I am trying to run a perl script on a set of files and I guess I am doing something wrong.
On one file the command is:
process.pl $infile.in $outfile.out
that works just fine for one file

When I try to run it on a set of files using this command:
for i in $(ls -1 *.sgm); do cat $i | utf82buck_linux.pl $(basename $i) $(basename $i).out; done
the output files are empty, no processing is done.

Please let me know how to fix this command

thanks

Go to any directory and do 'ls -l'. What do you see?

The filename is the last field in the output. If you want your script to work with 'ls -l' you will have to grep the filename from the output.

Or, make it easy on yourself and drop the '-l' from the command so that ls only shows the filename.

I tried that but the output files are still empty
Any other suggestions?
thanks
Rita

Hi.

I changed your script slightly to show you a debugging technique for problems like this. Essentially it is using echo to print what would have happened. This allows you to see the problem(s).

You may need to change this because it wasn't clear to me exactly what you wanted:
Which results in:
So it looks like the filenames (mine) are getting to the loop. I omitted the cat, because it seemed unlikely that you needed that.

What do you see for the commands? Did you really want $infile.in?, does the perl script really process two arguments, one for input and one for output? and so on ... cheers, makyo

I tried to omit cat by running:
for i in $(ls *.in); do process.pl $(basename $i) $(basename $i).out; done

but I still get the output files (*.out) empty

All I want to do is run :
process.pl $infile.in $outfile.out

on a set of files with the extention *.in
Thanks for the debugging commands, i was not sure how to use them.
I tried to explain what I want to do with the command, anything that looks incorrect in it.
thanks again
Rita

Hi, Rita.

Where do the files $infile.in come from? You do not have the ".in" extension in your script ... cheers, makyo

I thought the variable $i would have the filename with the extention .in
Am I wrong?
Rita

I tried to add the extention by using;
for i in $(ls *.in); do process.pl $(basename $i .in) $(basename $i).out; done

The output files are still empty.
I need some help here. I cannot see what s wrong with the command

thanks

The set of shell commands you have is very close. Here are the changes I would suggest.
You should be executing these in the directory where your .in files are located. There is no need to use the basename command since when you execute ls *.in in the current directory all you get is the filenames. Is process.pl in a directory that is in your $PATH?

With these commands an input file with the name stuff.in will have an output filename of stuff.in.out. If you want the filename to be stuff.out substitute ${i%.*}.out for $i.out.

If you are still not getting output files after this, maybe we should be looking at process.pl.

Bill

Hi Bill,
Thanks for your reply. The script you sent me might be working except that I get two error in processing a couple of file. THe error doesn't give me the name of the input file that gives the problem.
Is there a way to print the name of the input files being processed so I know which one causes the problem.
Thanks again
Rita

Do any of your input files have spaces in their names? The script could interpret these incorrectly. Try putting double quotes around "$i" and "$i.out".

If you want to see what files are being processed, replace process.pl with the echo command as Makyo suggested. This will give you a listing of what files are being fed into process.pl. Perhaps from there you can determine the input files causing the problem.

Bill

Bill, the command you sent me actually works. for the ls *.i, I was giving the path to a different directory. The output files were created in that directory, not the current directory.
I also figure what files give the errors!!
Thanks for your help.
Rita