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If you'd like to see the latest date first and the earliest date last, you need to sort in descending order. Use the DESC keyword in this case. ORDER BY ExamDate DESC ; Note that in T-SQL, NULL s are displayed first when sorting in ascending order and last when sorting in descending order.
We can group the resultset in SQL on multiple column values. When we define the grouping criteria on more than one column, all the records having the same value for the columns defined in the group by clause are collectively represented using a single record in the query output.
SQL Server comes with the following data types for storing a date or a date/time value in the database: DATE - format YYYY-MM-DD. DATETIME - format: YYYY-MM-DD HH:MI:SS. SMALLDATETIME - format: YYYY-MM-DD HH:MI:SS.
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)
GROUP BY DATE(date_time_column)
CAST datetime field to date
select CAST(datetime_field as DATE), count(*) as count from table group by CAST(datetime_field as DATE);
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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