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Full Discussion: Difference between 2 dates
Top Forums Shell Programming and Scripting Difference between 2 dates Post 302832725 by vsachan on Monday 15th of July 2013 03:02:50 PM
Old 07-15-2013
Thanks Joyeg...I had the script to calculate the difference but it is taking too much time..Let me explain you the logic that is being used by me:
Code:
for i in `cat calc_file_list.txt`
do
  echo "$i" >> newfile.txt
  while read line
  do
    sh timediffcalc.sh $line
  done<"$i"
done

where timediffcalc.sh is:
Code:
x="$2 $3"
y="$4 $5"
z="$1"
year1=`echo $x|cut -d- -f1`
month1=`echo $x|cut -d- -f2`
day1=`echo $x|cut -d- -f3|cut -d" " -f1`
lhr1=`echo $x|cut -d: -f1|cut -d" " -f2`
lmin1=`echo $x|cut -d: -f2`
lsec1=`echo $x|cut -d: -f3`
year2=`echo $y|cut -d- -f1`
month2=`echo $y|cut -d- -f2`
day2=`echo $y|cut -d- -f3|cut -d" " -f1`
lhr2=`echo $y|cut -d: -f1|cut -d" " -f2`
lmin2=`echo $y|cut -d: -f2`
lsec2=`echo $y|cut -d: -f3`

yr1=$(((year1 - 1970)*365.25*24*60*60))
mn1=$(((month1*30.5)*86400))
dy1=$(((day1-1)*86400))
hr1=$(((lhr1)*3600))
min1=$(((lmin1)*60))
secs1=$((yr1+mn1+dy1+hr1+min1+lsec1))
yr2=$(((year2 - 1970)*365.25*24*60*60))
mn2=$(((month2*30.5)*86400))
dy2=$(((day2-1)*86400))
hr2=$(((lhr2)*3600))
min2=$(((lmin2)*60))
secs2=$((yr2+mn2+dy2+hr2+min2+lsec2))

diffval=$((secs2-secs1))
difmin=$((diffval/60))
intervalobt=$((difmin/$z))

if [ "$intervalobt" -ge 700 ]; then
  echo $x,$y,$z,$intervalobt >> newfile.txt
fi

The file calc_file_list.txt is a list of files:
Code:
condition_1
condition_2
condition_3
condition_4

sample data for condition_1
Code:
5 2013-07-13 23:25:00 2013-07-14 19:40:00
5 2013-07-13 23:25:00 2013-07-14 19:40:00
5 2013-07-12 23:50:00 2013-07-13 20:30:00

But this logic is taking too much time and that is not desired...is there any way we can optimize it....
Last edited by Scott; 07-15-2013 at 04:27 PM.. Reason: Code tags
 

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bup-margin

bup-margin(1)						      General Commands Manual						     bup-margin(1)

NAME

       bup-margin - figure out your deduplication safety margin

SYNOPSIS

       bup margin [options...]

DESCRIPTION

       bup margin  iterates  through  all  objects  in	your  bup repository, calculating the largest number of prefix bits shared between any two
       entries.  This number, n, identifies the longest subset of SHA-1 you could use and still encounter a collision between your object ids.

       For example, one system that was tested had a collection of 11 million objects (70 GB), and bup margin returned 45.  That  means  a  46-bit
       hash  would be sufficient to avoid all collisions among that set of objects; each object in that repository could be uniquely identified by
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       with far fewer objects.

       If you're paranoid about the possibility of SHA-1 collisions, you can monitor your repository by running bup margin occasionally to see	if
       you're getting dangerously close to 160 bits.

OPTIONS

       --predict
	      Guess  the offset into each index file where a particular object will appear, and report the maximum deviation of the correct answer
	      from the guess.  This is potentially useful for tuning an interpolation search algorithm.

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	      don't use .midx files, use only .idx files.  This is only really useful when used with --predict.

EXAMPLE

	      $ bup margin
	      Reading indexes: 100.00% (1612581/1612581), done.
	      40
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	      4.19338e+18 times larger is possible

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	      $ bup margin --predict
	      PackIdxList: using 1 index.
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SEE ALSO

       bup-midx(1), bup-save(1)

BUP

       Part of the bup(1) suite.

AUTHORS

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Bup unknown-															     bup-margin(1)
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