees (ID, "DATE", WORKTIME) AS
SELECT 1, DATE '2014-09-01', 4 FROM DUAL UNION ALL
SELECT 2, DATE '2014-09-02', 6 FROM DUAL UNION ALL
SELECT 1, DATE '2014-09-03', 5.5 FROM DUAL UNION ALL
SELECT 1, DATE '2014-09-04', 7 FROM DUAL UNION ALL
SELECT 2, DATE '2014-09-04', 4 FROM DUAL UNION ALL
SELECT 1, DATE '2014-09-09', 8 FROM DUAL;
Outputs:
ID
DATE
WEEK
WORKTIME
WEEKLY_TIME
1
2014-09-01 00:00:00
2014-09-01 00:00:00
4
16.5
1
2014-09-03 00:00:00
2014-09-01 00:00:00
5.5
16.5
1
2014-09-04 00:00:00
2014-09-01 00:00:00
7
16.5
1
2014-09-09 00:00:00
2014-09-08 00:00:00
8
8
2
2014-09-04 00:00:00
2014-09-01 00:00:00
4
10
2
2014-09-02 00:00:00
2014-09-01 00:00:00
6
10
db<>fiddle here
edit: answer submitted without noticing "Oracle" tag. Otherwise, question answered here: Oracle SQL - Sum and group data by week
Select employee_Id,
DATEPART(week, workday) as [Week],
sum (worktime) as [Weekly Hours]
from WORK
group by employee_id, DATEPART(week, workday)
https://dbfiddle.uk/?rdbms=sqlserver_2019&fiddle=238b229156a383fa3c466b6c3c2dee1e
I have table in MS SQL Server like below:
ID_EVENT | PRODUCT| DATE
-------------------------------
123 | A | 2021-01-15
456 | A | 2021-01-22
789 | A | 2021-02-05
110 | B | 2021-01-18
124 | B | 2021-02-11
I need to calculate average monthly (for January and February) number of ID_EVENT per PRODUCT. So as a result I need something like below:
PRODUKT | AVG_PER_MNTH
-----------------
A | 1.5
B | 1
A = 1.5 because 3 / 2 = 1.5 --> (number of ID_EVENT for January + number of ID_EVENT for February) / number of months which we analyse (2 -> January and February)
B = 1 because 1 / 1 = 1 --> (number of ID_EVENT for January + number of ID_EVENT for February) / number of months which we analyse (2 -> January and February)
How can I do that in MS SQL Server ?
One option, aggregating first by month and year, then by product:
WITH cte AS (
SELECT PRODUCT, 1.0*COUNT(*) AS cnt
FROM yourTable
GROUP BY PRODUCT, FORMAT(DATE, 'MM.yyyy')
)
SELECT PRODUCT, AVG(cnt) AS AVG_PER_MNTH
FROM cte
GROUP BY PRODUCT;
Demo
WITH CTE AS
(SELECT PRODUCT
,COUNT(*) ac cnt,
,DATEADD(month, DATEDIFF(month, 0, Date), 0) AS StartOfMonth
FROM YourTable
GROUP BY PRODUCT
,DATEADD(month, DATEDIFF(month, 0, Date), 0)
)
SELECT
PRODUCT
,AVG(cnt) as MonthlyAverage
FROM CTE
GROUP BY PRODUCT
I'm trying to display a result set based on a min date value and today's date but can't seem to make it work. It's essentially a date sensitive price list.
Example Data
ID Title Value ExpireDate
1 Fred 10 2019-03-01
2 Barney 15 2019-03-01
3 Fred2 20 2019-06-01
4 Barney2 25 2019-06-01
5 Fred3 30 2019-07-01
6 Barney3 55 2019-07-01
Required Results:
Display records based on minimum date > GetDate()
3 Fred2 20 2019-06-01
4 Barney2 25 2019-06-01
Any assistance would be great - thank you.
Use where clause to filter all future rows and row_number() to find the first row per group:
SELECT *
FROM (
SELECT *, ROW_NUMBER() OVER (PARTITION BY Title ORDER BY ExpireDate) AS rn
FROM t
WHERE ExpireDate >= CAST(CURRENT_TIMESTAMP AS DATE)
) AS x
WHERE rn = 1
Based on your revised question, you can simply do this:
SELECT TOP 1 WITH TIES *
FROM t
WHERE ExpireDate >= CAST(CURRENT_TIMESTAMP AS DATE)
ORDER BY ExpireDate
I'm trying to figure out the count of users who listened to more than 10 mins each month in the last 6 months
We have this event: Song_stopped_listen and one attribute is session_progress_ms
Now I'm trying to see the monthly evolution of the count of this cohort over the last 6 months.
