shell script to auto process ten random files and generate logs

novice82 · 10-05-2009, 05:35 AM

Hello member's

I'm learning to script in the ksh environment on a Solaris Box.

I have 10 files in a directory that I need to pass, as input to a batchjob one by one. lets say, the files are named as follows;
abcd.txt ; efgh.bat ; wxyz.temp etc. (random filenames with varied extensions ).

How do I go about achieving the following :

I want my batch script to take one file at a time from the current directory and process it, redirect the console output to a logfile (log1.txt, log2.txt etc )

./batch_script.ksh > log1.txt -- likewise, create 10 log files and exit

for simplicity sake, contents of the script:

Code:

#! /bin/ksh

mv abcd.txt ./temp/
echo "test script"

### End of SCript ####

regards,

Kris

zhjim · 10-05-2009, 06:14 AM

I'd use a wrapper and do something like this

for i in $(ls); do
./$i
done

That should run every file in the current directory

catkin · 10-05-2009, 06:20 AM

Or maybe

Code:

for i in $(ls); do
      batchjob $i > log$i.txt 2>&1 
done

It will not work if you have any spaces in the file names though.

novice82 · 10-05-2009, 07:50 AM

Quote:

Originally Posted by catkin

Or maybe

Code:

for i in $(ls); do
      batchjob $i > log$i.txt 2>&1 
done

It will not work if you have any spaces in the file names though.

Thanks Catkin.

That was exactly what I was looking for.

Kris

konsolebox · 10-05-2009, 08:08 AM

concept:

Code:

FILES=(<put 10 files here>)
TARGETFILE="${FILES[$(( RANDOM % 10 ))]}"
: do something with "$TARGETFILE"