help using xargs to pass mulitiple filenames to shell script

Question

Can someone show me to use xargs properly? Or if not xargs, what unix command should I use?

I basically want to input more than (1) file name for input <localfile>, third input parameter.

For example:

1. use `find` to get list of files
2. use each filename as input to shell script

Usage of shell script:

test.sh <localdir> <localfile> <projectname>

My attempt, but not working:

find /share1/test -name '*.dat' | xargs ./test.sh /staging/data/project/ '{}' projectZ \;

Edit: After some input from everybody and trying -exec, I am finding that my <localfile> filename input with find is also giving me the full path. /path/filename.dat instead of filename.dat. Is there a way to get the basename from find? I think this will have to be a separate question.

Answer 1

I'd just use find -exec here:

% find /share1/test -name '*.dat' -exec ./test.sh /staging/data/project/ {} projectZ \;

This will invoke ./test.sh with your three arguments once for each .dat file under /share1/test.

xargs would pack up all of these filenames and pass them into one invocation of ./test.sh, which doesn't look like your desired behaviour.

Answer 2

If you want to execute the shell script for each file (as opposed to execute in only once on the whole list of files), you may want to use find -exec:

find /share1/test -name '*.dat' -exec ./test.sh /staging/data/project/ '{}' projectZ \;

Remember:

• find -exec is for when you want to run a command on one file, for each file.
• xargs instead runs a command only once, using all the files as arguments.