Query with group by with CustID and amounts daily and MTD - sql

I have the following data:
ID
username
Cost
time
1
test1
1
2021-05-22 11:48:36.000
2
test2
2
2021-05-20 12:55:22.000
3
test3
5
2021-05-21 00:00:0-0.000
I would like to count the costs for the username with a daily figure and a month to date figure in once table
I have got the following
SELECT
username,
COUNT(*) cost
FROM
dbo.sent
WHERE
time BETWEEN {ts '2021-05-01 00:00:00'} AND {ts '2021-05-22 23:59:59'}
GROUP BY
username
ORDER BY
cost DESC;
This will return the monthly figures and changing the time to '2021-05-22 00:00:00'} AND {ts '2021-05-22 23:59:59'} will give the the daily however I would like a table to show daily and MTD together
username
Daily
MTD
test1
1
1012
test2
2
500
test3
5
22
Any help or pointers would be fantastic, I am guessing that I need a temp table and then run again using the MTD range updating the temp table where the username is the same then to export and delete - but i have no idea where to start.

First, you need one row per user and date, not just one row per user.
Second, you should fix your date arithmetic so you are not missing a second.
Then, you can use aggregation and window functions:
SELECT username, CONVERT(DATE, time) as dte,
COUNT(*) as cost_per_day,
SUM(COUNT(*)) OVER (PARTITION BY username ORDER BY CONVERT(DATE, time)) as mtd
FROM dbo.sent s
WHERE time >= '2021-05-01' AND
time < '2021-05-23'
GROUP BY username, CONVERT(DATE, time)
ORDER BY username, dte;
You can learn more about window functions in a SQL tutorial (or, ahem, a book on SQL) or in the documentation.
EDIT:
If you only want the most recent date and MTD, then you can either filter the above query for the most recent date or use conditional aggregation:
SELECT username,
SUM(CASE WHEN CONVERT(DATE, time) = '2021-05-22' THEN 1 ELSE 0 END) as cost_per_most_recent_day,
COUNT(*) as MTD
FROM dbo.sent s
WHERE time >= '2021-05-01' AND
time < '2021-05-23'
GROUP BY username
ORDER BY username;
And, you can actually express the query using the current date so it doesn't have to be hardcoded. For this version:
SELECT username,
SUM(CASE WHEN CONVERT(DATE, time) = CONVERT(DATE, GETDATE()) THEN 1 ELSE 0 END) as cost_per_most_recent_day,
COUNT(*) as MTD
FROM dbo.sent s
WHERE time >= DATEFROMPARTS(YEAR(GETDATE()), MONTH(GETDATE()), 1) AND
time < DATEADD(DAY, 1, CONVERT(DATE, GETDATE()))
GROUP BY username
ORDER BY username;

Related

How to replace the loop in MsSQL?

For example
If I want to check in every day last week
select count(ID) from DB where date < "2019/07/01"
select count(ID) from DB where date < "2019/07/02"
select count(ID) from DB where date < "2019/07/03"
...
select count(ID) from DB where date < "2019/07/08"
like
0701 10
0702 15
0703 23
...
0707 45
How to do this without loop and one query?
You can generate the dates using a recursive CTE (or other method) and then run the query:
with dates as (
select convert(date, '2019-07-01') as dte union all
select dateadd(day, 1, dte)
from dates
where dte < '2019-07-08'
)
select d.dte,
(select count(*) from DB where DB.date < d.dte)
from dates d;
More efficient, though, is a cumulative sum:
select db.*
from (select date, count(*) as cnt, sum(count(*)) over (order by date) as running_cnt
from db
group by date
) d
where d.date >= '2019-07-01' and d.date < '2019-07-09';
Are you just counting the number by day?
Something like
SELECT MONTH(date), DAY(date), COUNT(ID)
FROM DB
GROUP BY MONTH(date), DAY(date);
(assuming date is a DATE or DATETIME)
Do it with window Count. range between current row and current row selects exactly this day rows.
select distinct date, count(1) over (order by Date) - count(1) over (order by Date range between current row and current row)
from DB
where date between '2019-07-01' and '2019-07-08';
I assume date column is exactly DATE.

