time a process within bash script

aubrey-calm2 · 06-14-2007, 01:40 PM

I'm trying to write a bash script that will copy files from old_server to new_server. I need to copy certain directories first (please don't ask why) and certain directories last.
I want to time each of these copies. I am currently using this configuration....

# First we sync the clock and record the time
/etc/rc.d/rc.clock sync
BEGIN_TIME="$(date +%A%l:%M%P)"

# Next we set the clock to zero and make that our begin time. Used for report
date 010100002000 >/dev/null 2>&1

# Begin the actual copy-over process
echo -e "\n\n"
echo Began copying ACSS to /$DEST/APPS/ACSS/
BEGIN_ACSS="$(date +%s)"
cp -rf $ACSS/ /$DEST/APPS/
END_ACSS="$(date +%s)"
echo -e "\n\n\n"

echo Began copying APPS/* without ACSS to /$DEST/APPS/
BEGIN_APPS="$(date +%s)"
cp -rf $APPS/*[!ACSS] /$DEST/APPS/
END_APPS="$(date +%s)"
echo -e "\n\n\n"

echo Began copying SOURCE/* without APPS or ACS to /$DEST/
BEGIN_SOURCE="$(date +%s)"
cp -rf $SOURCE/*[!APPS][!ACS] /$DEST
END_SOURCE="$(date +%s)"
echo -e "\n\n\n"
mkdir /$DEST/ACS

# Resync clock to reality
/etc/rc.d/rc.clock sync >/dev/null 2>&1
FINISH_TIME="$(date +%A%l:%M%P)"

# --Show time spent--
# first build the variables
ACSS_TOTAL="$(expr $END_ACSS - $BEGIN_ACSS)"
APSS_TOTAL="$(expr $END_APPS - $BEGIN_APPS)"
SOURCE_TOTAL="$(expr $END_SOURCE - $BEGIN_SOURCE)"
ACSS_TOTAL_MIN="$(expr $ACSS_TOTAL / 60)"
APPS_TOTAL_MIN="$(expr $APSS_TOTAL / 60)"
SOURCE_TOTAT_MIN="$(expr $SOURCE_TOTAL / 60)"

echo -e "\n\n\n\n\n\n"
echo This job started $BEGIN_TIME; echo
echo This job ended $FINISH_TIME; echo
echo Copying $ACSS took $ACSS_TOTAL_MIN minutes.; echo
echo Copying $APPS took $APSS_TOTAL_MIN minutes.; echo
echo Copying $SOURCE took $SOURCE_TOTAL_MIN minutes.; echo -e "\n\n"
echo This script saved you "$(expr $ACSS_TOTAL_MIN + $APPS_TOTAL_MIN + $SOURCE_
TOTAT_MIN)" minutes!

The reason that I set the clock to 12:00am is because the copy-over has to start usually around 11:30pm, and takes around 3 hours to complete. The problem is that during this time the clock for the system is wrong.

Is there any way to do this without using the system clock?

-=Aubrey=-

wjevans_7d1@yahoo.co · 06-14-2007, 02:28 PM

There are several ways to do this. I'm sure several people will be along to show you their favorite ways. Pay heed to each, and you'll learn some new stuff!

My favorite:

Code:

#!/bin/bash

# Instead of the sleep commands, substitute your cp commands.

cat > temp.sh <<EOD
/usr/bin/time -f '%e' -o temp1.dat sleep 5
/usr/bin/time -f '%e' -o temp2.dat sleep 10
/usr/bin/time -f '%e' -o temp3.dat sleep 15
EOD

chmod 700 temp.sh

/usr/bin/time -f '%e' -o temp0.dat ./temp.sh

echo first command took $(cat temp1.dat) seconds
echo second command took $(cat temp2.dat) seconds
echo third command took $(cat temp3.dat) seconds
echo the total is $(cat temp0.dat) seconds

rm temp.sh temp*.dat

Hope this helps.

aubrey-calm2 · 06-15-2007, 01:39 PM

That worked flawlessly. Thanks.