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bashcommand-substitution

bash doesn't run commands in sequential order

I have a bash script that contains the following:

MY_COMMAND="MY_PWD=`pwd`; export MY_PWD; MY_PWD_BASENAME=`basename $MY_PWD`; echo $MY_PWD_BASENAME"; export MY_COMMAND

When I source the script from the terminal, I get the following error:

basename: missing operand
Try `basename --help' for more information.

This indicates that the commands in MY_COMMAND are not executed in sequential order. What is happening here?

Solution

  • The following line:

    MY_COMMAND="MY_PWD=`pwd`; export MY_PWD; MY_PWD_BASENAME=`basename $MY_PWD`; echo $MY_PWD_BASENAME"

    will not execute the following commands (as you may think):

    MY_PWD=`pwd`
export MY_PWD
MY_PWD_BASENAME=`basename $MY_PWD`
echo $MY_PWD_BASENAME"

    Instead it will expand the command substitutions

    `pwd`
`basename $MY_PWD`

    and replace them with their output. Since $MY_PWD is not set, basename will get executed without the required argument, like:

    basename

    That leads to the error.

    Fix: I recommend to use $() instead of backticks for the command substitution. One benefit is that you can nest them:

    MY_COMMAND="MY_PWD=$(pwd); export MY_PWD; MY_PWD_BASENAME=$(basename "$(pwd)"); echo $MY_PWD_BASENAME"

    However, that's just the syntax fix. Generally I recommend to use a function like @chepner suggested

    lib.sh

    function basename_pwd() {
    basename "$(pwd)"
}

    Use the function:

    #!/bin/bash
source "lib.sh"
basename_pwd