I'm using bigquery and this is the query I tried, but I feel that something is off semantically, but I couldn't put my finger on:
SELECT
CONCAT(CAST(EXTRACT(YEAR FROM DATE (timestamp)) AS STRING),"-",CAST(EXTRACT(MONTH FROM DATE (timestamp)) AS STRING)) AS date
,SUM(absl.session_progress_ms/(1000*60*10)) as total_10_ms, COUNT(DISTINCT u.id) as total_10_listeners
FROM ios.song_stopped_listen as absl
LEFT JOIN ios.users u on absl.user_id = u.id
WHERE absl.timestamp > '2018-05-01'
Group by 1
HAVING(total_10_ms > 1)
Please help figure out what I'm doing wrong here.
Thank you.
data Sample:
user_id | session_progress_ms | timestamp
1 | 10000 | 2017-10-10 14:34:25.656 UTC
What I want to have:
||Month-year | Count of users who listened to more than 10 mins
|2018-5 | 500
|2018-6 | 600
|2018-7 | 300
|2018-8 | 5100
|2018-9 | 4500
|2018-10 | 1500
|2018-11 | 1500
|2018-12 | 2500
Use multiple levels of aggregation:
select user_id
from (select ssl.user_id, timestamp_trunc(timestamp, month) as mon,
sum(ssl.session_progress_ms/(1000*60)) as total_minutes
from ios.song_stopped_listen as ssl
where date(ssl.timetamp) < date_trunc(current_date, month) and
date(ssl.timestamp) >= date_add(date_trunc(current_date, month) interval 6 month),
group by 1, 2
) u
where total_minutes >= 10
group by user_id
having count(*) = 6;
To get the count, just use this as a subquery with count(*).
Let's assume I have one table in postgres with just 2 columns:
ID which is PK for the table (bigint)
time which is type of timestamp
Is there any way how to get IDs grouped by time BY YEAR- when the time is date 18 February 2005 it would fit in 2005 group (so result would be)
year number of rows
1998 2
2005 5
AND if the number of result rows is smaller than some number (for example 3) SQL will return the result by month
Something like
month number of rows
(February 2018) 5
(March 2018) 2
Is that possible some nice way in postgres SQL?
You can do it using window functions (as always).
I use this table:
TABLE times;
id | t
----+-------------------------------
1 | 2018-03-14 20:04:39.81298+01
2 | 2018-03-14 20:04:42.92462+01
3 | 2018-03-14 20:04:45.774615+01
4 | 2018-03-14 20:04:48.877038+01
5 | 2017-03-14 20:05:08.94096+01
6 | 2017-03-14 20:05:16.123736+01
7 | 2017-03-14 20:05:19.91982+01
8 | 2017-01-14 20:05:32.249175+01
9 | 2017-01-14 20:05:35.793645+01
10 | 2017-01-14 20:05:39.991486+01
11 | 2016-11-14 20:05:47.951472+01
12 | 2016-11-14 20:05:52.941504+01
13 | 2016-10-14 21:05:52.941504+02
(13 rows)
First, group by month (subquery per_month).
Then add the sum per year with a window function (subquery with_year).
Finally, use CASE to decide which one you will output and remove duplicates with DISTINCT.
SELECT DISTINCT
CASE WHEN yc > 5
THEN mc
ELSE yc
END AS count,
CASE WHEN yc > 5
THEN to_char(t, 'YYYY-MM')
ELSE to_char(t, 'YYYY')
END AS period
FROM (SELECT
mc,
sum(mc) OVER (PARTITION BY date_trunc('year', t)) AS yc,
t
FROM (SELECT
count(*) AS mc,
date_trunc('month', t) AS t
FROM times
GROUP BY date_trunc('month', t)
) per_month
) with_year
ORDER BY 2;
count | period
-------+---------
3 | 2016
3 | 2017-01
3 | 2017-03
4 | 2018
(4 rows)
Just count years. If it's at least 3, then you group by years, else by months:
select
case (select count(distinct extract(year from time)) from mytable) >= 3 then
to_char(time, 'yyyy')
else
to_char(time, 'yyyy-mm')
end as season,
count(*)
from mytable
group by season
order by season;
(Unlike many other DBMS, PostgreSQL allows to use alias names in the GROUP BY clause.)