[SOLVED] Using Awk to pass to Bash

Issue

Updated no longer using sed

I am trying to write a script that will process certain files types that have changed in a git branch. For instance, if someone changes a file with the ".text" extension, I want to pass that file name to some sort of bash script. I also need to remove part of the file name before passing it to the bash script.

I know that if I do the following command I can get the list of git files that have changed in a branch

git diff --name-only master

Combining that with grep, I can only retrieve the files with the special extension

git diff --name-only master | grep .text

Now here is the part I cannot get to work with command line, I want to to pass in each file individually to my bash script. The below only executes echo once with all the arguments. Is this possible?

git diff --name-only master | grep .text | awk -F"folder-name/" '{print $2} | xargs -t echo

Solution

You've asked sed not to print anything by default, and you don't print anything explicitly. Here's man sed:

   -n, --quiet, --silent
          suppress automatic printing of pattern space

If you want automatic printing of lines, remove -n.

Update:

Your awk attempt is missing a single quote:

awk -F"folder-name/" '{print $2} 
                                ^-- Here

If you want to call your script with only a single argument each time, you can use xargs -n 1. Here's man xargs:

   -n max-args, --max-args=max-args
          Use  at  most  max-args  arguments per command line.  Fewer than
          max-args arguments will be used if the size (see the -s  option)
          is  exceeded, unless the -x option is given, in which case xargs
          will exit.

Answered By - that other guy