Why does using a bash variable in find fail?

Question

I've got a find command in a bash script that works, but when I try to break it into variables that get added together it no longer works correctly.

I'm not really looking for a better way of doing this, I'd like to understand what Bash is doing in this case as I'm very stumped at this.

# Works, prints ./config
find . -type f -name 'config' ! -path './.git*'

echo
pathVar="! -path './.git*'"
# Doesn't correctly ignore './.git/config'
find . -type f -name 'config' $pathVar

echo
# Doesn't work 'find: ! -path './.git*': unknown primary or operator'
find . -type f -name 'config' "$pathVar"

No, you can't do it simply like that. $pathVar is assumed to be a single argument. you may either use eval to execute a string as a command or better yet, run the complete commands using conditional checks to decide which command the execute. — Anubis
– Anubis, Commented Apr 15, 2019 at 13:43
What defines a 'single argument'? Would it just be a string value with no spaces? — convergedtarkus
– convergedtarkus, Commented Apr 15, 2019 at 14:43

Anubis · Accepted Answer · 2019-04-15 13:53:51Z

1

As stated in the comments,

Option 1:

cmd="find . -type f -name 'config'"
if [[<condition to run long command>]]; then
    cmd="$cmd ! -path './.git*'"
fi
eval $cmd

Option 2:

if [[<condition to run long command>]]; then
    find . -type f -name 'config' ! -path './.git*'
    # ...
else
    find . -type f -name 'config'
    # ...
fi

answered Apr 15, 2019 at 13:53

Anubis

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2 Comments

convergedtarkus Over a year ago

What if the pathVar from my original question isn't something constant (user input, result from another process etc). How could I build up this find command dynamically?

convergedtarkus Over a year ago

Never mind, I can just cmd="$cmd $pathVar" to get that behavior with a dynamic variable.

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