Assume, I do archive several files with this functions:

gen_password () 
    gpg --gen-random 1 "$1" | perl -ne'
        print $_, "\n"'

archive () 
    PASSWORD=$(gen_password 32)
    7za a -p"$PASSWORD" -mhe -- "$ARCHIVE_NAME" "$@"
    echo "Created 7z archive with password '$PASSWORD'"

This works well and I tried to upload encrypted archive on file sharing server.

So there is the script that uploads content of file to the server(source):

    if [ $# -eq 0 ]; then echo "No arguments specified. Usage:\necho transfer /tmp/test.md\ncat /tmp/test.md | transfer test.md"; return 1; fi 
    tmpfile=$( mktemp -t transferXXX ); if tty -s; then basefile=$(basename "$1" | sed -e 's/[^a-zA-Z0-9._-]/-/g'); curl --progress-bar --upload-file "$1" "https://transfer.sh/$basefile" >> $tmpfile; else curl --progress-bar --upload-file "-" "https://transfer.sh/$1" >> $tmpfile ; fi; cat $tmpfile; rm -f $tmpfile;

So I'm trying to pipe the encrypted archive in naive way:

archive 1.rar pass.tar.gz d7432.png foo.7z | upload

But there is one problem - encrypted archive is unreacheable for upload and command exiting with no result.

So, the question is: how should I pipe the file to have it uploaded correctly?

  • it appears to me that your archive function is creating a filename (1.rar for example) and echos a string "Created 7z..."; that echoed string is the only thing that upload will see on its stdin. Maybe something like archive 1.rar ... && upload 1.rar ? – Jeff Schaller Feb 17 '16 at 20:21
  • @JeffSchaller seems like you're right. Is it possible to somehow pipe only name of resultiong foo.7z to upload function without deleting echo? – im_infamous Feb 17 '16 at 20:27
  • echo other stuff >&2; echo foo.7z – Jeff Schaller Feb 17 '16 at 20:29
  • @JeffSchaller unfortunately, still getting same result as initially. Seems like this functions could not be modularized(well, at least with my shell scripting skills) and I should consider upload handle the encryption too – im_infamous Feb 17 '16 at 20:38
  • well, upload() is getting stdin from your pipeline, but it seems to be expecting a parameter; either pass it a parameter or have it read the filename from stdin – Jeff Schaller Feb 17 '16 at 20:52

Since your upload() function is expecting a parameter ($1) to use as the archived filename, pass it along in your commandline:

archive foo.7z 1.rar pass.tar.gz d7432.png && upload foo.7z

If foo.7z is a variable parameter for archive() as well, simply pass the same variable to upload():

archive $archivename 1.rar pass.tar.gz d7432.png && upload $archivename

I would recommend the && glue, as you probably don't want to try to upload the archive file if the archive() function did not succeed.

Here is sample function for .bashrc:

    archive "$ARCHIVE_NAME" "$@" && upload "$ARCHIVE_NAME"
  • Seems like this works! Thanks a lot for your time! – im_infamous Feb 17 '16 at 21:41

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