Run command with args multiple time

Question

I got a file with some params of a script:

foo1 a b c
foo2 a b c
foo3 a b c

I need to run a script once for each line in this file passing the line as script arg, so it should do:

./MyScript.sh foo1 a b c
./MyScript.sh foo2 a b c
./MyScript.sh foo3 a b c

How can I achieve this using Bash?

Answer 1

xargs command, which is created for running a command with arguments that are read from stdin has a --arg-file parameter, which allows reading arguments from file. In combination with-L1 flag, it will read your arguments file line by line, and for each line execute the command. Here's an example:

$ cat args.txt
one two three
four file six

$ xargs -L1 --arg-file=args.txt echo                       
one two three
four file six

Replace echo with your script.

Alternatively, you can always redirect the file to be read by xargs from stdin stream like so:

$ xargs -L1  echo < args.txt                                                                                             
one two three
four file six

Answer 2

Using a while loop:

#!/bin/bash
while IFS= read -r line; do
   # if the line read is empty 
   # go to the next line. Skips empty lines
   if [ -z "${line}" ]
   then
       continue
   fi
  /path/to/MyScript.sh $line
done < "$1"

Then call this script anything.sh and run it like this:

anything.sh /path/to/file/with/foo

Remember to make both anything.sh and MyScript.sh executable