BASH: How to define a variable with another variable which is part of the strings name [duplicate]

Question

I am trying to create multiple variables but using a numeric counter as part of the variables sting name. All my attempts so far have been futile. Any help would be appreciated.

#!/bin/bash


Colors=( red    \
         blue   \
         green  \
         yellow )

declare i=0
declare j=1

while (( ${j} < $(( ${#Colors[*]} + 1 )) ))
do
  set Variable${j}=${Colors[i]}
  echo   " Tried to create a variable named [ Variable${j} ] and load it with the value [ ${Colors[i]} ]"
  printf " The value of [ Variable${j} ], is [ %s ]\n\n" Variable${j}
  (( j = j + 1 ))
  (( i = i + 1 ))
done

This is what the script currently outputs.

./test.sh

Tried to create a variable named [ Variable1 ] and load it with the value [ red ] The value of [ Variable1 ], is [ Variable1 ]

Tried to create a variable named [ Variable2 ] and load it with the value [ blue ] The value of [ Variable2 ], is [ Variable2 ]

Tried to create a variable named [ Variable3 ] and load it with the value [ green ] The value of [ Variable3 ], is [ Variable3 ]

Tried to create a variable named [ Variable4 ] and load it with the value [ yellow ] The value of [ Variable4 ], is [ Variable4 ]

Tried ( quotes, set variable ) and nothing so far worked. If I hard code the variable name, no problem.

Use an array or associative array. gnu.org/software/bash/manual/html_node/Arrays.html — Shawn
– Shawn, Commented Jul 19, 2023 at 16:19
You can use some eval in there, but it's not really recommended. I can drop an example, give me a sec. — Carl Norum
– Carl Norum, Commented Jul 19, 2023 at 16:28
Why do you want to end up with multiple scalars Variable1, Variable2, etc. instead of just one array Variable[1], Variable[2], etc.? — Ed Morton
– Ed Morton, Commented Jul 19, 2023 at 20:12

Carl Norum · Accepted Answer · 2023-07-19 16:32:51Z

0

Inside your loop:

eval Variable${j}=${Colors[i]}
echo   " Tried to create a variable named [ Variable${j} ] and load it with the value [ ${Colors[i]} ]"
eval echo " The value of [ Variable${j} ], is [ \$Variable${j} ]"

Be careful out there.

answered Jul 19, 2023 at 16:32

Carl Norum

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3 Comments

Bruce Ferjulian Over a year ago

Thank you Carl, not familiar with ( eval ) but you gave me a good starting point.

Gordon Davisson Over a year ago

Warning: eval has a tendency to interpret things you didn't expect, leading to really weird bugs. It's better to avoid it whenever possible, and in this case it's entirely possible; see "Dynamic variable names in Bash" and "BashFAQ #6: "How can I use variable variables (indirect variables, pointers, references) or associative arrays?"

pjh Over a year ago

Use of eval is generally dangerous, and it's unnecessary here. In addition, the way that it is used here is broken. To see why, try adding 'dark red' (for instance) to the Colors array. eval "Variable${j}=\${Colors[i]}" is safer, but still not a good option.

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