xargs with multiple arguments

Question

I have a source input, input.txt

a.txt
b.txt
c.txt

I want to feed these input into a program as the following:

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Answer 1

I was looking for a solution for this exact problem and came to the conclution of coding a script in the midle.

to transform the standard output for the next example use the -n '\n' delimeter

example:

 user@mybox:~$ echo "file1.txt file2.txt" | xargs -n1 ScriptInTheMiddle.sh

 inside the ScriptInTheMidle.sh:
 !#/bin/bash
 var1=`echo $1 | cut -d ' ' -f1 `
 var2=`echo $1 | cut -d ' ' -f2 `
 myprogram  "--file1="$var1 "--file2="$var2 

For this solution to work you need to have a space between those arguments file1.txt and file2.txt, or whatever delimeter you choose, one more thing, inside the script make sure you check -f1 and -f2 as they mean "take the first word and take the second word" depending on the first delimeter's position found (delimeters could be ' ' ';' '.' whatever you wish between single quotes . Add as many parameters as you wish.

Problem solved using xargs, cut , and some bash scripting.

Cheers!

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