Variables in shell script not working

[Jef Driesen]

Karl Auer wrote:
> On Sat, 2008-11-15 at 16:20 +0100, Jef Driesen wrote:
>> This is indeed the problem! At first, I only had the sed command,
>> which does accept a filename AND outputs to stdout. I simply didn't
>> think about this difference when adding the dos2unix command.
>>
>> Is it possible to have the pipe command in the variable, without cat 
>> interpreting both the pipe symbol "|" and the "dos2unix" as its
>> argument?
>>
>> SOURCE="cat $FILENAME | dos2unix"
> 
> Yes, that will work fine - but it doesn't run sed, and the output goes
> to stdout rather than being saved in the original file.
> 
> This will convert the filename and (if the conversion worked) run it
> through sed, sending the output to stdout:
> 
>    FILENAME=$1
>    dos2unix $FILENAME && cat $FILENAME | sed s/foo/bar/
> 
> That's basically what Carl was saying, too.
> 
> But what is it you actually want to achieve here? The above will send
> the sedded output to stdout after unixifying the specified file, but the
> sedded output will not be saved in the file.

What I want to achieve is process some data that is passed through stdin 
or a filename. Which source is used depends on whether a filename is 
passed on the commandline or not. That way I can run my script in two 
ways (similar to how you can use sed):

./myscript myfile
myprogram | ./myscript

Because sometimes my input is stored in a file and sometimes it is the 
output of another program.

The output should always go to stdout and when the script is used with a 
filename, the original should not be modified. So that excludes running 
dos2unix on it directly.