To find from a table if there is any nearer date values for the same key

I have an sql table (t_accountdetails) with an account column called AccountId and effective date column for that account. An account can have multiple effective date. I have a requirement to get all the entries for the accounts which has very close effective date entries.(an offset of +/-14 days)
Say for eg:
AccountId: 12345 has got 2 entries with effective date 12/11/2017 and 12/18/2017
So my query should return above case where we have an entry of effective dates within offset of +/-14days
Please note I am actually not looking for date +-14 from today. I am looking for effective date which +/- 14 days of another effective date for the same account
You want all records where exists another effective date within 14 days, so use WHERE EXISTS:
select *
from t_accountdetails t
where exists
(
select *
from t_accountdetails other
where other.accountid = t.accountid
and other.id <> t.id
and abs(datediff(day, other.effective_date, t.effective_date)) <= 14
)
order by accountid, effective_date;
You can use the DATEADD function to make it work
select * from t_accountdetails where AccountId = 12345 and effectiveDate >= DATEADD(day, -14, getdate()) and effectiveDate <= DATEADD(day, 14, getdate())
This will return all records with AccountID = 12345 and an effective date between today - 14 days and today + 14 days.
Note: if more than one record match the criteria then all matching records will be returned.
I would be inclined to use lag() and lead():
select ad.*
from (select ad.*,
lag(effective_date) over (partition by accountid order by effective_date) as prev_ed,
lead(effective_date) over (partition by accountid order by effective_date) as next_ed
from t_accountdetails ad
) ad
where effective_date <= dateadd(day, 14, prev_ed) or
effective_date >= dateadd(day, -14, next_ed);
It would be interesting to compare the performance of this version to the exists version with an index on t_accountdetails(accountid, effective_date).

SQL Count of date values that dont match hour and day in select statement

I have a problem unique to a business process. My user needs to know how many dates, counted, are before a specific end time that do not match on the hour or the day.
Here is an example.
AAA, 2016-03-15 16:00:28.967, 2016-03-15 16:02:58.487, 2016-03-17 14:01:24.243
In the example above id AAA has 3 entries. I need to count only the ones that don't have a matching hour and day. So the actual count should come out to be 2.
I have to do this all in SQL and can't use a CTE. It needs to be either a sub select or some type of join.
Something like this.
SELECT id, date, (
SELECT COUNT(*)
FROM x
WHERE day!=day
AND hour!=hour AND date < z
) AS DateCount
Results would be AAA, 2
I am thinking some type of recursive comparison but I am not sure how to accomplish this without a CTE.
In SQL Server you can try something like this:
SELECT id, CONVERT(VARCHAR(13), [date], 120) AS [Date], COUNT(*) AS DateCount
FROM YourTable
WHERE [date] < #ENDDATE
GROUP BY id, CONVERT(VARCHAR(13), [date], 120)
SELECT a AS current_a, COUNT(*) AS b,day AS day, hour as hour,
(SELECT COUNT(*)
FROM t
WHERE day != day
AND hour != hour
AND date < z ) as datecount
FROM t GROUP BY a ORDER by b DESC

How to count number of records per day?

I have a table in a with the following structure:
CustID --- DateAdded ---
396 2012-02-09
396 2012-02-09
396 2012-02-08
396 2012-02-07
396 2012-02-07
396 2012-02-07
396 2012-02-06
396 2012-02-06
I would like to know how I can count the number of records per day, for the last 7 days in SQL and then return this as an integer.
At present I have the following SQL query written:
SELECT *
FROM Responses
WHERE DateAdded >= dateadd(day, datediff(day, 0, GetDate()) - 7, 0)
RETURN
However this only returns all entries for the past 7 days. How can I count the records per day for the last 7 days?
select DateAdded, count(CustID)
from Responses
WHERE DateAdded >=dateadd(day,datediff(day,0,GetDate())- 7,0)
GROUP BY DateAdded
select DateAdded, count(CustID)
from tbl
group by DateAdded
about 7-days interval it's DB-depending question
SELECT DateAdded, COUNT(1) AS NUMBERADDBYDAY
FROM Responses
WHERE DateAdded >= dateadd(day,datediff(day,0,GetDate())- 7,0)
GROUP BY DateAdded
This one is like the answer above which uses the MySql DATE_FORMAT() function. I also selected just one specific week in Jan.
SELECT
DatePart(day, DateAdded) AS date,
COUNT(entryhash) AS count
FROM Responses
where DateAdded > '2020-01-25' and DateAdded < '2020-02-01'
GROUP BY
DatePart(day, DateAdded )
If your timestamp includes time, not only date, use:
SELECT DATE_FORMAT('timestamp', '%Y-%m-%d') AS date, COUNT(id) AS count FROM table GROUP BY DATE_FORMAT('timestamp', '%Y-%m-%d')
You could also try this:
SELECT DISTINCT (DATE(dateadded)) AS unique_date, COUNT(*) AS amount
FROM table
GROUP BY unique_date
ORDER BY unique_date ASC
SELECT count(*), dateadded FROM Responses
WHERE DateAdded >=dateadd(day,datediff(day,0,GetDate())- 7,0)
group by dateadded
RETURN
This will give you a count of records for each dateadded value. Don't make the mistake of adding more columns to the select, expecting to get just one count per day. The group by clause will give you a row for every unique instance of the columns listed.
select DateAdded, count(DateAdded) as num_records
from your_table
WHERE DateAdded >=dateadd(day,datediff(day,0,GetDate())- 7,0)
group by DateAdded
order by DateAdded
Unfortunately the best answer here IMO is a comment by #Profex on an incorrect answer , but the solution I went with is
SELECT FORMAT(DateAdded, 'yyyy-MM-dd'), count(CustID)
FROM Responses
WHERE DateAdded >= dateadd(day,datediff(day,0,GetDate())- 7,0)
GROUP BY FORMAT(DateAdded, 'yyyy-MM-dd')
ORDER BY FORMAT(DateAdded, 'yyyy-MM-dd')
Note that I haven't tested this SQL since I don't have the OP's DB , but this approach works well in my scenario where the date is stored to the second
The important part here is using the FORMAT(DateAdded, 'yyyy-MM-dd') method to drop the time without losing the year and month , as would happen if you used DATEPART(day, DateAdded)
When a day among last 7 days, has no record means, the following code will list out that day with count as zero.
DECLARE #startDate DATE = GETDATE() - 6,
#endDate DATE = GETDATE();
DECLARE #daysTable TABLE
(
OrderDate date
)
DECLARE #daysOrderTable TABLE
(
OrderDate date,
OrderCount int
)
Insert into #daysTable
SELECT TOP (DATEDIFF(DAY, #startDate, #endDate) + 1)
Date = DATEADD(DAY, ROW_NUMBER() OVER(ORDER BY a.object_id) - 1, #startDate)
FROM sys.all_objects a
CROSS JOIN sys.all_objects b;
Insert into #daysOrderTable
select OrderDate, ISNULL((SELECT COUNT(*) AS OdrCount
FROM [dbo].[MyOrderTable] odr
WHERE CAST(odr.[CreatedDate] as date) = dt.OrderDate
group by CAST(odr.[CreatedDate] as date)
), 0) AS OrderCount from #daysTable dt
select * from #daysOrderTable
RESULT
OrderDate     OrderCount
2022-11-22     42
2022-11-23     6
2022-11-24     34
2022-11-25     0
2022-11-26     28
2022-11-27     0
2022-11-28     22
SELECT DATE_FORMAT(DateAdded, '%Y-%m-%d'),
COUNT(CustID)
FROM Responses
GROUP BY DATE_FORMAT(DateAdded, '%Y-%m-%d');

