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 stumbled on a similar problem and found a solution which I think is nicer and cleaner than those presented so far.

The syntax for xargs that I have ended with would be (for your example):

xargs -I X echo --file=X

with a full command line being:

my-program $(cat input.txt | xargs -I X echo --file=X)

which will work as if

my-program --file=a.txt --file=b.txt --file=c.txt

was done (providing input.txt contains data from your example).

---

Actually, in my case I needed to first find the files and also needed them sorted so my command line looks like this:

my-program $(find base/path -name "some*pattern" -print0 | sort -z | xargs -0 -I X echo --files=X)

Few details that might not be clear (they were not for me):

• some*pattern must be quoted since otherwise shell would expand it before passing to find.
• -print0, then -z and finally -0 use null-separation to ensure proper handling of files with spaces or other wired names.

Note however that I didn't test it deeply yet. Though it seems to be working.