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Asked: 2026-05-26T18:39:14+00:00

I’m trying to build a command string based to pass in a -e flag

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I’m trying to build a command string based to pass in a “-e” flag and another variable into a another base script being call as a subroutine and have run into a strange problem; I’m losing the “-e” portion of the string when I pass it into the subroutine. I create a couple example which illustrate the issue, any help?

This works as you would expect:

$echo "-e  $HOSTNAME"

-e  ops-wfm

This does NOT; we lose the “-e” because it is interpreted as a special qualifier.

$myFlag="-e $HOSTNAME"; echo $myFlag

ops-wfm

Adding the “\” escape charactor doesn’t work either, I get the correct string with the “\” in front:

$myFlag="\-e $HOSTNAME"; echo $myFlag

\-e ops-wfm

How can I prevent -e being swallowed?

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1 Answer

  1. Editorial Team

    Use double-quotes:

    $ myFlag="-e $HOSTNAME"; echo "${myFlag}"
-e myhost.local

    I use ${var} rather than $var out of habit as it means that I can add characters after the variable without the shell interpreting them as part of the variable name.

    echo may not be the best example here. Most Unix commands will accept -- to mark no more switches.

    $ var='-e .bashrc' ; ls -l -- "${var}"
ls: -e .bashrc: No such file or directory
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