Variable assignment syntax error in shell [closed]

Question

VALUE=${VAR| |o}

I found this variable assignment in shell script. I can't understand why || sign is used here.

EDIT

Code given in a comment:

#!/bin/bash
source /opt/cpf/bin/cpf_logging_helper.sh
STATE=$1
DNS_MASTER=${TEST.DNS.DNS_MASTER:-o}
service status dns > /dev/null
rval=$?
if [ $rval -eq 0 ]
then
    if [ $DNS_MASTER == `hostname -s` ]
    then
        echo "1"
    else
        echo "2"
    fi
fi

Answer 1

It seems that the original coder meant to use the value of $VAR or o if it is unset. In bash this can be achieved by using:

VALUE=${VAR:-o}

See also man bash and search for Parameter Expansion or :-

Answer 2

The structure: VALUE=${VAR| |o} is invalid in dash, bash, ksh and zsh.

If the | | is actually a / / then it may(?) make sense in bash.

It would mean: replace the first occurrence of an space by an o.

Whether that is a reasonable replacement is another matter. I don't believe it is.

In the code added in the edit, the dots are not valid in a variable name neither in dash, bash or zsh and under very specific conditions in ksh.
This will fail in bash:

DNS_MASTER=${TEST.DNS.DNS_MASTER:-o}

Also, this test is incorrect, because of white space and because the hostname command is not being executed:

if [ $DNS_MASTER == hostname -s ]

Use:

if [ "$DNS_MASTER" == "$(hostname -s)" ]

In all, the code you present could not work correctly in bash, and is puzzling in most shells. Please edit your question to make it relevant.

Answer 3

The closest I can think of a shell-operator that looks like that, is the bash patch that William Park was maintaining in the years 2004-2008 that came with a few more or less useful extensions to bash3.

Among them was a series a parameter expansion operators that all started with ${var|...} some of them reminiscent of zsh's parameter flags (albeit with a completely different syntax).

In that shell, ${var| |o} would have been parsed as ${var|cmd}, (meant to be more or less the same as $(cmd "$var"}) but would have resulted in a invalid command error as that |o would not be seen as a valid command name (even if you created a command by that name).