Diff for "BashFAQ/050"

I'm trying to put a command in a variable, but the complex cases always fail!

Some people attempt to do things like this:

    # Non-working example
    args="-s 'The subject' $address"
    mail $args < $body

This fails because of WordSplitting. When $args is expanded, it becomes four words. 'The is the second word, and subject' is the third word.

The solution to the problem is in 90% of the cases a completely different approach. You generally do NOT want to put command names or command options in variables. Variables should contain the data you are trying to pass to the command, like usernames, hostnames, ports, text, etc. They should NOT contain options that are specific to one certain command. Those things belong in functions. What you probably should be doing, is this:

    # Send an email to someone.
    # Sends an email to the address specified as the first argument,
    # optionally with the subject specified by the second.
    # Reads the body of the mail from standard input.
    #
    # sendto address [subject]
    #
    sendto() {
        mail ${2:+-s "$2"} "$1"
    }

    sendto "$address" "The Subject" < "$bodyfile"

Unless you KNOW that this is not what you want, stop reading now.

The root of the issue described above is the fact that you need a way to maintain each argument as a separate word, even if that argument contains spaces. Quotes won't do it, but an array will.

As such, if you really wanted to create a command dynamically; put each argument in a separate element of an array, like so:

    # Working example
    args=(-s "The subject" "$address")
    mail "${args[@]}" < $body

Often, this question arises when someone is trying to use dialog to construct a menu on the fly. The dialog command can't be hard-coded, because its parameters are supplied based on data only available at run time. For an example of how to do this properly, see FAQ #40.

Another reason people attempt to do this is because they want their script to print each command before it runs it. If that's all you want, then simply use the set -x command, or invoke your script with #!/bin/bash -x or bash -x ./myscript. Note that you can turn it off and back on inside the script with set +x and set -x.

It's worth noting that you cannot put a pipeline command into an array variable and then execute it using the "${array[@]}" technique. The only way to store a pipeline in a variable would be to add (carefully!) a layer of quotes if necessary, store it in a string variable, and then use eval or sh to run the variable. This is not recommended, for security reasons. The same thing applies to commands involving redirection, if or while statements, and so on.

Some people get into trouble because they want to have their script print their commands including redirections before it runs them. set -x shows the command without redirections. People try to work around this by doing things like:

    # Non-working example
    command="mysql -u me -p somedbname < file"
    ((DEBUG)) && echo "$command"
    "$command"

(This is so common that I include it explicitly, even though it's repeating what I already wrote.)

Once again, this does not work. Not even using an array works here. The only thing that would work is rigorously escaping the command to be sure no metacharacters will cause serious security problems, and then using eval or sh to re-read the command. Please don't do that!

If your head is SO far up your ass that you still think you need to write out every command you're about to run before you run it, AND that you must include all redirections, then just do this:

    # Working example
    echo "mysql -u me -p somedbname < file"
    mysql -u me -p somedbname < file

Don't use a variable at all. Just copy and paste the command, wrap an extra layer of quotes around it (sometimes tricky), and stick an echo in front of it.

My personal recommendation would be just to use set -x and not worry about it.