convert number of seconds to date

I have an integer  1060128571 which is the number of seconds since
1/1/1970. The number evaluates to 
6/08/2003 10:09:30

So my question is given the start date is always 1970 how do I
calculate the end date e.g 6/08/2003 10:09:30.

if I divide 1060128571/60/60/24  I can get the number of days etc.

If anyone has an easy solutions,please send it on


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1/28/2004 5:16:36 AM
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Hi Gordon,

1/1/1970 12:00 AM equates to 25569. Divide the number of 
seconds by 86400 (number of seconds in a day). Subtract 
that from 25569 and format the cell as 
m/d/yyyy h:mm:ss AM/PM.

Based on your values, I got an answer of 
8/6/2003 12:09:31 AM ????

Biff

>-----Original Message-----
>I have an integer  1060128571 which is the number of 
seconds since
>1/1/1970. The number evaluates to 
>6/08/2003 10:09:30
>
>So my question is given the start date is always 1970 how 
do I
>calculate the end date e.g 6/08/2003 10:09:30.
>
>if I divide 1060128571/60/60/24  I can get the number of 
days etc.
>
>If anyone has an easy solutions,please send it on
>
>
>---
>Message posted from http://www.ExcelForum.com/
>
>.
>
0
biffinpitt (3171)
1/28/2004 5:59:21 AM
Take a look here:

   http://mcgimpsey.com/excel/unixtoxltime.html

In article <gordon_conroy.10q7pm@excelforum-nospam.com>,
 gordon_conroy <<gordon_conroy.10q7pm@excelforum-nospam.com>> wrote:

> I have an integer  1060128571 which is the number of seconds since
> 1/1/1970. The number evaluates to 
> 6/08/2003 10:09:30
> 
> So my question is given the start date is always 1970 how do I
> calculate the end date e.g 6/08/2003 10:09:30.
> 
> if I divide 1060128571/60/60/24  I can get the number of days etc.
> 
> If anyone has an easy solutions,please send it on
0
jemcgimpsey1 (104)
1/28/2004 6:08:31 AM
> 1/1/1970 12:00 AM equates to 25569 

Note that the above works only for the 1900 date system. Better to use 
something like

   = DATE(1970,1,1) + A1/86400


In article <53d501c3e563$dfb240b0$a101280a@phx.gbl>,
 "Biff" <biffinpitt@comcast.net> wrote:

> 1/1/1970 12:00 AM equates to 25569. Divide the number of 
> seconds by 86400 (number of seconds in a day). Subtract 
> that from 25569 and format the cell as 
> m/d/yyyy h:mm:ss AM/PM.
> 
> Based on your values, I got an answer of 
> 8/6/2003 12:09:31 AM ????
0
jemcgimpsey1 (104)
1/28/2004 6:12:46 AM
OOPS! I'm having a bad day.

Don't subtract, *ADD*.

25569 + (seconds/86400)

Formatted as: m/d/yyyy h:mm:ss AM/PM

Biff

>-----Original Message-----
>Hi Gordon,
>
>1/1/1970 12:00 AM equates to 25569. Divide the number of 
>seconds by 86400 (number of seconds in a day). Subtract 
>that from 25569 and format the cell as 
>m/d/yyyy h:mm:ss AM/PM.
>
>Based on your values, I got an answer of 
>8/6/2003 12:09:31 AM ????
>
>Biff
>
>>-----Original Message-----
>>I have an integer  1060128571 which is the number of 
>seconds since
>>1/1/1970. The number evaluates to 
>>6/08/2003 10:09:30
>>
>>So my question is given the start date is always 1970 
how 
>do I
>>calculate the end date e.g 6/08/2003 10:09:30.
>>
>>if I divide 1060128571/60/60/24  I can get the number of 
>days etc.
>>
>>If anyone has an easy solutions,please send it on
>>
>>
>>---
>>Message posted from http://www.ExcelForum.com/
>>
>>.
>>
>.
>
0
biffinpitt (3171)
1/28/2004 6:13:55 AM
thanks for that guys,it worked perfect

as a side note the reason for the 08/06/2003 and 06/08/2003 is because
of US date format - placing months first.The 08/06/2003 is 8/Jun

Cheers again


---
Message posted from http://www.ExcelForum.com/

0
1/29/2004 6:45:40 AM
Reply:

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