Shell Programming and Scripting

I've gone through bunch of threads on time calculations but none of them helps on my problem

I've to get the time difference in HHMM format from following inputs

Input 1 :

01/08/2010 01:30
01/08/2010 03:20

Input 2 :

01/06/2010 22:00
01/07/2010 16:00

First input is easy but the second one is weird one...

Any idea? logic? anything please throw up and I will use it.

Rolling your own date/time calculation is almost almost almost never a good idea. Either get the input in epoch seconds in the first place, or install something meant for the job.

Thanks for your advice but it's not my choice....

Please see -->
https://www.unix.com/shell-programmin...ss-format.html

And note the following command -->

Code:

date +%s -d"2009-01-03"

Perhaps this might set you onto a path to solve your issue?

That only works on GNU date, which Solaris is extremely unlikely to have.

I don't suppose there's mysql on your system? It has very nice date-handling abilities.

Thanks Joey,

You are right Carona, it does not work in Solaris.

But everyone is welcome to give some ideas ... you never know anything could work....

Meanwhile i am working on my own algo. and making the script to calculate.. would be done by tomorrow.

I had a function to do this but I've been unable to find it. Basically it went like this:

1. Convert each time into minutes (ex: 01:30 => 90)
2. Compare the start and end dates.
2a. If they are the same, then subtract the start minute from end minute.
2a. If not, then calculate the remaining minutes of the start time and call another
function that will find out how many days have passed inbetween. Keep track and
multiply that number by the number of minutes in a day (60*24). Then add in
the minutes from the end time to get the total number of minutes that elapsed.
3. Finally convert the total number of minutes back to HH:MM form.

Sorry I don't have the code for this but hopefully this helps with your logic.