How can I read out the values of dynamically declared bash variables? - bash

Question

I have some trouble with a bash script, maybe someone can help me.
Inside my script, I defined variables dynamically using a loop like this:
somecolors="red yellow green blue" # read out of a file, may vary
for color in $(echo $somecolors); do
# Actually, here is more code that generates the value I
# want to set for this variable, that is being written
# into "$value"
declare use_color_$color=$value
done
The result is that four variables have been defined:
use_color_red=1
use_color_yellow=1
use_color_green=1
use_color_blue=1
So far so good. But how can I dynamically read these? I thought of suing a "for" loop. For example:
for color in $(echo $colors); do
echo $use_color_${color}
done
But this does not work.
How can I compose two variable names to a single one?
Thanks in advance!

Answer 1

Better use indexed and associative arrays instead. Referencing and dereferencing variable variables that way is really wrong.
somecolors=(red yellow green blue)
declare -A use_color
for color in "${colors[#]}"; do
use_color[$color]=$value ## Or do you mean use_color[$color]=$color?
done
Granting $value == 1, when you do echo "${use_color[red]}" you'd get 1.
One variation:
declare colors=(red yellow green blue)
declare -A use_color
use_color[red]=1
use_color[yellow]=1
use_color[green]=1
use_color[blue]=1
for color in "${colors[#]}"; do
echo "use_color[$color]=${use_color[$color]}"
done
Output:
use_color[red]=1
use_color[yellow]=1
use_color[green]=1
use_color[blue]=1
Similarly:
declare -A use_color=([red]=1 [yellow]=1 [green]=1 [blue]=1)
for color in "${!use_color[#]}"; do
echo "use_color[$color]=${use_color[$color]}"
done
Output:
use_color[yellow]=1
use_color[red]=1
use_color[blue]=1
use_color[green]=1

Answer 2

The cleanest way is using variable substitution. A variable of the form ${!varabc} will match all previously defined variables beginning with varabc. In your case:
#!/bin/bash
use_color_red=1
use_color_yellow=1
use_color_green=1
use_color_blue=1
for i in ${!use_color#}; do
printf " name: %-16s value: %d\n" $i ${!i}
done
exit 0
output:
name: use_color_blue value: 1
name: use_color_green value: 1
name: use_color_red value: 1
name: use_color_yellow value: 1