Bash, read variable with variable inside

Question

I have a script.

style_header[true]="background-color: rgb(230,250,230);"
style_header[false]="background-color: rgb(250,230,230);"

if COMMAND; then
    export=true
else
    export=false
fi

echo "${style_header[$export]}"

COMMAND finished ok, so export=true, but it returns style_header[false] variable "background-color: rgb(250,230,230);".

background-color: rgb(250,230,230);

I need to return this.

background-color: rgb(230,250,230);

It works with number 0 or 1 as index, but I need 'true' or 'false' variable inside.

Is possible to do that? I mean set array index as variable.

By default, style_header is an indexed array, so things like style_header[true] are processed by treating the index as an arithmetic expression. Since neither true nor false is defined as a variable, they both evaluate to 0. So style_header[true] and style_header[false] are both equivalent to style_header[0]. — Gordon Davisson
– Gordon Davisson, Commented Jun 11, 2019 at 21:43

Barmar · Accepted Answer · 2019-06-11 21:12:06Z

3

Use declare -A style_array to declare it as an associative array. By default it's assumed to be an indexed array.

#!/bin/bash

declare -A style_header
style_header[true]="background-color: rgb(230,250,230);"
style_header[false]="background-color: rgb(250,230,230);"

if COMMAND; then
export=true
else
export=false
fi

echo "${style_header[$export]}"

DEMO

answered Jun 11, 2019 at 21:12

Barmar

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3 Comments

Stanislav Hosek Over a year ago

Perfect! Thank you very much! I tried it with declare before, but I did something wrong.

Barmar Over a year ago

Maybe you used -a instead of -A?

Stanislav Hosek Over a year ago

Yes, it could be the problem. Thanks.

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