G'day Derek, * [EMAIL PROTECTED] <[EMAIL PROTECTED]> [050419 11:31]: > Stewart, > 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 > do > echo $i > done > > /psofthr/hr88prd/intf/aflac/inbound/filename1 > /psofthr/hr88prd/intf/aflac/inbound/filename2 > > so I am ok there,
What shell are you using?
> 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
> /psofthr/hr88prd/intf/aflac/inbound/OHIO_HEALTH.gpg.319548
> /psofthr/hr88prd/intf/aflac/inbound/OHIO_HEALTH_20050324.TXT.pgp
> gpg --passphrase-fd 0 --decrypt --output
> /psofthr/hr88prd/intf/aflac/inbound/OHIO_HEALTH.gpg.319548
> /psofthr/hr88prd/intf/aflac/inbound/OHIO_HEALTH_20050408.TXT.pgp
>
> 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,
At the start of your program put
i=0
Inside the loop put something like
i=`expr $i + 1`
and then use ${i} as the unique filename addition.
However...
This is _not_ what the error was that you sent in your initial email:
> and the error I get is
> Sorry no terminal at all requested - cat get input.
The above fix won't fix this error unless I'm missing the point
entirely.
Cheers,
S.
P.S. Please don't bother replying to me and the mailing list...
pgpqJKRit9DDg.pgp
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