How to count number of records per day? - sql

Question

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?

Answer 1

select DateAdded, count(CustID)
from Responses
WHERE DateAdded >=dateadd(day,datediff(day,0,GetDate())- 7,0)
GROUP BY DateAdded

Answer 2

select DateAdded, count(CustID)
from tbl
group by DateAdded
about 7-days interval it's DB-depending question

Answer 3

SELECT DateAdded, COUNT(1) AS NUMBERADDBYDAY
FROM Responses
WHERE DateAdded >= dateadd(day,datediff(day,0,GetDate())- 7,0)
GROUP BY DateAdded

Answer 4

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 )

Answer 5

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')

Answer 6

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

Answer 7

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.

Answer 8

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

Answer 9

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)

Answer 10

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

Answer 11

SELECT DATE_FORMAT(DateAdded, '%Y-%m-%d'),
COUNT(CustID)
FROM Responses
GROUP BY DATE_FORMAT(DateAdded, '%Y-%m-%d');