Question

I read berfore SQL Count for a Date Column and SQL to get count by date but these does not point my problem.

I have the following table records (the table has only two columns)

| VALUE   | TIME               |
--------------------------------
| test    | 04.05.2012 11:46:46|
| test2   | 04.05.2012 11:46:49|
| test3   | 06.05.2012 11:47:46|
| test2   | 05.05.2012 11:47:46|
| test    | 04.05.2012 11:51:46|

I want to display like:

| VALUE   | COUNT   | NEWTIME  |
--------------------------------
| test    | 2       |04.05.2012|
| test2   | 1       |04.05.2012|
| test2   | 1       |05.05.2012|
| test3   | 1       |06.05.2012|

I counted VALUE by using day as datepart of TIME like:

select datepart(d, TIME), count(VALUE) from table1
group by datepart(d, TIME)

How to get also the additional NEWTIME column ? It is possible in SQL Server 2005 ?

Thank you

Answer 1

select VALUE,
       count(*) as COUNT,
       dateadd(day, datediff(day, 0, TIME), 0) as NEWTIME
from YourTable
group by VALUE, dateadd(day, datediff(day, 0, TIME), 0)

SE-Data