T-SQL recursive in order to find transactions for a period of time

I have tried but i am not able to figure this out. I have a table transactions (transaction_ID, transaction_Person_ID, Transaction_Date etc).
What i want is to return all the transaction_person_ID's that have more than 3 transactions per week for the last year. That means i have to check for 1-1-10 to 7-1-10 to see if someone had more than 3 transactions for that 7 day period, then for 2-1-10 to 8-1-10 then 3-1-10 to 9-1-10 etc etc.
I now i need to use a recursive select but i what i have writen does not produce the correct time frame.
What i have written so far is this
WITH Dates AS (
SELECT
[Date] = CONVERT(DATETIME,'01/01/2010')
UNION ALL SELECT
[Date] = DATEADD(DAY, 1, [Date])
FROM
Dates
WHERE
Date < '12/31/2010'
)
SELECT transaction_person_Id FROM transactions
JOIN DATES
ON transactions.transaction_date = dates.date
where transactions.Transaction_Date between dateadd(DAYOFYEAR,-7,dates.date) and dates.date
group by transaction_person_Id
having count(transaction_person_ID) >= 4
OPTION (MAXRECURSION 2000)
Thanks a lot
PS:
in simple words what i need to do is this
select transaction_person_ID from transactions
where Transaction_Date between '2010-01-01' and '2010-01-07'
group by transaction_person_Id
having count(transaction_person_ID) >= 4
then
select transaction_person_ID from transactions
where Transaction_Date between '2010-01-02' and '2010-01-08'
group by transaction_person_Id
having count(transaction_person_ID) >= 4
.
.
.
.
.
until it goes
select transaction_person_ID from transactions
where Transaction_Date between '2010-12-25' and '2010-12-31'
group by transaction_person_Id
having count(transaction_person_ID) >= 4
i need to have the results of these 365 queries
This will give one result set with person and weeks, rather then 360 results sets
WITH Weeks
AS (
SELECT
CONVERT(DATETIME,'01 Jan 2010') AS WeekStartMidnight,
CONVERT(DATETIME,'08 Jan 2010') AS WeekEndMidnight
UNION ALL
SELECT
DATEADD(day, 1, WeekStartMidnight),
DATEADD(day, 1, WeekEndMidnight)
FROM
Weeks
WHERE
WeekEndMidnight < '31 Dec 2010'
)
SELECT
t.transaction_person_Id,
w.WeekStartMidnight, w.WeekEndMidnight
FROM
weeks w
JOIN
transactions t ON t.Transaction_Date >= w.WeekStartMidnight AND t.Transaction_Date < w.WeekEndMidnight
GROUP BY
t.transaction_person_Id
HAVING
count(*) >= 4 --COUNT(t.transaction_person_Id) = same
OPTION
(MAXRECURSION 365)
If you want 360 results sets, it's a loop using WHILE or a CURSOR per row in the "weeks" derived table