multiple files for version below 1.2.5

Tue Apr 19 18:31:25 CEST 2005

I am trying to decrypt more than one file if ls -la |wc -l /dir is -gt one.
Thanks for the tips, but my $var variable actually contains both file names
like so:

      for i in $var
            echo $i


so I am ok there, but when I placed the echo infront of the gpg string I
found that the same PID is being used.

gpg --passphrase-fd 0 --decrypt --output
gpg --passphrase-fd 0 --decrypt --output

Finally, in the decryption process you need two uniq filenames b/c there
are two decryption processes which is why I used $$.

      gpg $p $de $outp $dec.$$ $i

Anyone more ideas?  thank you,

Derek B. Smith
OhioHealth IT
UNIX / TSM / EDM Teams

             "Stewart V.                                                   
             <swright at physics.                                          To 
   >          gnupg-users at               
             Sent by:                                                   cc 
             es at                                          Subject 
                                       Re: multiple files for version      
                                       below 1.2.5                         
             04/19/2005 11:47                                              


* DBSMITH at <DBSMITH at> [050419 10:37]:
> I am trying to decrypt 2 files on a UNIX machine with:
> var=/data/files
> file=file_that_has_my_passphrase
> p='- -passphrase-fd-0'
> outp='- -output'
> de='- -decrypt'
> dec=file
> for i in $var
> do
>       cat $file |gpg $p $de $outp $dec.$$ $i
>       sleep 1
> done
> and the error I get is
> Sorry no terminal at all requested - cat get input.

Hint 1: Use echo to help you debug - i.e.
    echo cat $file
    echo gpg ......
  That way you can see what your script is _trying_ to do, not what
  you want it to do.

  What are you trying to do?  Decrypt all the files in the directory
  "/data/files" ?  This script won't do that.  You are passing
  "/data/files" to GnuPG, not a list of the files IN that directory.
  (You will see this when you use the echo hint above.)

  If you want all of the files inside /data/files, set
  var=/data/files/* and that should work.

Hint 2: I find it useful (others will disagree) to put brackets around
  my variables to let the shell be sure what I want it to do...
  So the line:
    for i in $var
  would become
    for i in ${var}
  Then there is no confusion when you want to do something like
  However that is an aside.



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