Command line: piping find results to rm

Question

You are actually piping rm's output to the input of find. What you want is to use the output of find as arguments to rm:

find -type f -name '*.sql' -mtime +15 | xargs rm

xargs is the command that "converts" its standard input into arguments of another program, or, as they more accurately put it on the man page,

build and execute command lines from standard input

Note that if file names can contain whitespace characters, you should correct for that:

find -type f -name '*.sql' -mtime +15 -print0 | xargs -0 rm

But actually, find has a shortcut for this: the -delete option:

find -type f -name '*.sql' -mtime +15 -delete

Please be aware of the following warnings in man find:

  Warnings:  Don't  forget that the find command line is evaluated
  as an expression, so putting -delete first will make find try to
  delete everything below the starting points you specified.  When
  testing a find command line that you later intend  to  use  with
  -delete,  you should explicitly specify -depth in order to avoid
  later surprises.  Because -delete  implies  -depth,  you  cannot
  usefully use -prune and -delete together.

_{P.S. Note that piping directly to rm isn't an option, because rm doesn't expect filenames on standard input. What you are currently doing is piping them backwards.}

find /usr/www/bar/htdocs -mtime +15 -exec rm {} \;

Will select files in /usr/www/bar/htdocs older than 15 days and remove them.

Another simpler method is to use locate command. Then, pipe the result to xargs.

For example,

locate file | xargs rm

Assuming you aren't in the directory containing the *.sql backup files:

find /usr/www2/bar/htdocs/foo/rsync/httpdocs/db_backups/*.sql -mtime +15 -exec rm -v {} \;

The -v option above is handy it will verbosely output which files are being deleted as they are removed.

I like to list the files that will be deleted first to be sure. E.g:

find /usr/www2/bar/htdocs/foo/rsync/httpdocs/db_backups/*.sql -mtime +15 -exec ls -lrth {} \;