Help with subtracting time values?

[cyberkid]

Hi All,

Can some one help.
How do I go about writing a script to subtract 1 time value from another?
ie:

16:41:39 -
15:23:35
-----------
result =?


Thanks

Newbie.

[Jman]

I'm not experienced in script programming, but the standard date program may help. Give it the date, like this:

date -d 16:41:39 +%s

and it will output the date in seconds. In other words, it outputs the current date at that time in seconds.

Hope this gets you started.

[zyrotec]

This is an example, use it however you want. I am not sure how much you know so I will not bore you explaining what I did, unless you ask.

William

#variables
#ft% are from the first time
#st% are from the second time
#dt% difference time

#cut out the hours minutes seconds

fth=`echo "$ft" | cut -c1-2`
ftm=`echo "$ft" | cut -c4-5`
fts=`echo "$ft" | cut -c7-8`
sth=`echo "$st" | cut -c1-2`
stm=`echo "$st" | cut -c4-5`
sts=`echo "$st" | cut -c7-8`



# do the math for seconds

dts=`expr $sts - $fts`
if [ $dts -lt 0 ]
then
	dts=`expr $dts + 60`
	ftm=`expr $ftm + 1`
fi
for val in $dts
do
	if [ $val -lt 10 ]
	then
		dts=`echo "0$val"`
	else
		dts=`echo $val`
	fi
done



# do the math for minutes

dtm=`expr $stm - $ftm`
if [ $dtm -lt 0 ]
then
	dtm=`expr $dtm + 60`
	fth=`expr $fth + 1`
fi
for val in $dtm
do
	if [ $val -lt 10 ]
	then
		dtm=`echo "0$val"`
	else
		dtm=`echo $val`
	fi
done



# do the math for hours

dth=`expr $sth - $fth`



# put dtime back together

difftime=`echo $dth:$dtm:$dts`

[cyberkid]

@Jman

Thanks, but I'm using a minimal linux distro & it doesn't have this option.
Plus converting back, when finished with the calcs is a bit of a worry.

@zyrotec

Many thanks,
This is great saves me a lot of heart-ache & head scratching. It'll get me well on my way.


CK. :wink: