SQL group by day, with count

select cast(LogDate as date) as LogDate, count(refundId) as refundCount
from yourTable
group by cast(LogDate as date)

Depending on the dialect of SQL you're using, you may have to change the CAST to something else. The expression should convert the LogDate to a date-only value.

Also, if you say "different refundId" because there could be repeated values of refundId that you only want to count once, use count(DISTINCT refundId)

What database vendor are you using? Whichever it is, replace the "DateOnly(LogDate)" in the following with the appropriate construict to extract the date portion (strip off the time) from the logdate column value and then try this:

Select [DateOnly(LogDate)], Count Distinct RefundId
From RefundProcessLog
Group By [DateOnly(LogDate)]

In Sql server, for e.g., the appropriate construct would be:

Select DateAdd(day, 0, DateDiff(day, 0, LogDate)), Count(Distinct RefundId)
From RefundProcessLog
Group By DateAdd(day, 0, DateDiff(day, 0, LogDate))

SELECT COUNT(RefundId), DateOnly(LogDate) LoggingDate
FROM RefundProcessLog
GROUP BY DateOnly(LogDate)

"DateOnly" is specific to your SQL database, which you haven't specified.

For SQL Server you could use DateAdd(dd,0, DateDiff(dd,0,LogDate)) for "DateOnly"

I like this approach in (MS SQL):

SELECT 
  Convert(char(8), LogDate, 112),
  count(distinct RefundId)
FROM RefundProcessing
GROUP BY Convert(char(8), LogDate, 112)