Consider you have a file with some sort of terminal command. How might each line be executed? Can you pipe the output of more into an eval?

%> more ./foo.txt


After some help/guidance from the comments, I think I should also make note that it's been a while since I've done any shell scripting, so while I say eval I may mean echo or some other way of executing code.

If foo.txt contains various lines of code, but I want to execute the line that contains echo 'bar bar', I'm looking for a way to do that from, let's say a grep. Something logically similar to:

grep echo foo.txt | xargs echo

If you just need to evaluate every line of a file, you don't need a complicated eval or stdout redirection. Two easy options:

  • Source the file (source filename.sh or . filename.sh).
  • Make the file executable and run it (chmod +x filename.sh; ./filename.sh)

If you really need to eval each line of a file in a loop, do it with while:

while IFS= read -r line; do eval "$line"; done < filename.sh

You can also pipe the output of any command to while:

grep foo filename.sh | while IFS= read -r line; do eval "$line"; done

If you need to pass something to source (or .), which expects a file name as an argument, you can use process substitution:

source <( grep foo filename.sh )
  • I like, but is there a way to do this w/o using bash? I know I tagged it, so I'm just curious, since I don't always use it. – vol7ron Apr 15 '13 at 14:38
  • source isn't Bash-specific and should work in most other shells (Bourne Shell, (t)csh) as well. – slhck Apr 15 '13 at 14:48
  • Right now, source would work but I like to think ahead in the case that there are lines I wan't to strip out and possibly use the output of grep – vol7ron Apr 15 '13 at 14:59
  • 2
    Technically, source is a non-standard synonym for .. – chepner Apr 15 '13 at 15:08
  • @chepner: right, but can you pump more/grep into an eval (guessing echo) from the terminal command line? Perhaps into find -exec? – vol7ron Apr 15 '13 at 17:27

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