Shell Programming and Scripting

Hi,

I am passing the file filename to a program and I have to use the "<" sign:
program < filename

Inside the program I would like to assign the string "filename" to a variable var.
I can't do that with "var=$(echo $1)" because of the "<" sign.
Removing the "<" sign would work for echo, but not for the rest of the program.

Can someone help me sorting this out, using echo or whatever command is suitable?

Cheers

What is wrong with:

Code:

var="$1"

thanks for your reply
Using var="$1" I get an empty line when trying to display the variable (echo $var)
It works if I run the program with:
program filename
but as I said I have to use a "<" sign

if "$1" has some data in it - assume it is not unset - then var="$1" places data in to the variable "var". Okay?

You seem to want to be able to echo to the screen, not use echo to write to a variable from a child process - which is what I first gave you as a workaround.

Code:

echo "$1" > /dev/tty
# or
echo "$var" > /dev/tty

This writes to the console.
Is this what you want?

It appears that you want the contents of file name to be the stdin of the program (the < redirects the stdin of program and feeds the contents of the file into it), and you also want the name of the file to be a parameter available in program.

assuming that program expects a filename as the first parameter then

Code:

program filename <filename

should achieve what you want

The problem ist that $n (for n=1,2,...) denotes the first (second, ...) positional argument. In a construct like "program < file" "file" is not an argument but the "< file" creates an input stream consisting of the content of "file". A program presented with input this way will only recognize some input to its <stdin> but will not know where this is coming from. Otherwise one would have to make (different) provisions for the call "program < file" and "program < /dev/somedevice" or "program < <userinput>". But, contrary, the difference in working with a device, a file or whatever is taken care of by the operating system itself. It will take all these different data sources and turn them into "streams". Note, btw., that you can treat a file like a "clotted datastream" and a datastream like a file. This is one of the biggest avantages of UNIX designwise IMHO.

To the problem of the threadstarter:

You can always solve problems by adding another layer of indirection. ;-))

Seriously: create a small script which does only call the program itself. This script gets a positional parameter which would be the input files name:

Code:

#!/bin/ksh

fInput="$1"

print - "The input file is $fInput"

program < "$fInput"

exit $?

If your program awaits positional parameters too you have to modify this wrapper script a bit:

Code:

#!/bin/ksh

typeset fInput="$1"
typeset chOtherParms=""

shift
chOtherParms="$*"

print - "The input file is $fInput"

program $chOtherParms < "$fInput"

exit $?

If "program" is another script it would be even easier to modify it to accept an additional parameter "inputfile" and use it instead of using a pipeline. In this case post your script and we'll see how to do it if you can't do it by yourself.

I hope this helps.

bakunin

I am afraid I haven't given enough details about my case.
I will try to be more clear posting what I am doing.
I want to add the time information contained in the filename inside the file using the command:
sh AddTim < filename-0.5_s.dat > filename-0.5_s.out
where AddTim is:
#!/bin/sh
time=$(echo $1 | cut -d "-" -f2 | cut -d "_" -f1)
sed "s/X/X, TIME=$time/"

Unfortunately, the resulting time information is blank!