How can I group by date time column without taking time into consideration

I have a bunch of product orders and I'm trying to group by the date and sum the quantity for that date. How can I group by the month/day/year without taking the time part into consideration?

3/8/2010 7:42:00开发者_Go百科 should be grouped with 3/8/2010 4:15:00

Cast/Convert the values to a Date type for your group by.

GROUP BY CAST(myDateTime AS DATE)

GROUP BY DATEADD(day, DATEDIFF(day, 0, MyDateTimeColumn), 0)

Or in SQL Server 2008 onwards you could simply cast to Date as @Oded suggested:

GROUP BY CAST(orderDate AS DATE)

In pre Sql 2008 By taking out the date part:

GROUP BY CONVERT(CHAR(8),DateTimeColumn,10)

CAST datetime field to date

select  CAST(datetime_field as DATE), count(*) as count from table group by CAST(datetime_field as DATE);

GROUP BY DATE(date_time_column)

Here's an example that I used when I needed to count the number of records for a particular date without the time portion:

select count(convert(CHAR(10), dtcreatedate, 103) ),convert(char(10), dtcreatedate, 103)
FROM dbo.tbltobecounted
GROUP BY CONVERT(CHAR(10),dtcreatedate,103)
ORDER BY CONVERT(CHAR(10),dtcreatedate,103)

Here is the example works fine in oracle

select to_char(columnname, 'DD/MON/yyyy'), count(*) from table_name group by to_char(createddate, 'DD/MON/yyyy');

Well, for me it was pretty much straight, I used cast with groupby:

Example:

Select cast(created_at as date), count(1) from dbname.tablename GROUP BY cast(created_at as date)

Note: I am using this on MSSQL 2016.

I believe you need to group by , in that day of the month of the year . so why not using TRUNK_DATE functions . The way it works is described below :

Group By DATE_TRUNC('day' , 'occurred_at_time')