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

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

GROUP BY CAST(myDateTime AS DATE)

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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)

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In pre Sql 2008 By taking out the date part:

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

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CAST datetime field to date

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

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GROUP BY DATE(date_time_column)

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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)

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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');

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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.

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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')