Chris Moretti Chris Moretti - 8 months ago 99
Linux Question

Escaping quotes in bash (Embedded awk)

I have a complex command I am passing via ssh to a remote server. I am trying to unzip a file and then change its naming structure and extension in a second ssh command. The command I have is:

ssh [email protected] "gzip -d /tmp/file.out-20171119.gz; echo file* | awk -F'[.-]' '{print $1$3".log"}'"


Obviously the " around the .log portion of the print statement are failing me. The idea is that I would strip the .out portion from the filename and end up with file20171119.log as an ending result. I am just a bit confused on the syntax or on how to escape that properly so bash interprets the .log appropriately.

Answer Source

The easiest way to deal with this problem is to avoid it. Don't bother trying to escape your script to go on a command line: Pass it on stdin instead.

ssh [email protected] bash -s <<'EOF'
  gzip -d /tmp/file.out-20171119.gz
  # note that (particularly w/o a cd /tmp) this doesn't do anything at all related to the
  # line above; thus, probably buggy as given in the original question.
  echo file* | awk -F'[.-]' '{print $1$3".log"}'
EOF

A quoted heredoc -- one with <<'EOF' or <<\EOF instead of <<EOF -- is passed literally, without any shell expansions; thus, $1 or $3 will not be replaced by the calling shell as they would with an unquoted heredoc.


If you don't want to go the avoidance route, you can have the shell do the quoting for you itself. For example:

external_function() {
  gzip -d /tmp/file.out-20171119.gz
  echo file* | awk -F'[.-]' '{print $1$3".log"}'
}

ssh [email protected] "$(declare -f external_function); external_function"

declare -f prints a definition of a function. Putting that function literally into your SSH command ensures that it's run remotely.

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