How can I count the number of rows per hour in SQL Server with full date-time as result.
I've already tried this, but it returns only the hours
SELECT DATEPART(HOUR,TimeStamp), Count(*)
FROM [TEST].[dbo].[data]
GROUP BY DATEPART(HOUR,TimeStamp)
ORDER BY DATEPART(HOUR,TimeStamp)
Now the result is:
Hour Occurrence
---- ----------
10 2157
11 60740
12 66189
13 77096
14 90039
But I need this:
Timestamp Occurrence
------------------- ----------
2013-12-21 10:00:00 2157
2013-12-21 11:00:00 60740
2013-12-21 12:00:00 66189
2013-12-21 13:00:00 77096
2013-12-21 14:00:00 90039
2013-12-22 09:00:00 84838
2013-12-22 10:00:00 64238
Here is the SQL query to get data for every hour in MySQL. In the above query, we simply group by order_date using HOUR function and aggregate amount column using SUM function. HOUR function retrieves hour number from a given date/time/datetime value, which can be provided as a literal string or column name.
The SQL COUNT() function returns the number of rows in a table satisfying the criteria specified in the WHERE clause. It sets the number of rows or non NULL column values. COUNT() returns 0 if there were no matching rows.
Use the COUNT aggregate function to count the number of rows in a table. This function takes the name of the column as its argument (e.g., id ) and returns the number of rows for this particular column in the table (e.g., 5).
MySQL HOUR() Function The HOUR() function returns the hour part for a given date (from 0 to 838).
You actually need to round the TimeStamp
to the hour. In SQL Server, this is a bit ugly, but easy to do:
SELECT dateadd(hour, datediff(hour, 0, TimeStamp), 0) as TimeStampHour, Count(*)
FROM [TEST].[dbo].[data]
GROUP BY dateadd(hour, datediff(hour, 0, TimeStamp), 0)
ORDER BY dateadd(hour, datediff(hour, 0, TimeStamp), 0);
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