Handle invalid variable name error in bash indirect references

Question

Take a look at this simple example:

x="foo bar"
echo ${!x}

It erros: -bash: foo bar: invalid variable name.

I'm writing a script where a positional argument will indicate a "color name" and we use indirect references to fetch the hex code of the color name. I want the script to keep going if the user passes a color name with space in it or generally an invalid color name (which we will default to white).

I also tried this try/catch approach but to no avail:

x="foo bar"
echo ${!x} || echo error

The closes thing I could find was this, but didn't help:

Bash reference variable by string name

@Shayan If you MUST use indirection, you could abuse declare for that, e.g. declare -- "${x}" 2>/dev/null || echo error — pmf
– pmf, Commented Apr 26, 2024 at 18:48
It certainly was considered -- these names are explicitly illegal by specification. An associative array would avoid the issue -- key names are allowed to have spaces. — Charles Duffy
– Charles Duffy, Commented Apr 26, 2024 at 19:28
BTW, I'd strongly suggest declare -p instead of declare -- -- otherwise too much chance of that user input being something that has side effects when run. — Charles Duffy
– Charles Duffy, Commented Apr 26, 2024 at 19:33
you code snippets mention x='foo bar' while the description mentions processing colors; recommend you rewrite the question to focus on your actual (color) issue: a) show the setup of a couple colors, b) show your code working for a valid color and c) show your code's result with an invalid color (ie, show all stdout/stderr output); net result: provide a minimal reproducible example — markp-fuso
– markp-fuso, Commented Apr 26, 2024 at 20:36

Philippe · Accepted Answer · 2024-04-26 19:24:52Z

2

This may be the error handling you need :

test -v "$x" || echo error

answered Apr 26, 2024 at 19:24

Philippe

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