Use the output of a command as input of the next command

So I call this开发者_如何学Go PHP script from the command line:

/usr/bin/php /var/www/bims/index.php "projects/output"

and its output is:

file1 file2 file3

What I would like to do is get this output and feed to the "rm" command but I think im not doing it right:

/usr/bin/php /var/www/bims/index.php "projects/output" | rm 

My goal is to delete whatever file names the PHP script outputs. What should be the proper way to do this?

Thanks!

/usr/bin/php /var/www/bims/index.php "projects/output" | xargs rm

Simplest solution:

rm `/usr/bin/php /var/www/bims/index.php "projects/output"`

What is between the backticks (`` ) is run and the output is passed as argument torm`.

you can try xargs

/usr/bin/php /var/www/bims/index.php "projects/output" | xargs rm 

or just simply use a loop

/usr/bin/php /var/www/bims/index.php "projects/output" | while read -r out
do
  rm $out
done

I guess this could help>>

grep -n "searchstring" filename | awk 'BEGIN { FS = " " };{print $1}'