time it takes to complete script

[Ratzie]

I need to figure out how long it takes to complete a process.

Right now in my script I have...
echo "Such and such started at `date`"

I would also like it to at the end state
echo "It took ?? minutes to complete this process"

One more thing, I would also like to add...
The size of the file is ?? MB or GB. How do you convert an ls -l on a file?
Is ls -l in HP 11.00 showing it in kb, then I would times by 1024? for mb?

[Sridhar Bhaskarla · Accepted solution]

Hi,

If you want the time to be calculated within the script, you can use the built-in variable "SECONDS".

start=$SECONDS
your_script..
..
..
end=$SECONDS

(( TIME = ( $end - $seconds ) / 60 ))

echo "Time it took is $TIME minutes approximately"


Actually the size you get in ls -l is in bytes. You will need to devide it by 1048576 (1024*1024) to get it in MB.

SIZE=$(ls -al file|awk '{print $5/1048576}')
echo "size is $SIZE

TO calculate the size of a file, I would do the following

SIZE=$(du -sk file|awk '{print $1/1024}')
echo "Size of the file is $SIZE MB"

-Sri

[Sridhar Bhaskarla]

Hi (again),

I should have added. From externally you can use the command timex while running the script to get the total time it took.

timex your_script

real 0.02
user 0.01
sys 0.01

YOu should see something like the above. real is the real time it took. user is the time cpu spent to run the program in user mode and sys is the time cpu spent to run in kernel mode.

-Sri

[Sanjay Kumar Suri]

Hello

#ls -l show the size in bytes.

#du -ak can show the figures in Kbytes.

sks

[Steve Lewis]

This script will convert an ll into Kb/Mb/Gb as appropriate:

ll "$*" | awk '
function sv(val)
{
pv=sprintf("%3.3fKb",val/1024.0);
if (val>=1048576.0) pv=sprintf("%3.3fMb",val/1048576.0);
if (val>=1073741824.0) pv=sprintf("%3.3fGb",val/1073741824.0);
return pv;
}
{
if($1~"d.")
printf("%10s %-5d %9s %9s %3s %-2s %5s %s\n",$1,$2,$3,$4,$6,$7,$8,$9);
else printf("%10s %-5d %9s %9s %10s %3s %-2s %5s %s\n",$1,$2,$3,$4,sv($5),$6,$7,$8,$9);
}'