On 3 Dec 2002, at 10:45, Peter Abilla wrote:
> Column One Column Two
> 1999-09-17 16:30:18 1999-09-18 13:30:18
>
> I want to calculate the minutes like
>
> (Column Two - Column One) = Total Minutes
Something like ( UNIX_TIMESTAMP(column2) - UNIX_TIMESTAMP(column1) ) / 60 ?
> K
Try this:
select sec_to_time(unix_timestamp(column_two) - unix_timestamp(column_one))
Should give you the elapsed time in hh:mm:ss format.
--jeff
- Original Message -
From: "Peter Abilla" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Tuesday, December 03, 2002 8:45 AM
Subject: DateT
On 12/3/02 10:45 AM, Peter Abilla ([EMAIL PROTECTED]) wrote:
> Suppose I have a datetime in the following format:
>
> Column One Column Two
> 1999-09-17 16:30:18 1999-09-18 13:30:18
>
> I want to calculate the minutes like
>
> (Column Two - Column One) = Total Minutes
>
> I've
On Tue, 03 Dec 2002 10:45:57 -0600, Peter Abilla wrote:
>(Column Two - Column One) = Total Minutes
Assuming this doesn't work :) perhaps
select UNIX_TIMESTAMP(col2) - UNIX_TIMESTAMP(col1) as TimeDiff
This should give you an answer in seconds. [sql]
- Steve Yates
- File not found. S
Peter,
I would do something like this:
(UNIX_TIMESTAMP(Column Two) - UNIX_TIMESTAMP(Column One))/3600
to give you the hoursnot sure about the minutes, but this should get
you going!
HTH,
Cory
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On Tue, 2002-12-03 at 10:45, Peter Abilla wrote:
> Suppose I h
From: Peter Abilla <[EMAIL PROTECTED]>
> (Column Two - Column One) = Total Minutes
This is quite a hack, and there's probably a better way. But it works:
SELECT ROUND( (unix_timestamp(column_one) - unix_timestamp(column_two)) /60 )
as my_minutes;
> I've scoured the mysql site and haven't f
