Perl's diamond operator: can it be done in bash?

Question

Is there an idiomatic way to simulate Perl's diamond operator in bash? With the diamond operator,

script.sh | ...

reads stdin for its input and

script.sh file1 file2 | ...

reads file1 and file2 for its input.

One other constraint is that I want to use the stdin in script.sh for something else other than input to my own script. The below code does what I want for the file1 file2 ... case above, but not for data provided on stdin.

command - $@ <<EOF
some_code_for_first_argument_of_command_here
EOF

I'd prefer a Bash solution but any Unix shell is OK.

Edit: for clarification, here is the content of script.sh:

#!/bin/bash
command - $@ <<EOF
some_code_for_first_argument_of_command_here
EOF

I want this to work the way the diamond operator would work in Perl, but it only handles filenames-as-arguments right now.

Edit 2: I can't do anything that goes

cat XXX | command

because the stdin for command is not the user's data. The stdin for command is my data in the here-doc. I would like the user data to come in on the stdin of my script, but it can't be the stdin of the call to command inside my script.

Answer 1

Kind of cheezy, but how about

cat file1 file2 | script.sh

Answer 2

The Perl diamond operator essentially loops across all the command line arguments, treating each as a filename. It opens each file and reads them line-by-line. Here's some bash code that will do approximately the same.

for f in "$@"
do
   # Do something with $f, such as...
   cat $f | command1 | command2
   -or-
   command1 < $f
   -or-
   # Read $f line-by-line
   cat $f | while read line_from_f
   do
      # Do stuff with $line_from_f
   done
done

Answer 3

Also a little cheezy, but how about this:

if [[ $# -eq 0 ]]
then
  # read from stdin
else
  # read from $* (args)
fi

If you need to read and process line-by-line (which is likely) and don't want to copy/paste the same code twice (which is likely), define a function in your script and just pass the lines one-by-one to this function, and process them in said function.

Answer 4

Sure, this is totally doable:

#!/bin/bash
cat $@ | some_command_goes_here

Users can then call your script with no arguments (or '-') to read from stdin, or multiple files, all of which will be read.

If you want to process the contents of those files (say, line-by-line), you could do something like this:

for line in $(cat $@); do
    echo "I read: $line"
done

Edit: Changed $* to $@ to handle spaces in filenames, thanks to a helpful comment.

Answer 5

Why not use `cat @* ` in the script? For example:

x=`cat $*`
echo $x

Answer 6

You want to take the first argument and do something with it, and then either read from any files specified or stdin if no files?

Personally, I'd suggest using getopt to indicate arguments using the "-a value" syntax to help disambiguate, but that's just me. Here's how I'd do it in bash without getopts:

firstarg=${1?:usage: $0 arg [file1 .. fileN]}
shift
typeset -a files
if [[ ${#@} -gt 0 ]]
then
  files=( "$@" )
else
  files=( "/dev/stdin" )
fi
for file in "${files[@]}"
do
  whatever_you_want < "$file"
done

The ?: operator will die if there are no args specified, since you seem to want at least one arg either way. After grabbing that, shift the args over by one, and then either use the remaining args as your file list, or the bash special filehandle "/dev/stdin" if there were no other args.

I think that the "if no files are specified, use /dev/stdin - otherwise use the files on the command line" piece is probably what you're looking for, but the rest of the code is at least useful for context.

Answer 7

I am (like everyone else, it seems) a bit confused about exactly what the goal is here, so I'll give three possible answers that may cover what you actually want. First, the relatively simple goal of getting the script to read from either a list of files (supplied on the command line) or from its regular stdin:

if [ $# -gt 0 ]; then
  exec < <(cat "$@")
fi
# From this point on, the script's stdin is redirected from the files
# (if any) supplied on the command line

Note: the double-quoted use of $@ is the best way to avoid problems with funny characters (e.g. spaces) in filenames -- $* and unquoted $@ both mess this up. The <() trick I'm using here is a bash-only feature; it fires off cat in the background to feed data from files supplied on the command line, and then we use exec to replace the script's stdin with the output from cat.

...but that doesn't seem to be what you actually want. What you seem to really want is to pass the supplied filenames or the script's stdin as arguments to a command inside the script. This requires sort of the opposite process: converting the script's stdin into a file (actually a named pipe) whose name can be passed to the command. Like this:

if [[ $# -gt 0 ]]; then
  command "$@" <<EOF
here-doc goes here
EOF
else
  command <(cat) <<EOF
here-doc goes here
EOF
fi

This uses <() to launder the script's stdin through cat to a named pipe, which is then passed to command as an argument. Meanwhile, command's stdin is taken from the here-doc.

Now, I think that's what you want to do, but it's not quite what you've asked for, which is to both redirect the script's stdin from the supplied files and pass stdin to the command inside the script. This can be done by combining the above techniques:

if [ $# -gt 0 ]; then
  exec < <(cat "$@")
fi

command <(cat) <<EOF
here-doc goes here
EOF

...although I can't think why you'd actually want to do this.