How to prevent sourcing a part of a bash script?

Question

I have a shell script with a syntax compatible to both bash and zsh, except for a section that has zsh specific syntax. that throws syntax errors if sourced from a bash is there an easy way to escape such section when using bash.

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The script is a bash function that sources all the files in a directory. it works fine form zsh (and it is irrelevant to the question)

#!/usr/bin/env bash

shell=$(ps -p $$ -oargs=)

if [ $shell = "bash" ]; then
    for f in ~/.functions.d/*.sh; do source $f; done
elif [ $shell = "zsh" ]; then
    for f (~/.functions.d/**/*.sh) source $f
fi

the error is produced by the 7th line, when sourcing it in bash

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relevant links

• https://unix.stackexchange.com/questions/242854/source-only-part-of-a-script-from-another-script
• Using source to include part of a file in a bash script

Answer 1

The problem is that the entire if/elif/else statement is parsed as a unit, so it can't contain invalid syntax.

What you could do is exit the sourced script before executing the zsh-specific code:

shell=$(ps -p $$ -oargs=)

if [ $shell = "bash" ]; then
    for f in ~/.functions.d/*.sh; do source $f; done
    return
fi

if [ $shell = "zsh" ]; then
    for f (~/.functions.d/**/*.sh) source $f
fi

However, a more general solution is to extract the bash-specific and zsh-specific code into separate scripts.

shell=$(ps -p $$ -oargs=)

if [ $shell = "bash" ]; then
    source load_functions.bash
fi

if [ $shell = "zsh" ]; then
    source load_functions.zsh
fi

load_functions.bash would contain the first for loop, while load_functions.zsh would have the second one.